Weatherwax, John - A Solution Manual for An Introduction to Mathematical Statistics and Its Applicat

SOLUTION MANUAL

Solutions to the Questions in An Introduction to Mathematical Statistics and Its Applications

by Richard J. Larsen and Morris L. Marx John Weatherwax Text copyright © 2018 John L. Weatherwax All Rights Reserved Please Do Not Redistribute Without Permission from the Author To my family.

Introduction This is a solution manual all of the questions in the excellent statistical textbook: An Introduction to Mathematical Statistics and Its Applications

by Richard Larsen and Morris Marx This solution manual was prepared form the fifth edition of the textbook. I have looked at other editions of the book and have had a hard time finding any significant differences between the editions.

I’ve tried to provide much more detail than is normally found in a typical solutions manual. Thus rather than just give the solution I try to explain in great detail how I arrived a the solution. This will further reinforce the main ideas from class and better teach the material. The manual stands at over 400 pages.

One of the benefits of this manual is that I heavily use the R statistical language to perform any of the needed numerical computations (rather than do them ”by-hand”). Thus if you use this manual you will be learning some R programming at the same time as you learn statistics. The R programming language is one of the most desired skills for anyone who hopes to use data/statistics in their future career. All the R code can be found at the following location: http://waxworksmath.com/Authors/G_M/Larsen/larsen.html

As a final comment, I’ve worked hard to make these notes as good as I can, but I have no illusions that they are perfect. If you feel that that there is a better way to accomplish or explain an exercise or derivation presented in these notes; or that one or more of the explanations is unclear, incomplete, or misleading, please tell me. If you find an error of any kind – technical, grammatical, typographical, whatever – please tell me that, too. I’ll gladly add to the acknowledgments in later printings the name of the first person to bring each problem to my attention.

Chapter 2 (Probability)

In working some of these questions I have chosen to use R or python to explicitly enumerate the spaces requested rather than to derive them by hand. While many of these questions could no doubt be done “by hand” it is very easy to change a problem statement in such a way that will make that impossible. Using a computer is a more robust (with respect to problem specification) way to work problems like these and is a helpful skill to have.

Question 2.2.1

The full sample space in this case would be

Event A would be the set

Event B would be the set

Question 2.2.2

We can use R to enumerate the possible outcomes when we toss three die. We have D = 1:6 outcomes = expand.grid(D, D, D) colnames(outcomes) = c(’R’, ’B’, ’G’) mask = apply(outcomes, 1, sum) == 5 print(outcomes[mask, ])

Running the above we get > print(outcomes[mask, ]) R B G 3 3 1 1 8 2 2 1 13 1 3 1 38 2 1 2 43 1 2 2 73 1 1 3

for the members of the event A .

Question 2.2.3

We can generate this sample space with the following python code from itertools import combinations for it in combinations(range(1, 7), 3): if sorted(it)[1]==3: print it

Running the above python code gives (1, 3, 4) (1, 3, 5) (1, 3, 6)

(2, 3, 4) (2, 3, 5) (2, 3, 6)

for the elements of the sample space.

Question 2.2.4

We can pick the suit of the first card in four ways. We can pick the number on the first card to be one of the numbers 1, 2, 3, 4, 5, 6, 7. For all of these choices (excluding 4) the number on the other card required that the two cards sum to 8 is determined and different than the number on the first card. Ignoring the choice of the number 4 (for the time being) on the first card we have 7 − 1 = 6 ways to pick the number. We then have four ways to pick the suit of the second card. This gives

total ways to draw this type of pair.

If we pick the number 4 on the first card the number 4 must be picked on the second card (so the sum is eight) and we thus can’t select the same suit for the second card as we did on the first card. This means that the number of ways we can draw two cards that sum two eight in this way is

The total number of ways to pick this type of pairs is the sum of the two above numbers or 96 + 12 = 108.

Question 2.2.5

One would expect that this would have the sum of the two die to be eight and is perhaps the special case where both die show four.

Question 2.2.6

There are

= 1287 ways to draw five cards in the suit of hearts (i.e. a flush). The “straight flushes”

are

of which there are ten such hands. If we let N F be the number of flushes of the form we are interested in then

Question 2.2.7

This would be the set of triangles that satisfy a 2 + b 2 = 5 2 .

Question 2.2.8

We can generate this sample space using the following python code

from itertools import permutations for it in permutations(’B B B S S’.split(), 5): print it

Running the above give the following output (partial) (’B’, ’B’, ’B’, ’S’, ’S’) (’B’, ’B’, ’B’, ’S’, ’S’) (’B’, ’B’, ’S’, ’B’, ’S’) (’B’, ’B’, ’S’, ’S’, ’B’) (’B’, ’B’, ’S’, ’B’, ’S’) (’B’, ’B’, ’S’, ’S’, ’B’) Question 2.2.9

Part (a): We can generate this sample space using the following R code R = 0:1 outcomes = expand.grid(R, R, R, R) colnames(outcomes) = c(’P1’, ’P2’, ’P3’, ’P4’)

Running the above give the following output (partial) > outcomes P1 P2 P3 P4 1 0 0 0 0 2 1 0 0 0 3 0 1 0 0 4 1 1 0 0 5 0 0 1 0 6 1 0 1 0

Part (b): To find the events where exactly two phones were used we could do mask = apply(outcomes, 1, sum) == 2 outcomes[mask, ]

which gives > outcomes[mask, ] P1 P2 P3 P4 4 1 1 0 0 6 1 0 1 0 7 0 1 1 0 10 1 0 0 1 11 0 1 0 1 13 0 0 1 1

Part (c): We will have at most space for one more call if there is only one unused phone (or no unused phones). As the one free phone can be in any of the k spaces we have k outcomes. The stated event will also occur if all the phones are in use which can happen in only one way. The the total number of outcomes for this event is k + 1.

Question 2.2.10

Part (a): We can generate this sample space using the following R code

R = c(2, 4, 1) outcomes = expand.grid(R, R) colnames(outcomes) = c(’u’, ’v’)

This gives the following sample space > outcomes u v 1 2 2 2 4 2 3 1 2 4 2 4 5 4 4 6 1 4 7 2 1 8 4 1 9 1 1

Part (b): The outcomes in this sample space can be listed as u_plus_v = outcomes$u + outcomes$v possible_sums = sort( unique( u_plus_v ) ) for( s in possible_sums ){ mask = u_plus_v == s print(sprintf(’Events where the sum= %d’, s)) print(outcomes[mask, ]) }

This gives [1] "Events where the sum= 2" u v 9 1 1 [1] "Events where the sum= 3" u v 3 1 2 7 2 1 [1] "Events where the sum= 4" u v 1 2 2 [1] "Events where the sum= 5" u v 6 1 4 8 4 1 [1] "Events where the sum= 6" u v 2 4 2 4 2 4 [1] "Events where the sum= 8" u v 5 4 4 Question 2.2.11

We think of the suspects shown to her as the elements of the set { I 1 ,I 2 ,I 3 ,P 1 ,P 2 } . Here I 1 , I 2 , and I 3 are the three innocent people and P 1 and P 2 are the two perpetrators. The sample space in this case will be a selection of two of these five people. The outcomes in the event A would be

Question 2.2.12

This would be values of a , b , and c that satisfy

Question 2.2.13

To win with an initial draw of a nine we must get on of the following sequence of rolls

Here X i is a number from { 2, 3, … , 11, 12} that is not a seven or a nine.

Question 2.2.14

The sample space of the prisoner’s possible allocations can be denoted (w 1 ,b 1 ) and (w 2 ,b 2 ) where w 1 + w 2 = 10, b 1 + b 2 = 10, w i ≥ 1, and b i ≥ 1. Here w 1 represents the number of white chips in the first urn, b 1 are the number of black chips in the first urn etc.

Intuitively the best chance of survival consists of placing one white chip in one of the urns (say urn one) and then all the other chips in the other urn. Then the probability that we survive (assuming the urns are picked with equal probability) is given by

Question 2.2.15

From the description of the question at the k step (for k = 1, 2, 3, … ) we will have

chips and be then infinite.

k −1

minutes from midnight. As we let k →∞ I would conclude that the number of chips is

Figure 1: Left: A ∪ B for Question 2.2.16. Right: A ∩ B for Question 2.2.16.

Question 2.2.16

The region A ∪ B is plotted in Figure 1 (left) and the region A ∩ B is plotted as the doubly hashed region in Figure 1 (right).

Question 2.2.17

The first region is the set of points where x 2 + 2x − 8 ≤ 0 or (x + 4)(x − 2) ≤ 0. This is the region − 4 ≤ x ≤ 2.

The second region is the set of points where x 2 + x − 6 ≤ 0 or (x + 3)(x − 2) ≤ 0. This is the region − 3 ≤ x ≤ 2.

The union of these two regions is − 4 ≤ x ≤ 2 while the intersection of these two regions is − 3 ≤ x ≤ 2.

Question 2.2.18

This would be the set { 2, 3, 4} .

Question 2.2.19

The system will fail if the first or the second pair fail. The first pair will fail if both components fail or

and similarly for the second pair. Thus the total system will fail if

Question 2.2.20

We can draw the diagrams and then represent these regions mathematically.

Part (a-b): The region B ∩ C is 0 ≤ x ≤ 2. Intersecting this with A c gives 1 ≤ x ≤ 2. Taking the union of this with A c gives the entire real line.

Part (c): This would be the empty set.

Part (d): Now (A ∪ B ) ∩ C c will be the region 2 ≤ x ≤ 3 so the region we want is the complement of this or −∞ < x < 2 with 3 < x < ∞ .

Question 2.2.21

The intersection are the set of hands that have consecutive cards of the same suit. There are 4 × 9 = 36 such hands. The four is to select the suit and the nine represents the nine consecutive hands

This is a different answer than in the back of the book. If anyone sees anything wrong with this answer please contact me.

Question 2.2.22

From the question statement we have that the events defined will happen if we draw the following letters F = { A,E,I,L } R = { E,I,L,S,T } V = { A,E,I,O } .

Part (a): We would have that

Part (b): We have

Part (c): We have

Question 2.2.23

Part (a): Both sets are the elements in A or the elements in both B and C .

Part (b): Both sets are the elements in A and the elements in both B or C .

Question 2.2.24

As

the total set we are considering is the full space S .

Question 2.2.25

These both can be seen by using the Venn diagram (see the next section) or by taking what it means to be a member of the left set and showing that that means that same element is also a member of the right set and vice versa i.e. showing that a member of the right set is also a member of the left set.

Question 2.2.26

Part (a): This would be (A ∪ B ∪ C ) c .

Part (b): This would be A ∩ B ∩ C .

Part (c): This would be A ∩ B c ∩ C c .

Part (d): This would be (A ∩ B c ∩ C c ) ∪ (A c ∩ B ∩ C c ) ∪ (A c ∩ B c ∩ C ).

Part (e): This would be (A ∩ B ∩ C c ) ∪ (A ∩ B c ∩ C ) ∪ (A c ∩ B ∩ C ).

Question 2.2.27

Part (a-b): We need to have A be a subset of B .

Question 2.2.28

Part (a): This would be 5 ≤ x ≤ 10.

Part (b): This would be 3 ≤ x < 5.

Part (c): This would be 0 < x ≤ 7.

Part (d): This would be

Part (e): This would be

Part (f): This would be A c ∩ B c = { x : 5 ≤ x ≤ 10}∩ ({ x : 0 ≤ x ≤ 3}∪{ x : 7 < x ≤ 10} ) = { x : 7 < x ≤ 10} .

Question 2.2.29

Part (a): B and C are mutually exclusive.

Part (b): A is a subset of C and B is a subset of A .

Question 2.2.30

Part (a): This would be A = A 1 ∩ A 2 ∩ A 3 .

Part (b): This would be A = A 1 ∪ A 2 ∪ A 3 .

Question 2.2.31

Let N (A ) be the number that saw the movie the first time so N (A ) = 850 and let N (B ) be the number that saw the movie the second time so N (B ) = 690. Then we are also told that

which is equal to 6000 − N (A ∪ B ) so that N (A ∪ B ) = 1300. Then using these results and the fact that

we find N (A ∩ B ) = 240.

Question 2.2.32

These are both apparent by drawing the Venn diagrams.

Question 2.2.33

These are both apparent by drawing the Venn diagrams.

Question 2.2.34

These are both apparent by drawing the Venn diagrams.

Question 2.2.35

Using DeMorgan’s laws we know that (A c ∪ B c ) c = A ∩ B which is already listed. We have the following subsets The sets A , B , A ∩ B , A c ∩ B , and A ∩ B c are subsets of A ∪ B .

The set A ∩ B and A ∩ B c are subsets of A .

The set A ∩ B and A c ∩ B are subsets of B .

Question 2.2.36

I will use DeMorgan’s laws to suggest alternative representations.

Part (a): This would be

Part (b): This would be

Part (c): This would be A ∩ (A ∩ B ) c = A ∩ (A c ∪ B c ) = (A ∩ A c ) ∪ (A ∩ B c ) = ϕ ∪ (A ∩ B c ) = A ∩ B c .

Question 2.2.37

Let M be the event that the MCAT score was larger than 27 and G be the event that the GPA was larger than 3.5. Then we want to know N (M c ∩ G c ). As this event is equal to M ∪ G ) c we have that N (M c ∩ G c ) = N ((M ∪ G ) c ) = N (S ) − N (M ∪ G ) = 1200 − (N (M ) + N (G ) − N (M ∩ G )) = 1200 − (1000 + 400 − 300) = 100.

The proportion of the number of graduates in this group is then .

Question 2.2.38

This is in general called the inclusion-exclusion identity and we will have

Question 2.2.39

Let T and H be the events that a person supports the taxes and homeland security positions respectively.

Then from the question we are told that N (S ) = 1200 N (T ) = 600 N (H ) = 400 N (T c ∩ H ) = 300.

Here S is the full space of responses. We want to compute N (T ∩ H c ). Now as T ∪ H = T ∪ (T c ∩ H ) and the two sets in the union are disjoint we have that

Then as T ∪ H = H ∪ (H c ∩ T ) and the two sets in the union are disjoint we have that

Question 2.2.40

We are told that N (A ∩ B c ) = 15 N (A ∩ B ) = 2 N (B ∩ A c ) = 50.

Then using a Venn Diagram we can conclude that

Question 2.3.1

Using set notation we would have P (L ) = 0. 42 P (V ) = 0. 27 P (L ∩ V ) = 0. 1.

Then

We want to know P ((L ∪ V ) c ) = 1 − 0. 59 = 0. 41.

Question 2.3.2

Using what we are given we have

Then since we can write

and the three sets on the right are mutually exclusive so we have

Next using A = (A ∩ B c ) ∪ (A ∩ B ) and B = (B ∩ A c ) ∪ (A ∩ B ), so we have 0. 4 = (A ∩ B c ) + 0. 1 and 0. 5 = (B ∩ A c ) + 0. 1.

Thus we have P (A ∩ B c ) = 0. 3 and P (B ∩ A c ) = 0. 4. Using these the probability we want is given by

Question 2.3.3

Part (a): This would be P (A c ∪ B c ) = P ((A ∩ B ) c ) = 1 − P (A ∩ B ).

Part (b): First note that we can write

Then note that as we can write B as the union of two disjoint sets as

the probability we seek can be written as

Question 2.3.4

We can write what we are told in the question as

As each of the three sets in the above union are disjoint we have

Now B can be written as B = (B ∩ A c ) ∪ (B ∩ A ) which are again disjoint. As that is the probability we computed above we have shown that the desired probability is given by 0.2.

Question 2.3.5

No the event A 1 ∪ A 2 ∪ A 3 is the event that we get at least one six on three tosses of the die which is equal to

Question 2.3.6

Note that

Thus we have that P (A ∪ B ) = 0. 5. Next using the mutually exclusive decomposition P (A ∪ B ) = P ((A ∩ B c ) ∪ (B ∩ A c ) ∪ (A ∩ B )) = P ((A ∩ B c ) ∪ (B ∩ A c )) + P (A ∩ B ), we see that

The set expression above is what we are looking for.

Question 2.3.7

This would be B = ∪ i =1 n (B ∩ A i ).

Question 2.3.8

Part (a): We will have this if B is a subset of A .

Part (b): We will have this if A is a subset of B .

Question 2.3.9

No one will be eliminated if all coins are heads or all coins are tails. This will happen with a probability of 2 3 = . Thus someone will be eliminated with a probability of 1 − = .

Question 2.3.10

Using P (A ∪ B ) = P (A ) + P (B ) − P (A ∩ B ) = In the above we compute P (A ∩ B ) to be are both divisible by two and three.

Question 2.3.11

+

−

=

= .

as there are four elements i.e. the numbers { 6, 12, 18, 24} that

Let S be the event that we win the football game this Saturday and N the event that we win the game two weeks from now. Then we are told that P (S ) = 0. 1 P (N ) = 0. 3 P (S c ∩ N c ) = 0. 65.

(1)

We want to know P ((S ∩ N c ) ∪ (N ∩ S c )). Note that this is equal to P (S ∪ N ) − P (S ∩ N ). From Equation 1 we have that

This means that P (S ∪ N ) = 0. 35. Next using

we can compute P (S ∩ N ) = 0. 05. Thus the desired probability is 0. 35 − 0. 05 = 0. 3.

Question 2.3.12

We know that

This with the expression 3p 1 − p 2 = are two equations for p 1 and p 2 . Solving them we find p 1 = and p 2= .

Question 2.3.13

Let F be the event the employee is a female, M = F c the event that the employee is a male, and B the event that the employee is a minority. Then from the question we are told that P (F ) = 0. 6 P (B ) = 0. 3 c P (B ∩ M ) = 0. 25, and we want to know P (B ∩ F ). Note that we can write

and the two sets on the right-hand-side are mutually exclusive. This means that

Next note that we can write

and again the two sets on the right-hand-side are mutually exclusive. This means that

so P (B ∩ F ) = 0. 45.

Warning: This is different than the answer in the back of the book. If anyone sees anything wrong with what I have done please contact me.

Question 2.3.14

As

to make the left-hand-side as large as possible we want to make P (A ∪ B ∪ C ) as large as possible. As we can write (2)

we will do that by taking A ∩ B = A ∩ C = B ∩ C = ∅ . Then we find that the maximum possible value for P ((A ∪ B ∩ B ) c ) is 1 − 0. 2 − 0. 1 − 0. 3 = 0. 5.

Question 2.3.15

From the question statement we can compute that P (X ) = P (Y ) = P (X ∩ Y ) =

2

2

=

=

P (X ∪ Y ) = + − = .

Part (a): As we can write

and the two sets on the right-hand-side are mutually exclusive we have

Thus we have P (X c ∩ Y ) = .

Part (b): As we can write

and the two sets on the right-hand-side are mutually exclusive we have

Thus we have P (X ∩ Y c ) = .

Question 2.3.16

By counting we can determine that P (A ) =

(as there are five pairs that sum to six). In the same way

as there are six pairs i.e. { (1, 2), (2, 1), (2, 4), (3, 6), (4, 2), (6, 3)} that have the desired property. Now

As we can write

and the two sets on the right-hand-side are mutually exclusive we have

Solving we have P (A ∩ B c ) =

=

.

Question 2.3.17

As (A ∩ B ) ∪ (A ∩ C ) will be a “larger set” than A ∩ B the ordering is given by

Question 2.3.18

Let C 1 and C 2 be the events that the scam in the first and second chatrooms are “discovered”. Then we are told that P (C 1 ) = P (C 2 ) = P (C 1 ∩ C 2 ) = 0. 0025.

We then want to compute P ((C 1 ∪ C 2 ) c ) = 1 − P (C 1 ∪ C 2 ) = 1 − P (C 1 ) − P (C 2 ) − P (C 1 ∩ C 2 ) =1−

−

+ 0. 0025 = 0. 8691667,

Question 2.4.1

By counting we have P (B ) = P (sum exceeds eight) =

Question 2.4.2

and P (A ∩ B ) =

. This means that

From

we can write

Using the fact that we know P (A ) = 0. 2 and P (B ) = 0. 4 we can use the above to find P (A ∩ B ) = 0. 1.

Question 2.4.3

From

we have that P (A ∩ B ) < P (A )P (B ). This means that

Question 2.4.4

First we have that

Next note that the set E is given by E = (A ∩ B c ) ∪ (B ∩ A c ). From this we know that A ∪ B = E ∪ (A ∩ B ) and as these sets are all mutually exclusive we have that

Thus P (E ) = 0. 3. Now

as E ∩ (A ∪ B ) = E and E is a subset of A ∪ B .

Question 2.4.5

Let A be the event that both children are boys and B the event that at least one child is a boy. Then we have P (B ) = + + = P (A ∩ B ) = so P (A | B ) = which is the same answer as the example in the book.

= ,

Question 2.4.6

We are told that P (A | B ) = 0. 6 = P (A ∪ B ) = 0. 8 P ((A ∩ B ) ∪ (B ∩ A c )) = 0. 6. c

Using the fact that

and the sets on the right-hand-side are disjoint we have

Then we have

which is one of the desired answers.

Next note that we can write

and the two sets on the right-hand-side are mutually exclusive. This means that

Now from Equation 3 we have

so that we can compute

Finally then from

we have

which is the other desired answer.

Question 2.4.7

(3)

Let A be the event that the first chip selected is red and B the event that the second chip selected is red.

Then we want to compute

Question 2.4.8

Given the question statement we can write P (A | B ) = =

= .

Now we know that P (A ∪ B ) ≤ 1 so that − P (A ∪ B ) ≥− 1 which in the above we can take to mean

Question 2.4.9

I was not sure I fully understood what this question was asking. If anyone understands it better would they please contact me.

Question 2.4.10

Now

Note that A ∩ B ⊂ A ∪ B so that

which means that

Thus we have P (A ∩ B | (A ∪ B ) c ) = 0.

Question 2.4.11

We are told (or can compute) that P (A ) = 0. 65 P (B ) = 0. 55 P (A ∩ B ) = 0. 25 P (A ∪ B ) = P (A ) + P (B ) − P (A ∩ B ) = 0. 65 + 0. 55 − 0. 25 = 0. 95.

Using A = (A ∩ B ) ∪ (A ∩ B c ) we can compute

Using B = (B ∩ A ) ∪ (B ∩ A c ) we can compute

Part (a): We would have

Part (b): We would have

Part (c): This would be P (A ∪ B ) = 0. 95.

Part (d): This is P ((A ∩ B c ) ∪ (B ∩ A c ) ∪ (A c ∩ B c )) = P ((A ∩ B c )) + P (B ∩ A c ) + P (A c ∩ B c )) = 0. 4 + 0. 3 + 0. 05 = 0. 75, as all the sets are mutually exclusive.

Part (e): This is P ((A ∩ B c ) ∪ (B ∩ A c )| A ∪ B ) = =

=

=

.

Part (f): This is

Part (g): This is

Question 2.4.12

Let A be the event that at least two heads occur and let B be the event that at most two heads occur. We will use

Now we have

The event A ∩ B is the event that exactly two heads occur so

This means that

Question 2.4.13

Let A be the event that the number on the first die is greater than or equal to four and B be the event that the sum of the two die equals eight. Now by enumeration we have

and the intersection event A ∩ B = { (4, 4), (5, 3), (6, 2)} thus

Using these two we have

Question 2.4.14

Let A be the event that all four cards are aces and B the event that three of the four cards are aces. Then we have

Question 2.4.15

Thus P (A ∪ B ) = 0. 8. Now using the fact that

and all sets on the right-hand-side are disjoint that

Now using this we have

Thus

Question 2.4.16

From the given statements we have P (A | B ) =

= 0. 5 so P (B ) =

P (B | A ) =

= 0. 4 so P (A ) =

.

Then as P (A ) + P (B ) = 0. 9 and using the above we can solve for P (A ∩ B ) to get P (A ∩ B ) = 0. 2.

Using this we then have

Question 2.4.17

For this question we want to compute P (( A ∩ B c ) ∪ ( B ∩ A c ) | ( A ∩ B c ) ∪ ( B ∩ A c ) ∪ ( A c ∪ B c )) = = = Question 2.4.18

By making a grid and “counting” we have that P (A ) = P (B ) = P (A ∩ B ) =

.

The last calculation is from the fact that the event A ∩ B is the set

Thus we have that

Question 2.4.19

For this question we want to compute

=

.

which is equal to (winner ∈{ Outandout} ) (winner { Australian Doll, Dusty Stake} )

Note that

Thus the probability we want to compute is given by Question 2.4.20

No his chances stay the same. Let A , B , and C be the event that person A , B , and C will be released. Then P (A | B ∪ C ) =

=

=

=

= ,

which is the same as before. Here the notation (1, 1, 0) means that A and B are freed while C is not.

Question 2.4.21

Using conditional probability we would have P (B 1 B 2 R 3 W 4 W 5 ) = P (W 5 | B 1 B 2 R 3 W 4 )P (W 4 | B 1 B 2 R 3 )P (R 3 | B 1 B 2 )P (B 2 | B 1 )P (B 1 ) =

=

.

Here I have performed the multiplications above “backwards” (i.e. the first factor is P (B 1 ) = ) as it is easier to reason about the state of the urn evaluating the product on the first line from right-to-left.

In the second case, we have to specify a specific number on each draw so our probability becomes

Question 2.4.22

Let K i be the event that the key works on the i th attempt then we have P (K 3 ) = P (K 1 c K 2 c K 3 ) = P (K 3 | K 1 c K 2 c )P (K 2 c | K 1 c )P (K 1 c ) = Question 2.4.23

= .

Drawing this specific hand would happen with a probability of

Question 2.4.24

Using conditional probability this would be P (W 1 W 2 B 3 B 4 B 5 ) = P (B 5 | W 1 W 2 B 3 B 4 )P (B 4 | W 1 W 2 B 3 )P (B 3 | W 1 W 2 )P (W 2 | W 1 )P (W 1 ) =

=

.

Question 2.4.25

Let S i be the event we use supplier i and D the event that the toy is defective. Then we have

Question 2.4.26

Let E be the event in question. Then we have

Question 2.4.27

Let W be the “war” event and T the event of escalating terrorist activity then we have

Question 2.4.28

Let D be the event the solicitor gets a donation then we have P ( D ) = P ( D | Belle Meade ) P ( Belle Meade ) + P ( D | Oak Hill ) P ( Oak Hill ) + P ( D | Antioch ) P ( Antioch ) =0. 6

+ 0 . 55

+ 0 . 35

= 0 . 4625 .

Question 2.4.29

Let T be the “truth” event and M the “male” event. Then we have

Question 2.4.30

Let R 1 be the event that a red chip was drawn from urn 1, R 2 be the event that a red chip was drawn from urn 2, and R (with no subscript) be the event that a red chip was drawn from urn 1 after the other two have been transferred. Then we have P (R ) = P (R | R 1 ,R 2 )P (R 1 ,R 2 ) + P (R | R 1 ,R 2 c )P (R 1 ,R 2 c ) + P (R | R 1 c ,R 2 )P (R 1 c ,R 2 ) + P (R | R 1 c ,R 2 c )P (R 1 c ,R 2 c ) =

+

+1

+

= 0. 59375.

Question 2.4.31

Let T be the event the test is positive for HIV and H the event a person has HIV. Then conditioning on the event H we have

or 0.019%.

Question 2.4.32

This is worked in a previous question (see Page § ).

Question 2.4.33

Let B be the event that Backwater A&M wins the the conference football championship and H the event that their quarter back is healthy. Then conditioning on the event H we have

Question 2.4.34

Let D i be the event that after all draws we have drawn i red chips and R the event that the seventh chip is red. Then we have

We can evaluate the above with the following R code is = 0:6 pt1 = ( choose(40, is) * choose(60, 6-is) ) / choose(100, 6) pt2 = ( 40 - is ) / 94 sum( pt1 * pt2 )

when we do that we get 0.4.

Question 2.4.35

No. Let B be the event that the person calling heads is correct and H the event that the coin lands head.

Then conditioning on the event H we have

Question 2.4.36

Let G be the event that the jury returns a guilty verdict and D the event that the defense can discredit the police department. Then conditioning on the event D we have

Question 2.4.37

Let M be the event that Markus gets into medical school. Then conditioning on his GPA we can compute

Question 2.4.38

Let E be the event that a prisoner gets early release and G the event that the prisoner is related to a member of the governor’s staff. Then conditioning on the event G we have

for the probability of early release.

Question 2.4.39

Let M be the event that the student is male. Conditioning on the division of study we have P (M ) = P (M | Humanities)P (Humanities) + P (M | Natural Science)P (Natural Science) + P (M | History)P (History) + P (M | Social Sciences)P (Social Sciences) = 0. 4(0. 4) + 0. 85(0. 1) + 0. 55(0. 3) + 0. 25(0. 2) = 0. 46.

Question 2.4.40

Let W be the event that the chip transferred over is white so that W c is the event that the chip transferred over is red. Let D be the event that we draw a red chip from urn number two after this initial transfer. Then we want to compute P (W | D ). We find P (W | D ) = = Question 2.4.41

=

= .

Let R be the event a red chip is drawn and let U i be the event we draw from the i th urn for i = 1, 2, 3.

Then P (U 3 | R ) = =

=

=

.

Question 2.4.42

Let O be the event that the oil pressure is too low and L the event that the dashboard warning light is flashing. Then from the question statement we have P (L | O ) = 0. 99 P (L | O c ) = 0. 02 P (O ) = 0. 1 P (O c ) = 0. 9.

Using these we have

Question 2.4.43

Let L be the event that the basement is leaky. Let T , W , H be the event the house was build by the contractor Tara, Westview, and Hearthstone respectively. Then using the statements from the question we have P (L | T ) = 0. 6 P (L | W ) = 0. 5 P (L | H ) = 0. 4 P (T ) =

=

P (W ) = P (H ) =

.

Using these we can compute

Then we have P (T | L ) =

= 0. 2352941

P (W | L ) =

= 0. 2941176

P (H | L ) =

= 0. 4705882.

Thus the most likely contractor is Hearthstone.

Question 2.4.44

Let C be the event that Francesca passes the course, X the event that she has professor X , and Y the event she has professor Y . Then in the question statement we are told that P (C | X ) = 0. 85 P (C | Y ) = 0. 6 P (X ) = 0. 4 P (Y ) = 0. 6.

Using these we are asked for

Question 2.4.45

Let S be the event that a person wears sunglasses and B be the even that a cashed check bounces. Then from the question statement we have P (B | S ) = 0. 5 P (B | S c ) = 0. 02 P (S ) = 0. 1.

Using these we can compute

Question 2.4.46

Let F be the event that Uncle Basil will die, C be the event that Uncle Basil chooses the cherries flambe for dessert, M the event he chooses the chocolate mousse, and N the event he chooses no dessert. Then from the question statement we are told that P (F | C ) = 0. 6 P (F | M ) = 0. 9 P (C ) = 0. 5 P (M ) = 0. 4 P (N ) = 0. 1.

Using these we can compute P (C | F ) = =

= 0. 4545455,

and P (M | F ) = =

= 0. 5454545.

Then as P (M | F ) > P (C | F ) we must conclude that the chocolate mousse is the most likely (and thus Margo) suspect.

Question 2.4.48

Let A be the event that a child is abused and D be the even that this abuse is discovered. Then from the question statement we have P (A ) = P (D | A ) = 0. 9 P (D | A c ) = 0. 03.

Using these we can compute

Question 2.4.49

Let F be the event that a person is female, H the event a person studies humanities, C the event the person is studying history and culture, and S the event a person is studying science. Then from the question statement we have P (C | F ) = =

= 0. 4411765.

Question 2.4.50

Let N be the event of a one and Z be the event of a zero. Then from the question statement we can compute P (N | Z ) = =

= 0. 1147541.

Question 2.4.51

Let E be the event Zac communicates via email, M the event he leaves a phone message, and R the event his girlfriend responds within two hours. Then from the question statement we have P (R | E ) = 0. 8 P (R | M ) = 0. 9 P (E ) = 2P (M ).

The last equation implies that P (E ) = and P (M ) = . Using these we can compute

Question 2.4.52

Let F be the event that the product is faulty and let A , B , and C be the events corresponding to which warehouse the product is shipped from. Then from the question statement we have that P (F | A ) = 0. 03 P (F | B ) = 0. 05 P (F | C ) = 0. 02 P (A ) = 0. 3 P (B ) = 0. 2 P (C ) = 0. 5.

Using these we can compute

Question 2.4.53

Let D i be the event we draw from the drawer i for i ∈{ 1, 2, 3} , and S the event that a silver coin is drawn. Then from the question statement we have that P (S | D 1 ) = 0 P (S | D 2 ) = 1 P (S | D 3 ) = .

The other coin will be gold if we have drawn from drawer three. Using the above we can compute P (D 3 | S ) = =

= .

Notes on the Text: The Independence of A and B c In this section we are asked about the independence of A and B c when A and B are independent. To show that these two events are independent note that we can write A = (A ∩ B ) ∪ (A ∩ B c ) and the sets on the right-hand-side are mutually exclusive. Because of this we have

or

showing the independence of A and B c .

Question 2.5.1

Part (a): As P (A ∩ B )≠ 0 these two events are not mutually exclusive.

Part (b): As P (A )P (B ) = 0. 3≠ P (A ∩ B ) = 0. 2 these two events are not independent.

Part (c): For this we have

Question 2.5.2

Let C be the event that Spike passes chemistry and M that he passes mathematics. Then we wan to see if

as these are not equal the two events are not independent. The probability that he fails both subjects is P (C c ∩ M c ) = P ((C ∪ M ) c ) = 1 − P (C ∪ M ) = 1 − (P (C ) + P (M ) − P (C ∩ M )) = 0. 37.

Question 2.5.3

By enumerating the various choices we find this probability to be

= .

Question 2.5.4

Let C i be the chip color drawn from the first urn and C 2 the same thing for the second urn. Then we want to compute P (C 1 = C 2 ) = P (C 1 = C 2 | C 1 = red)P (C 1 = red) + P (C 1 = C 2 | C 1 = black)P (C 1 = black) + P (C 1 = C 2 | C 1 = white)P (C 1 = white) =

+

+

=

.

Question 2.5.5

The probability that Dana looses both games is then 1 − 0. 3 = 0. 7. This means that

Thus P (D ) = 0. 16334. For Dana to win at least one out of four is the complement that Dana looses all of the four games which is P (D c ) 4 = 0. 49. Thus the probability we want is 1 − 0. 49 = 0. 51.

Question 2.5.6

From all samples of three numbers I would expect that one-third of the time the second random variable Y 2 would be in the middle (one-third of the time the first Y 1 would be and one-third of the time the last Y 3 would be). As events A and B are independent we would have P (A ∩ B ) = P (A )P (B ) = .

Question 2.5.7

Part (a): For these two cases we would have

P (A ∪ B ) = P (A ) + P (B ) − 0 = P (A ∪ B ) = P (A ) + P (B ) − P (A )P (B ) =

.

Part (b): For these two cases we would have P (A | B ) =

=0

P (A | B ) =

= P (A ) = .

Question 2.5.8

Part (a): This would be given by Equation 2 and if the events are independent all intersections turn into products to give (4)

Part (b): For this we have

Question 2.5.9

Let X be the number of heads on the first two tosses and Y the number of heads on the second two tosses.

Then we want to compute P (X = Y ) = ∑ x =0 2 P (X = Y | X = x )P (X = x ) = ∑ x =0 2 P (Y = x )P (X = x ) = ∑ x =0 2

2

= 0. 375.

Here we have used the fact that P (X = Y | X = x ) = P (Y = x ) by independence Question 2.5.10

The probabilities can be computed as

P (A ) = P (B ) = P (A ) P (A ∩ B ) = 0, as A ∩ B = ∅ . As P (A ∩ B )≠ P (A )P (B ) these events are not independent.

Question 2.5.11

Lets consider the probabilities of the different events. We have P (A ) = = P (B ) = = P (C ) =

=

P (A ∩ B ) = P (A | B )P (B ) =

=

≠ P (A )P (B )

P (A ∩ C ) = P (A | C )P (C ) =

=

= P (A )P (C )

P (B ∩ C ) = P (B | C )P (C ) =

=

≠ P (B )P (C )

P (A ∩ B ∩ C ) =

= P (A )P (B )P (C ),

which shows the desired equities.

Question 2.5.12

Lets consider the probabilities of the different events. We have P (A ) =

=

P (B ) = P (C ) = P (A ∩ B ) =

= = P (A )P (B )

P (A ∩ C ) =

= = P (A )P (C )

P (B ∩ C ) =

= = P (B )P (C )

P (A ∩ B ∩ C ) =

= ≠ P (A )P (B )P (C ) = ,

which shows the desired equities.

Question 2.5.13

The number of intersections of a set of four events would be 2 4 but one of these is the “no intersection” set and four of these are the Singleton sets. Thus the number of probability equations is 2 4 − 1 − 4 = 11.

Question 2.5.14

Lets consider the probabilities of the different events. We have P (A ) =

=

P (B ) =

=

P (C ) =

=

P (A ∩ B )

=

= = P (A )P (B )

P (A ∩ C ) =

=

= P (A )P (C )

P (B ∩ C ) =

=

= P (B )P (C )

P (A ∩ B ∩ C ) =

= P (A )P (B )P (C ),

which shows needed equities for independence.

Question 2.5.15

Lets consider the probabilities of the different events. We have P (A ) =

=

P (B ) =

=

P (C ) =

=

P (A ∩ B ) =

= = P (A )P (B )

P (A ∩ C ) =

= = P (A )P (C )

P (B ∩ C ) =

= = P (B )P (C )

P (A ∩ B ∩ C ) =

= 0≠ P (A )P (B )P (C ),

which shows that these events are pairwise independent but not mutually independence.

Question 2.5.16

The complement of this event is the event that the commuter has to stop at most one time. If we let E i be the event that the commuter has to stop at the i th light, and E the event we want to compute the probability for then we have P (E ) = 1 − P (E 1 c ∩ E 2 c ∩ E 3 c ∩ E 4 c ) − P (E 1 ∩ E 2 c ∩ E 3 c ∩ E 4 c ) − P (E 1 c ∩ E 2 ∩ E 3 c ∩ E 4 c ) − P (E 1 c ∩ E 2 c ∩ E 3 ∩ E 4 c ) − P (E 1 c ∩ E 2 c ∩ E 3 c ∩ E 4 ).

From the question statement we have that P (E 1 ) = P (E 2 ) =

=

P (E 3 ) = P (E 4 ) = .

Then using independence we can compute the above expression to find

Question 2.5.17

Let M , L , and G be the events that a student fails mathematics, language, or general knowledge respectively. Then from the question statement we have P (M ) = P (L ) = P (G ) =

.

Now a student won’t get a diploma with a probability of P (M ∪ L ∪ G ). Then using Equation 4 we can write this as a sequence of independent intersections. Evaluating this we get P (M ∪ L ∪ G ) = 0. 558.

Independence is probability not a good assumption as students who fail one of these areas will likely fail in another.

Question 2.5.18

The switch will work if the event

is true. Using independence, we can write the probability of this event as P (E ) = P (A 1 ∩ A 2 ) + P (A 3 ∩ A 4 ) − P (A 1 ∩ A 2 ∩ A 3 ∩ A 4 ) = P (A 1 )P (A 2 ) + P (A 3 )P (A 4 ) − P (A 1 )P (A 2 )P (A 3 )P (A 4 ) = 2p 2 − p 4 .

Question 2.5.19

Let p be the probability that the person does not win on a purchase. Then assuming independence and from what we are told we have

Solving for p we find p = 0. 9257672. The probability that a person wins on a single purchase is then 1 − p = 0. 07423282.

Question 2.5.20

Person A will win with a probability of P (A ) = +

3

= ∑ k =0 ∞

+

6

3k

+

=

= .

Person B will win with a probability of P (B ) =

+

3

= ∑ k =0 ∞

3k

Person C will thus win with a probability of

+ =

6

+ = .

P (C ) =

+

2

= ∑ k =0 ∞

2

3k

+

3

=

2

6

+

= .

Note that we have P (C ) = 1 − P (A ) − P (B ) as we must. In the above derivations we have assumed that to win some number of people who flip before must have lost.

Question 2.5.21

Let p =

be the survival probability. Then we want to know n such that

Solving for n in the above we find n ≥ 6. 212567. As n must be an integer we should take n = 7.

Question 2.5.22

Let p be the probability that a given person can identify a person’s voice so from the given question we are told that p = 0. 15. Then we want to compute

Next we want to compute

Question 2.5.23

If we let E be the event in question and D n be the event that we get a n on the die roll. Then we can compute this probability by conditioning as P (E ) = ∑ n =1 6 P (E | D n )P (D n ) = ∑ n =1 6 P (E | D n ) = ∑ n =1 6 =

0

=

n−0

= ∑ n =1 6

n

=

∑ n =1 6

n−1

=

∑ n =0 5

n

= 0. 1640625.

Question 2.5.24

As each urn has the same distribution of red and white chips the probability of drawing a red chip on a single draw from any urn is given by p = . The probability we draw no red chips from the r draws on a single urn is

The probability that we then draw no red chips from any of the r draws from each of the m urns is then

The probability that we draw at least one red chip is then

Question 2.5.25

If two fair dice are tossed the probability of getting a double six is p =

. Then in n tosses

For the question we want n such that

Solving for n we get n ≥ 24. 6051 thus we should take n = 25.

Question 2.5.26

If two fair dice are tossed the probability of getting a seven is p 7 = = , the probability of getting an eight is given by p 8 = , and the probability of getting anything else is p e = 1 − − = . Then a sum of eight will appear first with a probability of

Question 2.5.27

The probability of getting a white chip is p w = , the probability of getting an black chip is given by p b = , and the probability of getting a red chip is p r = . Then the probability that a white appears before a red is given by

Question 2.5.28

Lets first evaluate the probability that we don’t find the ship when searching. Then from the question if we send m helicopter crews to area I (and n − m crews to area II) we would have P (not found) = P (not found| A 1 )P (A 1 ) + P (not found| A 2 )P (A 2 ) = (1 − (1 − r ) m )p + (1 − (1 − r ) n − m )(1 − p ).

Now as B be the event that the ship is found. The probability of this event is P (B ) = 1 − P (not found) = 1 − (1 − (1 − r ) m )p − (1 − (1 − r ) n − m )(1 − p ).

To find the number m to send to area I that will maximize the value of P (B ) above we need to take the derivative of P (B ) with respect to m . Doing this we find

If we set this equal to zero we find that

Taking the logarithm of both sides gives

or

This is the argument to basically send one-half of the crews into each area but with more crews sent to the the region with a larger likelihood of having the ship found.

Question 2.5.29

The probability of any digit being a four is fours is then

. The probability that a sequence of n digits does not have any

The probability that a sequence of length n will have at least one four is then

For the question we want n such that P n ≥ 0. 7. Solving for n we find n = 11. 42717. As n must be an integer we would take n = 12.

Question 2.5.30

Let B be the event that we have selected the biased (two-headed) coin when we select our initial coin. Let H n be the event that when flipping this coin n times we get n heads. Then using Bayes’ rule we have P (B | H n ) =

=

=

=

.

Note that in the above as n →∞ we have that P (B | H n ) → 1 and thus that P (B c | H n ) → 0.

Question 2.6.1

She would need to run 2 × 3 × 2 = 12 unique experiments and then double this number (for 24) to have each experiment duplicated twice.

Question 2.6.2

There are 26 − 6 = 20 consonants (if we count Y), nine digits between one and nine, and six vowels (from A, E, I, O, U, and Y). The total number of different ciphers is then 20 × 9 × 6 × 20 = 21600.

Question 2.6.3

We have to select one of the terms in each factor to give a total of 3 × 3 × 5 = 45. The terms that we get will have to have one term from each factor. Thus we will have terms like aeu and cdx .

Question 2.6.4

Part (a): This would be 26 2 × 10 4 = 6760000.

Part (b): This would be 26 2 × 10 × 9 × 8 × 7 = 3407040.

Part (c): This would be 26 2 × (10 4 − 1) = 6759324.

Question 2.6.5

The number of distinct digit numbers would be 9 × 9 × 8 = 648. Now that number is composed of even numbers and odd numbers that satisfy the distinct digits requirement. To compute the number of odd numbers we will compute the number of even numbers and subtract that from this total. An even number has to end in a digit of zero, two, four, six, or eight. Thus there are five choices for the ones digit. There are nine choices for the tens digit (the digits 0-9 excluding the digit picked in the ones place). Finally there are seven choices for the hundreds digit (the digits 1-9 but excluding the two digits selected in the ones and tens place). This gives or 7 × 9 × 5 = 315 even numbers and thus

odd numbers.

Warning: this result does not match that in the back of the book. If anyone sees anything wrong with my logic please contact me.

Question 2.6.6

This would be 2 8 = 256 because each topping can be added (or not) for two choices for each topping.

Question 2.6.7

Each “base” can have a runner on it or not to give 2 b possible base configurations. We can have from one to t “outs” giving a total of 2 b t different “base-out” configurations. If we check this number for the normal game of baseball we get 2 3 × 3 = 24 (which is correct). For sleazeball this number would be 2 7 × 5 = 640.

Question 2.6.8

The first number can be any of the digits in { 6, 7, 8, 9} (or four choices). The last number must be one of zero, two, four, six, or eight (or five choices). The third number must be the value of seven (or one choice).

All other digits can be between zero and nine (or ten choices). This gives

possible zip codes that meet this criterion.

Question 2.6.9

We first have to decide what three items the patron will choose and then count the number of different dinners the patron can order using them as components. A patron can make his meal selecting from an appetizer, an entree, and a dessert an appetizer, an entree, and a beverage an entree, a dessert, and a beverage an appetizer, a dessert, and a beverage The number of ways to make meals of this form is then

Question 2.6.10

We can start with a black key and then alternate to get

We can also start with a white key and alternative to get

which is the same number as before. Adding these two, in total we have 3001250 combinations.

Question 2.6.11

If we assume that a button can be pushed (or not) we have 2 8 combinations. One of these is “don’t push any buttons”. Thus we can have 2 8 − 1 = 255 combinations. If we have 250 families already we have “space” for 255 − 250 = 5 more.

Question 2.6.12

If we have n “spaces” then each space can be a dot or a dash. This gives a total of 2 n unique symbols that can be represented. As 2 4 = 16 < 26 < 32 = 2 5 we need at least five spaces to represent the English alphabet.

Question 2.6.13

The only sequence of coin flips that have a value smaller than 256 are the ones where the value of a 8 is zero.

Thus all numbers larger than (or equal to) 256 are the ones that can have a 0 ,a 1 , … ,a 6 ,a 7 take any value and a 8 = 1. There are 2 8 possible ways to assign binary 0-1 digits to a 0 ,a 1 , … ,a 6 ,a 7 but one of these is the all zero assignment. This will give a binary number equal to 256. Thus there are 2 8 − 1 numbers of the desired form.

Question 2.6.14

There are three vowels to choose from in ZOMBIE (O, I, and E). There are four consonants (Z, M, B, and S).

Thus we have 3 × 4 = 12 ways to select a vowel and a consonant. If the order is different we have two ways to order them giving a total of 24 orderings.

Question 2.6.15

There are thirteen ways the first card can be a club one of these is the ace of clubs. If we have picked the ace of clubs there are three other ways we can draw the second card and have it be another ace. If we have not picked the ace of clubs for the first card there are four other ways we can draw the second card and have it be an ace. This gives

ways.

Question 2.6.16

On her departing trip she has 3 × 5 × 2 = 30 ways it can be scheduled. On her returning trip she has the same number so the total number of different round-trip options she has is 30 2 .

Question 2.6.17

This would be 6 P 3 = 6(5)(4) = 120.

Question 2.6.18

For the first part, this would be 4 P 0 = 1. For the second part, we can select the location of the first snow tire in four ways and the location of the second in three ways to make a total of 12 ways.

Question 2.6.19

We will use (5) Taking the logarithm to the base ten of both sides we get

(6) Taking n = 30 the right-hand-side gives 32.42245. This means that n ! ≈ 2. 645148 × 10 32 .

Question 2.6.20

This would be 9!.

Question 2.6.21

To have our three digit number be less than 289 we have to have our first digit be a one or a two. If its a one we can use any of the other digits in the other two places to give

numbers of this form starting with a one. If the first digit is a two then the same logic above still holds and we have

numbers of this form starting with a two. The total number of numbers of this form then is 60.

Question 2.6.22

Part (a): We can pick the location of the man in 8 × 7 × 6 × 5 = 1680 ways and have 4! ordering of the men.

We can also have 4! ordering of the women in the remaining locations. This gives

arrangements.

Part (b): We can start with a man at position one (or at position two) and there are 4! orderings of the men and the woman to give

arrangements.

Question 2.6.23

The engineer can select an ordering of his three electives in 10 P 3 = 720 ways. He can select the semester he does not want to take an elective in four ways. Thus he can complete his requirements in 4(720) = 2880 ways.

Question 2.6.24

We can order the six men (and then the women) in 6! ways each to get a total of 6! 2 pairings. We can order these six pairs in 6! ways to get a total of 6! 3 ordered pairings along one sideline.

The number 6! 2 2 6 might represent the number of ways the six male-female teams could be partitioned on one of two of the sidelines.

The number 6! 2 2 6 2 12 might represent the number of ways the six male-female teams could be partitioned on one of two of the sidelines (the 2 6 factor) and then whether or not the person is doing a cartwheel (the 2 12 factor).

Question 2.6.25

There are 6! ways to play the six different sides of the album of which only one ordering is fully correct.

Thus there are

ways to play the album incorrectly.

Question 2.6.26

We can order the n families in n ! ways. In each family we can order the m people in m ! ways as we have n families each to order the total number of arrangements is

Question 2.6.27

There are 8! ways to order everyone but you and your friend. Once these people are ordered you can go before the first person (your friend is then between the third and fourth person) go after the first person and before the second person (your friend is then between the fourth and fifth person) etc. go after the fifth person and before the sixth person (your friend is then after the eighth person) This gives 6 possible locations for you to go (and then your friends location is determined). This gives 6 × 8! ordering of this type. As we can switch the location of you and your friend we have an additional factor of two to get 12 × 8! possible orderings.

Question 2.6.28

Note that by selecting one of the n items to be at the first position and recalling that the definition of n P k is the number of ways of selecting k ordered items from n we have that

Thus we have selected one item (from n ) and then we have to select k − 1 more items from n − 1 items. Note that our boundary condition is n P 1 = n . This is the basis of the induction proof.

Question 2.6.29

Start by laying out all 13 cards of the same suit. There are 13! ways to order these cards. If we do this for each of the four suits there are (13!) 4 ways to give each player one card from each suit.

Question 2.6.30

From the definition of n ! we note that we have

As 1! = 1 if we let n = 1 in the above becomes 1 = 1 × (0!). To be consistent means that 0! = 1.

Question 2.6.31

We can pick two aviators from the nine in

= 36 ways and order them (to get a pilot and a copilot) in

two ways. We can pick one geologist from the four in four ways. This gives

different crews.

Question 2.6.32

We can put Harry and then Minnie at the positions { (1, 4), (2, 5), (3, 6), (4, 7)} and they will have two people between them. { (1, 5), (2, 6), (3, 7)} and they will have three people between them. { (1, 6), (2, 7)} and they will have four people between them. { (1, 7)} and they will have five people between them.

Note that for each of the above sets we can put Minnie at the first location and Harry at the second location and get a valid seating. Once we have selected Harry and Minnie’s seats we can order the other five people in 5! ways. Thus the number of seats we have are given by

ways.

Question 2.6.33

The even numbers are { 2, 4, 6, 8} of which there are four of them. The odd numbers are { 1, 3, 5, 7, 9} of which there are five of them.

Part (a): We can order the even digits in 4! ways and the odd digits in 5! ways to give a total of 4! × 5! = 2880 ways.

Part (b): We can order the even digits in 4! ways and then place the four digits considered as a single unit with the odd numbers if 6! ways. This gives 4! × 6! = 17280 orderings.

Part (c): We have four choices for the first even digit, three choices for the second even digit, two choices for the next to last even digit and one choice for the last even digit. Then we can order the five odd digits in 5! ways to give a total of 4! × 5! = 2880 orderings.

Part (d): There are two orderings of the even digits in ascending or descending order either 2, 4, 6, 8 or 8, 6, 4, 2. Once we pick an ordering we have five ways to pick an odd digit and then five places we could “put it down”. That is, there are five places (from before the first even digit to after the last even digit) where we can put it. Then there are four choices for the next odd digit and six places we can put it. This pattern continues and we have

Warning: Note that this result is different than the one given in the back of the book. If anyone sees anything wrong with what I have done please contact me.

Question 2.6.34

We would need to compare

As these two numbers are 3780 and 5040 the second (FLORIDA) has more permutations.

Question 2.6.35

To be larger than four million we must have a four or a five be the numbers first digit. If a four is the first digit then we can place the other digits in

ways. If the five is the first digit then we can place the other digits in

ways. Adding these two gives a total of 300 ways.

Question 2.6.36

Part (a): This would be

Part (b): This would be 8! = 40320.

Part (c): This would be

= 6720.

Question 2.6.37

Part (a): This would be 4! × 3! × 3! = 864.

Part (b): For each of the orderings above we can pick to order the Nigerians, the Chinese, and the Greeks in 3! ways to have 864 × 3! = 5184 orderings.

Part (c): This would be (4 + 3 + 3)! = 10! = 3628800.

Part (d): This would be

Question 2.6.38

There are eleven letters in the word SLUMGULLION and we can use the following python code to count the number of each letter from collections import Counter word = ’SLUMGULLION’ print Counter(word)

This gives Counter({’L’: 3, ’U’: 2, ’G’: 1, ’I’: 1, ’M’: 1, ’O’: 1, ’N’: 1, ’S’: 1})

The consonants (excluding L) are { G,M,N,S } . We can arrange these in 4! ways. The three L’s must then be placed to the left of the specific arrangement of these consonants. We still have to place the four vowels. To think of how to do this note that the final “word” will occupy eleven spaces and we can select the vowel locations in

ways. Once these locations are specified we can put the previously created “word” (the

one with three L’s followed by a permutation of { G,M,N,S } in the other 11 − 4 = 7 blanks. Finally in the four spaces for the vowels { U,U,I,O } we can order them in ways. Thus in total we have

ways Question 2.6.39

Imagine ordering all 2n people linearly i.e. along a line. We can do that in 2n ! ways. Then take the last onehalf of the people and pair them with the first one-half of the people. We can order these pairs then in n ! ways and not have a different total pairing. Finally for each of these n pairs we have two ways to order the players and each of which gives the same pair. Thus if N p are the number of possible pairing we have that

Solving for N p gives the desired answer.

Question 2.6.40

We can form x 12 from the product of two terms like x 6 and sixteen ones. This can be done in

ways. We can also form x 12 from the product of two terms like x 3 , one term like x 6 , and then fifteen ones.

This can be done in

ways. We can also form x 12 from the product of four x 3 and then fourteen ones. This can be done in

The total of these three numbers is 5661.

Question 2.6.41

Now without the letter S there are eleven letters in this “word” that can be ordered in

ways. For each of these ordering as there are two Y’s we can place the S in front of either one to get a desired ordering of the letters. This means we have

valid orderings.

Question 2.6.42

We start by ordering the letters of this puzzle in a square with the first A in the lower-left and the last A in the upper right (this is basically rotating the page counterclockwise a little less than 180 degrees). Then to spell the word ABRACADABRA we need to trace a path in this grid from corner to corner which we do by moving in steps that move either right or up. As we move five times rightwards and five times upwards this question is equivalent to the number of ways we can order five R’s and five U’s. We can to that in

ways.

Question 2.6.43

The last pitch must be a ball or a strike. If the last pitch is a ball then the first four could be 3 balls and 1 strike 2 balls and 2 strikes These can be ordered in

ways. If the last pitch is a strike then the first four could be 3 balls and 1 strike 2 balls and 2 strikes These can be ordered in four and six ways as above. Thus the total number of ways is 4 + 6 + 4 + 6 = 20.

Question 2.6.44

We would need to select w twice, x three times etc. Thus the total number of ways we can get this monomial is

and will be the coefficient of w 2 x 3 yz 3 .

Question 2.6.45

Following the hint we the number we seek is the number of permutation of 3 A’s and 3 B’s which is

ways.

Question 2.6.46

If we consider Theorem 2.6.2 and take n = k ! and r = (k − 1)! with n i = k for i = 1, 2, … ,r then using the results of that theorem we have that

is the number of ways to arrange these n objects with n i into each group and is thus an integer. Note that there are (k − 1)! items in the product in the denominator so the above fraction is

As this fraction is an integer we have that the denominator divides the numerator.

Question 2.6.47

The number of ways we can order all the letters in this word is (7) Looking only at the vowels A, A, I, I, O in this word they can be uniquely ordered in

ways.

In any of the letter orderings given by Equation 7 above the vowels can be in many different orders only one of which is the desired O, I, A, I, A ordering. This means that from the 30 vowel orderings only one satisfies the requirement of the question so we need to reduce the number above by to get

Question 2.6.48

Counting the number of letters and the number of each duplicates the number of different orderings we could make with the letters would be

Question 2.6.49

The number of ways we can assign three A’s and two B’s in using the following python code

= 10 different ways. We can enumerate these

from itertools import combinations for it in combinations(’EMFPH’, 3): print it

Running the above we get (’E’, ’M’, ’F’) (’E’, ’M’, ’P’) (’E’, ’M’, ’H’) (’E’, ’F’, ’P’) (’E’, ’F’, ’H’) (’E’, ’P’, ’H’) (’M’, ’F’, ’P’) (’M’, ’F’, ’H’) (’M’, ’P’, ’H’) (’F’, ’P’, ’H’)

for the list of subjects that Linda gets A’s in (the other subjects she gets B’s in).

Question 2.6.50

This would be

= 10.

Question 2.6.51

If we pick m marginally acceptable candidates we must then pick 2m highly desirable candidates to have the ratio be two-to-one. But as we are only picking nine total candidates this means that

Thus the number of ways we can select m and 2m from the choices is then

.

Question 2.6.52

This would be

= 560.

Question 2.6.53

Part (a): Assume we will choose k women (from the four) and then 4 − k men (from the five). Then the number of ways we can select the summer interns is

Note that using the result from Question 2.6.63 the above is equal to

Part (b): If we have to have two men and also two women then the number of ways is

Part (c): This would be

Question 2.6.54

This would be

= 21 and order does not matter.

Question 2.6.55

Now = 252 would be the number of ways to select a group of five where the order of the players in the group selected is not important. For one of the given selections we get one group (say A ) of five players.

Of course the players that are not selected form the other (say B ) team. Now the above number also includes as one of its counts the selection of the players that ended up in the B team. Thus double counts the number of distinct teams that we can have. The number of teams that we can have is thus

Question 2.6.56

If we let x i be the number of pages read on each of the three days we must have x 1 + x 2 + x 3 = 20 with x i ≥ 1. Now for x 1 we can choose any number in the range [1, 18] for 18 choices. Once x 1 is selected then we have x 2 + x 3 = 20 − x 1 . This means that x 2

which gives (20 − x 1 − 1) − 1 + 1 = 20 − x 1 − 1 choices for x 2 . Once x 2 has been specified x 3 is determined. Thus the number of ways to complete the reading assignment is x 1 = 1 then x 2 has 18 choices x 1 = 2 then x 2 has 17 choices x 1 = 3 then x 2 has 16 choices etc. x 1 = 18 then x 2 has one choice This gives a total of

Counting the number of solutions to equations of this type is discussed in [2 ] where the number of solutions is shown to be . In this case n = 20 and r = 3 so the number of solutions is = 114 the same as we got using the logic above.

Question 2.6.57

In the word MISSISSIPPI There are one M, four I’s, four S’s, and two P’s. Thus there are four Is and seven other characters. We can place these seven characters down in

unique orderings. Once these letters are placed we can place an I in the “spaces” between the letters of the word just constructed. There are eight “spaces” (counting the front and back of the word) where we could (or not) put a single I. This will result in no to adjacent Is. This can be done in = 70 ways. Thus the total number of spellings is 105(70) = 7350.

Question 2.6.58 (sums of binomial coefficients)

Recall the binomial theorem which is

If we let x = 1 and y = 1 then x + y = 2 and the sum above becomes

which is the desired identity.

Question 2.6.59

We can use the identity = to write the left-hand-side in the form suggested by the hint. Now in that form the right-hand-side is the number of ways to select a set of n items from 2n items. For the left-hand-side we can imagine that the set of 2n items have n of them colored red and the other n of them colored blue. Then we can select n items from the 2n by deciding how many blue items to select (say k ) and then n − k red items. This is what the expression on the left-hand-side represents when we sum over all of the values k can take i.e. 0 ≤ k ≤ n .

Question 2.6.60 (more sums of binomial coefficients)

Again recall the binomial theorem which is

If we let x = − 1 and y = 1 then x + y = 0 and the sum above then becomes

If we move the odd (or even) terms to the other side of the equation we get the desired expression.

Question 2.6.61 (maximum of

as a function of k )

Because the binomial coefficients satisfy

as a function of k the values of the binomial coefficients repeat once k ≥ . If we can show that

is an

increasing function of k , the k value that will maximize the binomial coefficient will be the one corresponding to k ≈ and the binomial coefficient will increase until k ≈ and then must decrease. If n is even this maximum will occur the value floor( ) while if n is odd this maximum will be found at floor( ) and ceiling( ).

All one has to do is show that

is an increasing function of k , which means proving the following chain

of reasoning is reversible

which says that

is increasing as a function of k as long as k <

, which equals the floor and ceiling

expressions above if n is odd or even respectively.

Question 2.6.62

Let J be the number of jokes that Mitch has. The number of ways to draw sets of four jokes is . To have repeating sets of jokes this number must be larger than four months times 30 days per month or 120 performances. If J = 8 we find = 70 while if J = 9 we find = 126. Thus Mitch needs nine jokes.

Question 2.6.63

We can expand each of the terms in the product (1 + t ) d (1 + t ) e = (1 + t ) d + e to get (1 + t ) d = ∑

i =0

(1 + t ) e = ∑

j =0

(1 + t ) d + e = ∑

k =0

d

tl

e

tj

d+e

tk.

The product of (1 + t ) d (1 + t ) e will have the coefficient of its t k th term given by all products t i and t j that will sum to k or

This must equal the coefficient

which is the coefficient of t k in the expansion of (1 + t ) d + e above.

Question 2.7.1

This would be

Question 2.7.2

As there are two pairs { (1, 4), (2, 3)} that sum to five the probability of this event would be

Question 2.7.3

There are

ways to draw two chips. There are the nineteen pairs { (1, 2), (2, 3), … , (19, 20)} where

the two chips differ by one. There are the eighteen pairs { (1, 3), (2, 4), … , (18, 20)} where the two chips differ by two. Thus the probability that the two chips differ by more than two is

Question 2.7.4

Using the notation from this question we can evaluate P (A ∪ B ) as P(A∪B) =P(A)+P(B)−P(A∩B) =

+

−

= 0 . 005280403 .

Question 2.7.5

Let A be the event that we picked one of the first nine urns, B the event that we picked the tenth urn and E the event that the three drawn chips are white. Then we want to compute P (B | E ) which we will do using Bayes’ rule as

From the question statement we know

P (E | B ) =

=

P (E | A ) =

=

=

P (A ) = P (B ) =

.

This means that

Question 2.7.6

We can choose our fifty politicians in

ways. In each state we can choose one representative in

2 ways to give the probability P of

Question 2.7.7

There are

ways to select the fact that be duplicated n times and each sequence of that form has a

chance of occurring. This gives our total probability P of

Question 2.7.8

This probability would be

Here

= 6 is the number of ways we can pick the face on the die that will repeat three times (call that

number x ) and

= 5 is the number of ways we can pick the face on the die that repeats two times (call

=

that number y ), and

is the number of ways that we can select three spots to place the x ’s (the other two

spots get the y ).

Question 2.7.9

If we think of the items as distinguishable then we have (2n )! ways to order all of them. In the configuration where all of the white items are on the left we have n ! ways to order them and n ! ways to order the n black items on the right. As we could switch and place the white items on the right and the black items on the left we have 2(n !) 2 valid separate orderings of white and black grains of sand. Thus our probability P is given by

Question 2.7.10

There are

unique permutations of CALCULUS and

unique permutations of ALGEBRA. As

it is more likely that a monkey would randomly spell ALGEBRA. Question 2.7.11

From all of the seven people there will be one floor that at least one person gets off on. Then a second person will get off on a different floor with a probability of . This event happening will leave two of the seven floors “taken”. A third person will get off on a different floor than the first two with a probability of .

Repeating this argument for all people gives

for the probability that all people want to get off on different floors.

As the first man must get off on one floor we then need to have the six get off on the same floor which will 6 happen with a probability of .

Question 2.7.12

This “word” has 26 total letters in it of which four are the letter S. Counting the numbers of the other letters we have

unique permutations of all of the letters. Of these one subset are the ones that have all of the S’s adjacent. If we think of the four S’s as one “unit” we can order all of these objects (letters and the S unit) in

unique ways. The difference between these two numbers would be the number of ways we can order the letters where not all S’s will be adjacent.

Question 2.7.13

We can choose 10 pairs from 20 in

ways. The number of ways all broken parts can be recombined and

stay matched is 2 10 . Thus the probability P is

Question 2.7.14

There are 6 6 possible outcomes when we roll six die. There are 6! outcomes where we have one outcome from each face. Thus the probability we want is given by

Question 2.7.15

The number of sequences of birthdays that k people can have is 356 k . If we desire to have exactly one pair from these people share the same birthday then the pair be can be selected in

ways. Once that pair is

selected to have the same birthday the number of unique sequences is for the k − 1 unique birthdays is

Thus the probability P we seek is

Question 2.7.16

The number The number

is the number of subsets of size three we can form from to make the three single cards. is a permutation which implies that the order of the cards is important

(which it is not). For example this number would include both “hands” of three single cards like 3, 4, 6 and 6, 4, 3 which differ only in the order of the numbers given.

Question 2.7.17

Recall that a flush is a hand with all cards of the same suit but not necessarily in sequential rank. There are 52 − 5 = 47 cards she has not been drawn of which 13 − 2 = 11 are diamonds (the suit she needs to get a flush). Thus the probability P is given by

Question 2.7.18

Recall that a full house is three matching cards of one rank and two matching cards of another rank. Thus to draw a full house the player needs another ace either an: ace of diamonds, an ace of clubs, or an ace of spades. Thus

as there are 47 cards left after removing the five the player originally took. A hand is called “four of a kind” is four cards of the same rank and one card of another rank. Then to draw “four of a kind” the player must draw the queen of spades. Thus

As these events are mutually exclusive the probability of drawing either a full house or a four of a kind is

Question 2.7.19

To draw a straight flush Tim needs to draw a five and a seven of hearts or a seven and a ten of hearts which are two of the possible pairs he could draw. This event happens with a probability P of

To draw flush Tim needs any two hearts. As there are ten hearts he has not drawn this probability P is

as two of these pairs (from the Question 2.7.20

possible pairs) give the straight flush from before.

If the king, queen, and jack all count for ten then there are a total of sixteen cards that have a value of ten. The ways we can get a sum of 48 or more are five cards with value of 10 four cards with a value of 10 and one with a value of nine four cards with a value of 10 and one with a value of eight three cards with a value of 10 and two with a value of nine There are

ways to draw five cards and

ways to draw the four ways we can get a sum of 48 (or more). The ratio of these two numbers gives the probability in question.

Question 2.7.21

We can pick the three even numbers in

ways and the suit for each of them in

. This gives

as the partial “count” of the number of hands of this type to this point. Next we can pick the face card in ways and its suit in

ways. This gives

as the count to this point. Finally, we can pick the second face card in

ways and its suit in

ways.

Thus the probability P we seek is given by

Question 2.7.22

In a coke hand no cards are the 10, the ace, the jack, the queen, or the king which as each has four suits removes 5 × 4 = 20 cards from the deck leaving 52 − 20 = 32 remaining. Then the probability P of being dealt such a hand is given by

Question 2.7.23

There are two jacks, two kings, and two queens of each suit in the pinochle deck. Then the probability P must be

Chapter 3 (Random Variables) Question 3.2.1

Let p be the probability that the stock goes up. From the question we estimate p = . In four days the stock can go up 0, 1, 2, 3, 4 times (which means it has gone down 4, 3, 2, 1, 0 times). It will be at the same price it is now if it goes up and down the same number of times. This only happens if the number of times it goes up is two. The probability that this happens is (in R notation) dbinom(2, 4, 0.25) which equals 0.2109375.

Question 3.2.2

We must have at least one-half of the rods inserted successfully. This means that the system will fail if less than one-half of the rods are inserted successfully or

Here p = 0. 8 when we evaluated the above.

Question 3.2.3

We find 0. 23 12 = 2. 191462 × 10 8 which is close to the given number.

Question 3.2.4

This would be given by

with p = 0. 153. Evaluating the above we get 0.230583.

Question 3.2.5

This would be

with p = 1 − 0. 1 = 0. 9. Evaluating the above we get 0.01853476.

Question 3.2.6

This would be

with p = . This is one minus the probability that from all 10000 sightings none are genuine. Evaluating the above we get 0.09516303.

Question 3.2.7

The plane with two engines will fly safely with a probability of

while the plane with four engines will fly safely with a probability of

where p = 1 − 0. 4 = 0. 6. Evaluating these two we get 0.84 and 0.821 respectively. Thus we would prefer the two engine plane.

Question 3.2.8

The probability that a system will need k bulbs replaced is given by a binomial random variable. The probability that we will need to replace at least one bulb is then given by

with p the burn out probability. Evaluating this for the two systems gives the numbers [1] 0.9230550 0.8673804

Thus we would prefer the first (the 50 bulb system).

Question 3.2.9

Each of these can be evaluated using a binomial variable. To roll a single six on a die will happen with a probability of 1⁄ 6. Let N be the number of sixes rolled. Then N is a binomial random variable with p = 1⁄ 6 and n different depending on the experiment. We want to evaluate the following (all with p = 1⁄ 6)

In R we compute these to be [1] 0.6651020 0.6186674 0.5973457 Question 3.2.10

In R the probability that the boat (plane) destroy the other can be computed using p_boat = sum( dbinom( 2:6, 6, 0.2 ) ) p_plane = sum( dbinom( 1:10, 10, 0.05 ) ) print( c(p_boat, p_plane) )

These give [1] 0.3446400 0.4012631

Thus one would prefer to be on the plane.

Question 3.2.11

We can evaluate these using p_2_n_2 = dbinom( 2, 4, 0.5 ) p_3_n_1 = dbinom( 1, 4, 0.5 ) + dbinom( 3, 4, 0.5 ) print( c( p_2_n_2, p_3_n_1 ) )

These give [1] 0.375 0.500 Question 3.2.12

We can evaluate this using sum( dbinom( 0:6, 12, 0.5 ) )

which gives 0.612793.

Question 3.2.13

If we have d drivers and assume that we could call each one and ask if he/she could drive the probability that we have k drivers that can drive is

p k (1 − p ) d − k with p = 0. 8. We need to have 4 ≤ k ≤ d to be sure

that all the buses are working. Thus we want to know the smallest value of d such that

We can find the smallest value of d using the following R code for( d in 4:10 ){ p = sum( dbinom( 4:d, d, 0.8 ) ) if( p >= 0.95 ){ print(c(d, p)) break } }

which gives d = 7 and a probability above of 0.966656.

Question 3.2.14

The probability that the Navy gunboat hits the bunker on one shot is p = three hits. We can compute the probability that this happens with

=

and we need to have at least

sum( dbinom( 3:25, 25, 3/50 ) )

which gives 0.1871054.

Question 3.2.15

We can compute these two with p_a = dbinom( 3, 7, 0.5 ) p_b = sum( dbinom( 0:2, 7, 0.25 ) ) print( c( p_a, p_b ) )

which gives [1] 0.2734375 0.7564087 Question 3.2.16

The world series is a “best of seven” event so we can follow the Stanley Cup example from this section of the book to work this problem. We would compare the expected frequencies (under the binomial model) with the actual frequencies observed. We can do that in the following R code P_4 = 2*0.5^4 P_5 = 2*dbinom(3, 4, 0.5)*0.5 P_6 = 2*dbinom(3, 5, 0.5)*0.5 P_7 = 2*dbinom(3, 6, 0.5)*0.5 expected_freq = 58 * c( P_4, P_5, P_6, P_7 ) obs_freq = c( 12, 10, 12, 24 ) print( data.frame( expected_freq, obs_freq ) )

This gives the following 1 2 3 4

expected_freq obs_freq 7.250 12 14.500 10 18.125 12 18.125 24

This does not seem to fit the data very well.

Question 3.2.17

The given sum, using the binomial expansion of (x + y ) n with x = p and y = 1 − p can be seen to be equal to

as claimed.

Question 3.2.18

This would be the expression

Question 3.2.19

We would use the formula above as p1 = 0.5 p2 = 0.3 n = 10 k1 = 3 k2 = 5 c = factorial(n) / (factorial(k1) * factorial(k2) * factorial(n-k1-k2)) c * p1^k1 * p2^k2 * (1-p1-p2)^(n-k1-k2)

which gives 0.030618.

A note on question with the hypergeometric probability density

In each of the following questions we will associate a “red” class and a “white” class to the objects under discussion. Once we have done that we can answer the probability questions more easily using the hypergeometric distribution.

Question 3.2.20

Let the accountants be the “red” class and the non-accountants be denoted the “white” class then we want to evaluate a hypergeometric distribution with k = 2. Using R we find this to be 0.4242424.

Question 3.2.21

We break the six bears into two groups. We can let the black bears be members of the “red” class and the tan bears members of the “white” class. Let X be the random variable denoting the number of black bears observed. Then we will have at least twice as man black bears as tan bears if X ∈{ 6, 5, 4} . Using the hypergeometric density the probability that this happens is given by (in R ) as dhyper(6, 6, 3, 6) + dhyper(5, 6, 3, 6) + dhyper(4, 6, 3, 6)

This gives the number 0.7619048.

Question 3.2.22

If we let the children that have not been vaccinated be the “red” class then the probability that we have k children not vaccinated is given by a hypergeometric probability or

for 0 ≤ k ≤ 514.

Question 3.2.23

If we let the six nuclear warheads be the “red” class and the four other nonnuclear missiles be the “white” class. Then when country B launches its seven missiles it will only be safe if it hits the six nuclear ones. In R the probability country B will not be safe is then 1-dhyper(6, 6, 4, 7)

This gives the value of 0.9666667.

Question 3.2.24

Let the “red” class be the eight topics that Anne studies. Then the “white” class are the two topics that Anne did not study. She will have an 85% chance of getting at least four of the five questions correct if from the five that her professor selects four (or five) of them are red. In R this happens with a probability of dhyper(4, 8, 2, 5) + dhyper(5, 8, 2, 5)

This gives the number 0.7777778 which is not larger than 0.85 so she does not have the required chance.

Question 3.2.25

If we let the men be the “red” class and the women the “white” class then we will not have a man and women represented if the selection is all men or all women. The complement of this event is the one that we want to know the probability of. In R we can compute this with 1 - dhyper(5, 8, 10, 5) - dhyper(5, 10, 8, 5)

This gives the value 0.9640523.

Question 3.2.26

Let the “red” class be the numbers that will be called by the “caller”. Then the “white” class are all the other numbers. We want to know from the draw of size ten (the number of numbers the player picked) what is the probability six are “red”. In R this is computed with dhyper(6, 10, 70, 20)

This is the number 0.01147939.

Question 3.2.27

The burglar must first draw a single diamond in the first set of three gems removed. Then the burglar must draw a second diamond on his fourth draw (after removing one real diamond and two fake diamonds). This happens with probability of dhyper(1, 10, 25, 3) * (9/32)

which gives the number 0.1289152.

Question 3.2.28

In this problem the student must first draws one sock. The probability that the sock is black, brown, or blue respectively are given by

The color of the first sock drawn will affect his chances of getting a matched pair. Specifically the color of the second sock will match that of the first with a hypergeometric distribution with the appropriate parameters. In R this would be (2/10) * dhyper(1, 1, 8, 1) + (6/10) * dhyper(1, 5, 4, 1) + (2/10) * dhyper(1, 1, 8, 1)

This gives the number 0.3777778.

Question 3.2.29

Using the binomial expansion we can write the left-hand-side as

Using the same expansion we can write the right-hand-side as

If we multiply these two series together we get

As n + m = k in this sum (or m = k − n ) the above can be written as

If we equate coefficients of μ k with the expansion we had for the left-hand-side we see that

If we then divide both sides of the above by

we get

Recalling the hypergeometric probability density function we see that the resulting expression shows that the sum of all possible values of the hypergeometric probability mass function gives one.

Figure 2: The OC curve for Question 3.2.32.

Question 3.2.30

For this we can expand the given expression. We have that it is equal to

or

If we cancel common factors this becomes

the expression we were to show.

Question 3.2.31

To solve this question we will condition on the result of the draw from the first urn. That draw can results in two reds, one red and one white, or two whites. The probability of each of these is given by a hypergeometric random variable. Once we know what the first draw was, the probability we draw a white chip is given by another hypergeometric random variable. Combining these results by conditioning in R code this probability is given by dhyper(2, 5, 4, 2) * dhyper(1, 5, 6, 1) + dhyper(1, 5, 4, 2) * dhyper(1, 6, 5, 1) + dhyper(0, 5, 4, 2) * dhyper(1, 7, 4, 1)

Evaluating this we get 0.5353535.

Question 3.2.32

An operating characteristic curve is a plot of the acceptance probability as a function of the percent defective. In Figure 2 I plot this curve. We see that the probability that an item is defective should be about 16% for us to accept shipments 50% of the time.

Question 3.2.33

This would be given by the value of

with r = r 1 + r 2 + r 3 .

Question 3.2.34

The complement of the probability in question is the probability that there is at least one person in each group that is a carrier of the disease. If there is at least one carrier in each group then we have one group with two carriers and the other groups with one. This division happens with a probability (see the previous question) of

As the two person group could be any of the three we need to multiply this probability by three to get the value 0.5157895. The probability we seek is the complement of this or 0.4842105.

Question 3.2.35

This would be a generalization of what we have seen before (like in question 3.2.34) for example we would have

with n = k 1 + k 2 +

+k t.

Question 3.2.36

Using the previous question this would be given by

which evaluates to 0.08391608.

B 1 B 2 Max Sum 1 1 1 1 2 2 2 3 3 4

2 3 4 5 3 4 5 4 5 5

2 3 4 5 3 4 5 4 5 5

3 4 5 6 5 6 7 7 8 9

Table 1: Possible values for the maximum X and the sum V for Question 3.3.1.

Question 3.3.1

Part (a): We begin by constructing the sample space of possible outcomes. These numbers are displayed in Table 1 where the first two numbers in each row corresponds to the numbers drawn and the third column then computes their maximum. Notice that there are a total of ten possible outcomes (some values are duplicated) and for each outcome we can count how many times it appears. This gives P (X = 2) = P (X = 3) =

=

P (X = 4) = P (X = 5) = , for the probability density for X .

Part (b): As in the previous part we begin by constructing the sample space of possible outcomes. These numbers are displayed in Table 1 where the first two numbers in each row corresponds to the numbers drawn and the fourth column is then computes their sum. Notice that there are a total of ten possible outcomes (some values are duplicated) and for each outcome we can count how many times it appears. This gives P (V = 3) = P (V = 4) = P (V = 5) = P (V = 6) = P (V = 7) = P (V = 8) = P (V = 9) =

,

for the probability density for V .

1 2 3 4 5

1 1 2 3 4 5

2 2 2 3 4 5

3 3 3 3 4 5

4 4 4 4 4 5

5 5 5 5 5 5

Table 2: The possible values for the maximum X when the draws are done with replacement Question 3.3.2

Part (a): This problem is just like the previous one but now the table of possible outcomes is larger. In Table 2 we present these outcomes for X the maximum of the two draws. The first column is the first value drawn and the first row is the value of the second value drawn and their intersection holds the value for the maximum. Notice that there are twenty five equally likely outcomes so that the probability density for X looks like P (X = 1) = P (X = 2) = P (X = 3) = P (X = 4) = P (X = 5) =

.

1 2 3 4 1 2 3 4 5 2 3 4 5

3 4 5 6

4 5 6 7

5 6 7 8

5 6

6 7 7 8 8 9 9 10

Table 3: The possible values for the sum V when the draws are done with replacement Part (b): This problem is just like the previous one but now the table of possible outcomes is larger. In Table 3 we present these outcomes for V the sum of the two draws. The first column is the first value drawn and the first row is the value of the second value drawn and their intersections holds the value for the sum.

Notice that there are twenty five equally likely outcomes so that the probability density for V looks like

P (V = 2) = P (V = 3) = P (V = 4) = P (V = 5) = P (V = 6) = P (V = 7) = P (V = 8) = P (V = 9) = P (V = 10) =

.

Question 3.3.3

Below I present a way to do this problem analytically but here I simply note that we can also consider enumerating the possible values found on each die and computing the value of X under each case. Using the values of X in all possible situations we can compute a probability density function for X . In R we can do that with the following simple program D = 1:6 outcomes = expand.grid(D, D, D) colnames(outcomes) = c(’X1’, ’X2’, ’X3’) X = apply(outcomes, 1, max) T = table(X) print(T/sum(T))

This gives the output X

1 2 3 4 5 6 0.00462963 0.03240741 0.08796296 0.17129630 0.28240741 0.42129630

for the probabilities that X takes on the values 1… 6.

To do this problem analytically we imagine a 3d “cube” with its three dimensions denoting the value of the i th die rolled for i = 1, 2, 3. Thus the location indexed by (1, 2, 5) would mean that the first die showed a one, the second showed a two, and the third showed a five. For each of these 6 3 = 216 possible rolls we could compute the maximum value observed from the three numbers. To obtain the value of P (X = k ) we will first compute P (X ≤ k ). To do the later we need to know the number of cells (in the cube described above) where X ≤ k . This number is given by the volume of the cube below k or

This means that

Using that expression we have

for 1 ≤ k ≤ 6.

Question 3.3.4

We can use a program like in the previous question to compute the answer to this question. The only thing that needs to change from the previous statements is the how we compute the X variable. The following will do this change n_unique = function(row){ length(unique(as.double(row))) } X = apply(outcomes, 1, n_unique) T = table(X) print(T/sum(T))

This gives X

1 2 3 0.02777778 0.41666667 0.55555556

for the probabilities that X takes on the values 1, 2, 3.

Question 3.3.5

If we let H be the number of heads and T be the number of tails then the random variable X is given by X = H − T . As we toss the coin three times we have H + T = 3 so that T = 3 − H and X can be written as X = 2H − 3. The number of heads H in three flips is given by a binomial random variable so in R notation we have P (X = − 3) = dbinom(0, 3, 1⁄ 2) P (X = − 1) = dbinom(1, 3, 1⁄ 2) P (X = 1) = dbinom(2, 3, 1⁄ 2) P (X = 3) = dbinom(3, 3, 1⁄ 2).

Question 3.3.6

We will just enumerate all the possible outcomes for the variable X using R and compute the density function from them. We have D1 = c( 1, 2, 2, 3, 3, 4 ) # vs. 1:6 D2 = c( 1, 3, 4, 5, 6, 8 ) outcomes = expand.grid(D1, D2) colnames(outcomes) = c(’X1’, ’X2’) X = apply(outcomes, 1, sum) T = table(X) print(T/sum(T))

This gives the output X

2 3 4 5 6 7 8 0.02777778 0.05555556 0.08333333 0.11111111 0.13888889 0.16666667 0.13888889

9 10 11 12 0.11111111 0.08333333 0.05555556 0.02777778

for the probabilities that X takes on the values 2… 12. This is to be compared with the result if we take the outcome from each die from the normal 1… 6 range. Changing the D1 and D2 variables in the above and running the code we see that we get exactly the same probably distribution as above.

Question 3.3.7

Its position will be the number or right-facing steps minus the number of left facing steps. This is X = R − L .

As we take a total of four steps we have R + L = 4 or L = 4 − R . This means that X = R − (4 − R ) = 2R − 4.

The number of right facing steps is then given by a binomial random variable with n = 4 and p = 1⁄ 2. Thus the pdf for X is given by (in R notation) P (X = − 4) = dbinom(0, 4, 1⁄ 2) P (X = − 2) = dbinom(1, 4, 1⁄ 2) P (X = 0) = dbinom(2, 4, 1⁄ 2) P (X = 2) = dbinom(3, 4, 1⁄ 2) P (X = 4) = dbinom(4, 4, 1⁄ 2).

Question 3.3.8

The logic used in the previous question is exactly the same but the value of p (for the probability of a right step) in the binomial random variable is now p = 2⁄ 3.

Question 3.3.9

To solve this problem imagine that you and then your friend stand next to each other. We then want to place the remaining three people either to the left of this group of two, to the right of this group of two, or in between this group of two. There is a 1⁄ 3 of a chance that any given one of these three people will randomly be placed between you and your friend. Thus the probability that k for 0 ≤ k ≤ 3 of these people are placed between you and your friend is a binomial random variable with n = 3 and p = 1⁄ 3.

Note that this answer does not agree with the one given in the back of the book. If anyone sees anything wrong with my logic please contact me.

Question 3.3.10

The pdf for X i is a hypergeometric random variable with X i ∈{ 0, 1, 2} . As X 1 and X 2 are independent we can compute the probability of a pair of values of (X 1 ,X 2 ) by multiplication of their corresponding probability. This can be done in R with the following code PX1 = dhyper(0:2, 2, 2, 2) PX2 = dhyper(0:2, 2, 2, 2) X_sum = outer( 0:2, 0:2, ’+’ ) # the value of the sum X_1 + X_2 P_X1_X2 = outer( PX1, PX2, ’*’ ) # the probability of the combination (X_1, X_2) dim(X_sum) = NULL # flatten these matrices dim(P_X1_X2) = NULL DF = data.frame( X=X_sum, P=P_X1_X2 ) library(plyr) # the split-apply-combine paradigm for R. print(ddply(DF, ~X, summarise, sum=sum(P)))

Running the above gives X sum 1 0 0.02777778 2 1 0.22222222 3 2 0.50000000 4 3 0.22222222 5 4 0.02777778 Question 3.3.11

The pdf for V = 2X + 1 is given by (in R notation) P (V = 1) = dbinom(0, 4, 2⁄ 3) P (V = 3) = dbinom(1, 4, 2⁄ 3) P (V = 5) = dbinom(2, 4, 2⁄ 3) P (V = 7) = dbinom(3, 4, 2⁄ 3) P (V = 9) = dbinom(4, 4, 2⁄ 3).

Question 3.3.12

This is given by the cumulative sum of the values of p X (k ) given in the stated question.

Question 3.3.13

The number of sixes observed in four roles of a die is a binomial random variable with n = 4 and p = 1⁄ 6.

The cdf would be the cumulative sum of these values.

Question 3.3.14

The pdf is the difference between subsequent values of the cdf. For example

Question 3.3.15

As in the previous question the pdf is the difference between subsequent values of the cdf. For example

Question 3.4.1

This would be given by

Question 3.4.2

This would be given by

which we could further simplify if desired.

Question 3.4.3

Note that

is equivalent to

so the probability we want is given by

which can be simplified further if needed.

Question 3.4.4

Let p be this probability. Then we have

Question 3.4.5

Part (a): Here we compute P (Y > 10) = ∫ 10 ∞ 0. 2e − 0 . 2 y dy =

10

∞

= − (0 − e − 2 ) = e − 2 = 0. 1353353.

Part (b): Here X is a binomial random variable with n = 2 and p = P (Y > 10) which we computed above. From this we have p X (2) =

p 2 (1 − p ) 0

p X (1) =

p (1 − p ) = 0. 2340393

p X (0) = (1 − p ) 2 .

Question 3.4.6

Note that f Y (y ) ≥ 0 for all y and that ∫ 0 1 y n (1 − y )dy = ∫ 0 1 y n dy − ∫ 0 1 y n +1 dy =

−

=

.

Using this result we see that ∫ 0 1 f Y (y )dy = 1 as it must for a probability density.

Question 3.4.7

We would have

Using this we then have

Question 3.4.8

This would be given by

Question 3.4.9

This pdf is a triangle with its peak at y = 0. The cdf is then given by

when y ≤ 0. Then if y > 0 we have

when 0 ≤ y ≤ 1.

Question 3.4.10

For the first way using the cdf given we have are

For the second way need to compute the pdf where we find

so that

when we simplify.

Question 3.4.11

Part (a): This is F Y (2) = ln(2) = 0. 6931472.

Part (b-c): These would be F Y (2. 5) − F Y (2) = 0. 2231436.

Part (d): This is f Y (y ) = for 1 ≤ y ≤ e .

Question 3.4.12

Because

from the fundamental theorem of calculus we would have

which we could simplify if desired.

Question 3.4.13

This would be

Question 3.4.14

We have

when we integrate and simplify.

Question 3.4.15

We find

for the pdf.

Question 3.4.16

We can compute this using

Using this we have

We will have f W (w )≠ 0 if 0 ≤ w⁄ 2 ≤ 1 or 0 ≤ w ≤ 2.

Question 3.4.17

To start we have

Now in this second integral take f Y (y ) = f Y (− y ) and we write it as

Let v = − y so that dv = dy and we can write it as

Using this result we have that P (− a ≤ Y ≤ a ) = ∫ −∞ a f Y (y )dy − ∫ a ∞ f Y (y )dy = F Y (a ) −

= F Y (a ) − (1 − F Y (a )) = 2F Y (a ) − 1, as we were to show.

Question 3.4.18

Using the pdf and cdf for an exponential random variable we have

a constant.

Question 3.5.1

We can compute this using R as payoff = c( -1, 2, 18, 180, 1300, 2600, 10000 ) prob = c( 0.935, 0.0514, 0.0115, 0.0016, 1.35*10^(-4), 6.12*10^(-6), 1.12*10^(-7) ) sum( payoff * prob )

This gives -0.144668.

Question 3.5.2

We are told that in Monte Carlo the roulette wheel has the numbers 0, 1, 2, … , 36. Each number has an equal chance of appearing or a probability of . The expectation of betting on red would then be

In the book they computed the expected winnings from a roulette wheel in Las Vegas to be -0.05263158. The difference between these two numbers is 0.02560455. To pay for the cost of the travel a player would have to bet at least D dollars such that

since after betting this much in Monte Carlo he would have “saved” enough for the trip there.

Question 3.5.3

Table 3.3.4 gives the profit and probabilities needed to compute the expectation. Using R as profit = c( 30000, 18000, 6000, -6000 ) prob = c( 0.857375, 0.135375, 0.007125, 0.000125 ) sum( profit * prob )

we get the expected profit to be given by 28200.

Question 3.5.4

We would want to select the rule that has the largest expectation. The number of red balls drawn is given by a hypergeometric random variable so we can compute the expected payoff in R as prob = dhyper( 0:2, 2, 4, 2 ) rule_a = sum( prob * c( 0 2, 10 ) rule_b = sum( prob * c( 0 1, 20 ) print( c( rule_a, rule_b ) )

This gives [1] 1.733333 1.866667

Thus one should choose Rule B. Question 3.5.5

I found this question somewhat confusing. This how I can best understand it given what I think it is trying to say. If the woman dies during the first year she has paid P and the company must payout $50000 so the company has a loss of 50000 − P . This event happens with a probability of 0.00051 (and thus does not happen with a probability of 1 − 0. 00051 = 0. 99949). If she dies during the second year she has paid a total of 2P and the company has a loss of 50000 − 2P . For this to happen means that the woman must have lived the first year and then died in the second which happens with a probability of 0. 99949(0. 00052) = 0. 0005197348. Continuing this logic we can compute the probabilities that the woman dies in year i as prob_die = c( 0.00051, 0.00052, 0.00054, 0.00056, 0.00059 ) prob_live = 1 - prob_die # i.e. the probability the woman dies in year n prob_payout_in_year_n = c( 1, cumprod( prob_live ) ) * c( prob_die, 1 ) print(prob_payout_in_year_n)

Running the above code we get [1] 0.0005100000 0.0005197348 0.0005394439 0.0005591213 0.0005887443 [6] 0.9972829557

for the probability that the woman dies during year i . Call these numbers p i . The losses to the company in year i are

for 1 ≤ i ≤ 5 and − 5P for the sixth year (which is when the policy expires and the company gets to keep all of the premiums paid). To have the expected profit requirement hold true means that

We can use the numbers above and this equation and solve for P where we find P = 227. 4086 for the yearly premium.

Question 3.5.6

If 4% are defective from 100 then 4 are defective. The number of defective chips is given by a hypergeometric random variable (which we could approximate as a binomial random variable if desired). Using the formula for the expectation of a hypergeometric random variable we have

Approximating this as a binomial random variable would give the same expectation.

Question 3.5.7

Of the 125 non vaccinated students 12% are likely to be absent. This means that 0. 12(125) = 15 will still not be vaccinated.

Question 3.5.8

Part (a): This would be given by

Part (b): This would be given by

Part (c): This would be given by

Part (d): This would be given by

Question 3.5.9

This would be given by

Question 3.5.10

This would be given by

As the center of gravity of an interval is the midpoint the expectation should be the midpoint of the interval (which it is).

Question 3.5.11

This would be given by E (Y ) = ∫ 0 ∞ λye − λy dy = −

0

∞

= − (0 − 1) = .

Question 3.5.12

We have that f (y ) ≥ 0 for all y . Next we check that the integral of f Y (y ) is one. We have

as it must. Computing the expectation we would have to be able to compute

which does not converge. This means that Y does not have a finite expectation.

Question 3.5.13

As the number of cars expected to pass is a hypergeometric (or a binomial) random variable depending on assumptions. The expectation under either case is given by np = 200(0. 8) = 160.

Question 3.5.14

The probability that any one observation Y falls in that interval is

The expected number is then np = 15(7⁄ 8) = 13. 125.

Question 3.5.15

In the month of November approximately 1⁄ 12 of the automobiles will need to renew and each will generate $50.

Thus we should expect to have

Note that this answer does not agree with the one given in the back of the book. If anyone sees anything wrong with my logic please contact me.

Question 3.5.16

The proportion of companies that failed due to bankruptcy is

. In a new set of nine companies we would expect

to fail due to bankruptcy.

Question 3.5.17

See the Table 1 for the possible values for the larger of the two draws. In that table there are draws with a value of five. If we drop these (since they are not possible in this question) we get six equally likely outcomes. This means that the expectation of X is given by

Question 3.5.18

Let the digit of 1 denote a toss that gives a head and a digit of 0 a toss that gives a tail. Then we can enumerate all possible outcomes from this experiment and the value of the random variable X using the following R code T = 0:1 outcomes = expand.grid(T, T, T) colnames(outcomes) = c(’T1’, ’T2’, ’T3’) X = apply(outcomes, 1, sum) * outcomes[, 1] * outcomes[, 3] mean(X)

Running this we find the expected value of X to be 0.625.

Question 3.5.19

For a fair game the ante must be the expectation of the winnings. The probability that the game ends in k steps for 1 ≤ k ≤ 9 is 2 − k as before. This means that the probability that the game ends on a toss for k > 9 is given by

when we simplify. This means that the expectation of the payoff can be computed using xs = 1:9 ps = c( 2^(-xs), 2^(-9) ) EX = sum( ps * c( 2^xs, 1000 ) )

Running the above we get 10.95312.

Question 3.5.20

Part (a): For this we would find E (X ) = ∑ k =1 ∞ =

= ∑ k =0 ∞ −1=

k

−1

,

which is valid as long as < 1 or c < 2.

Part (b): For this we would find

when use an identity to evaluate the sum.

Question 3.5.21

In Question 3.3.3 on Page § we explicitly enumerated the outcomes when we toss three die. In Question 3.3.5 on Page § we computed the probability that there are be 1, 2, or 3 different faces showing (i.e. the probability the random variable X takes on the values one, two, or three). Using the output from these two programs we can compute the expectation for this question using probs = T/sum(T) sum( probs * 1:3 )

This gives the value 2.527778.

Question 3.5.22

We can do this using the following R code T = 1:5 outcomes = expand.grid(T, T) colnames(outcomes) = c(’D1’, ’D2’) outcomes = outcomes[outcomes$D1 != outcomes$D2, ] # drop any duplicate entries X = abs(outcomes$D1 - outcomes$D2)

probs = table(X)/length(X) EX = sum( probs * 1:4 )

This gives the value two.

Question 3.5.23

If the teams are evenly matched each has a probability of of either team winning. Lets consider the case where 4 team A wins the World Series. Team A will win in four games with a probability of . Team A will win in five games with a probability of

since in the first four games team B must win once. In six games with a probability of

Finally team A will win in seven games with a probability of

As team B has exactly the same ways they can win the World Series i.e. in four, five, six, or seven games we must double the above numbers to get the total probability that the series lasts that many games. This means that if X is the number of games played its expectation can be computed using probs = 2 * c( 1/2^4, 4/2^5, 10/2^6, 20/2^7 ) sum( probs * 4:7 )

Running the above we get the output of 5.8125 for the expectation.

Question 3.5.24

Let Y be the number of draws needed to get a white a chip. Then we have P { Y = 1} = P { Y = 2} =

=

P { Y = 3} =

=

P { Y = 4} =

=

.

It looks like the pattern is

Lets consider the sum of N terms of P { Y = k } . We have S N = ∑ k =1 N P { Y = k } = ∑ k =1 N = ∑ k =1 N − ∑ k =1 N

= ∑ k =1 N

= ∑ k =1 N − ∑ k =2 N +1

=1−

.

Note that as N →∞ then S N → 1 as it must to be a density. Using this lets compute

does not converge thus E (Y ) is not finite.

Question 3.5.25 (deriving the expectation of a hypergeometric random variable)

For this problem we will use a slightly different notation (than the book uses) to prove the expectation of a hypergeometric random variable. As such the “new” notation in terms of the old notation is N = r + w , M = r , and N − M = w . This notation is also standard when discussing the hypergeometric random variable.

Since X is a hypergeometric random variable we have

Now the limits of X are max(0,n − N + M ) ≤ x ≤ min(n,M ). We will assume that n < M so we know that the upper limit of our summation is min(n,M ) = n . Thus we have

Expanding the binomial coefficients this becomes E (X ) = ∑ x = max (0 ,n − N + M ) n = ∑ x = max (1 ,n − N + M ) n =

∑ x = max (1 ,n − N + M ) n

.

Let y = x − 1 then the above is E (X ) = =

∑ y = max (1 ,n − N + M ) − 1 n − 1 ∑ y = max (0 ,n − 1 − ( N − 1)+( M − 1)) n − 1

=

,

since the sum above is the sum of another hypergeometric density h (y ; n − 1,M − 1,N − 1) over all possible y values (and hence sums to one as all probability densities must).

Question 3.5.26

We can show this with the following

We will now use

in the above to get E (X ) = ∑ k =1 ∞ kP (X ≥ k ) − ∑ k =1 ∞ kP (X ≥ k + 1) = ∑ k =1 ∞ kP (X ≥ k ) − ∑ k =2 ∞ (k − 1)P (X ≥ k ) = ∑ k =1 ∞ kP (X ≥ k ) − ∑ k =1 ∞ (k − 1)P (X ≥ k ) = ∑ k =1 ∞ (k − (k − 1))P (X ≥ k ) = ∑ k =1 ∞ P (X ≥ k ), as we were to show.

Question 3.5.27

Part (a): We would need to solve

Doing this integral and solving for m we get that m =

1 ⁄ ( 𝜃 +1)

.

Part (b): We would need to solve

If we integrate we get the equation m 2 + m − 1 = 0 which has solutions m = 0. 618034 and m = − 1. 618034. Only the first of which is in the interval 0 ≤ y ≤ 1 so that is the median.

Question 3.5.28

This would be

Question 3.5.29

This would be 12(0. 11)(100) = 132.

Question 3.5.30

For this we would have

when we integrate and simplify. As computing this in another way we have

when we integrate and simplify.

Question 3.5.31

For this we have

E [Q (Y )] = ∫ 0 ∞ 2(1 − e − 2 y )6e − 6 y dy = 12 ∫ 0 ∞ e − 6 y − 12 ∫ 0 ∞ e − 8 y dy = , when we integrate and simplify. The above is in units of hundreds of thousands dollars.

Question 3.5.32

If we let V = 5Y 2 we find

when we integrate and simplify.

Question 3.5.33

This would be

when we integrate and simplify.

Question 3.5.34

This would be

when we integrate and simplify.

Question 3.5.35

In the notation of the question the area is given by A = a 2 . In terms of Y the variable a is given by a = Y cos . This means that A = . This means the expectation of A is given by

when we integrate and simplify.

Question 3.5.36

Let p i = ki then to be normalized we must have

Using the above expression for p i and the sum of the first n integers we find that

=

Then the expectation we want to evaluate is given by

Question 3.6.1

We have E (X ) = 2 E (X 2 ) = 2 2

2

+1⋅2

+0

2

= 1. 2

+12⋅ 2

+02

E (Y ) = ∫ 0 1 y

dy + ∫ 2 3 y

dy = + = 1

E (Y 2 ) = ∫ 0 1 y 2

dy + ∫ 2 3 y 2

2

2

= 1. 92.

This means that

Question 3.6.2

We have

dy = +

=

,

when we integrate and simplify. Using these we have that

Question 3.6.3

The number of men hired is given by a hypergeometric random variable with parameters r = 6, w = 4, and n = 3.

From the book we know that

Using the same logic used there to compute E (X ) we find E (X 2 ) = 3. 8. This means that

The standard deviation is the square root of this number or 0.7483315.

Question 3.6.4

The expectation of a [0, 1] uniform random variable is . The expectation squared is given by

This means that the variance is given by

Question 3.6.5

We have E (Y ) = ∫ 0 1 y (3(1 − y ) 2 )dy = E (Y 2 ) = ∫ 0 1 y 2 (3(1 − y ) 2 )dy =

,

when we integrate and simplify. Using these we have that

Question 3.6.6

We have E (Y ) = ∫ 0 k

dy =

E (Y 2 ) = ∫ 0 k

dy =

,

when we integrate and simplify. Using these we have that

If this is to equal two then we must have k 2 = 36 so as k > 0 this means that k = 6.

Question 3.6.7

For the given function we see that the pdf is given by

We first check that this is a valid pdf. We see that f Y (y ) ≥ 0 for all y . Next consider ∫ f Y (y )dy = ∫ 0 1 (− y + 1)dy + ∫ 2 3 dy = =

+ = 1,

as it should. Next we find

when we integrate and simplify. Next we find

when we integrate and simplify. These together mean that

The square root of this or 1.114924 is the standard deviation.

0

1

+ (3 − 2)

Question 3.6.8

Part (a): We have

as we were to show.

Part (b): For this we have

as we were to show.

Part (c): For this we have

because of this Var

cannot be finite.

Question 3.6.9

Lets assume that Johnny will pick the value j . Then his loss will be (j − F ) 2 where F is the random value that Frankie selected. We want to minimize Johnnies expected loss L (j ) or

To find the minimum of the above we taking the j derivative, set it equal to zero, and solve for j . We find the derivative set equal to zero given by

This means that

Thus Johnny should always pick the mean of the numbers Frankie would pick. For a uniform distribution this number is .

Question 3.6.10

We have E (Y ) = ∫ 0 1 5y 5 dy =

0

E (Y 2 ) = ∫ 0 1 5y 6 dy =

0

when we integrate and simplify. Using these we have that

1

=

1

= ,

Question 3.6.11

In a previous question we have shown that E (Y ) = . Next we compute

when we integrate and simplify. Using this we have that

as we were to show.

Question 3.6.12

Using what we know about the exponential distribution we have P (Y > E (Y ) + 2

) =P =∫

∞

λe − λy dy = −

3 ⁄λ

∞

= − (0 − e − 3 ) = e − 3 .

Question 3.6.13

We can write g (a ) as g (a ) = E [(X − a ) 2 ] = E [(X − μ + μ − a ) 2 ] = E [(X − μ ) 2 + 2(X − μ )(μ − a ) + (μ − a ) 2 ] = E [(X − μ ) 2 ] + (μ − a ) 2 , using the fact for E (X − μ ) = 0. Notice that the minimum of the above expression with respect to a will happen when a = μ . In that case g (a ) = g (μ ) = E ((X − μ ) 2 ) = Var .

Question 3.6.14

We would have

Question 3.6.15

We are told that σ Y = σ F = 15. 7 then we have that

This means that σ c = 8. 722222.

Question 3.6.16

To show these two expressions note that we have

E

= E (W − μ ) = (E (W ) − μ ) = (μ − μ ) = 0

Var

=

Var

=

= 1.

Question 3.6.17

Part (a): Note that if U = 0 then we get Y = a and if U = 1 we have Y = b with Y linear in between.

Part (b): We can compute this using

Question 3.6.18

In the following R code I compute the expectation and variance of the two estimators W 1 and W 2 .

E_W1 = 0.2281 + 0.9948 * (11/2) V_W1 = 0.9948^2 * (9/12) + 0.0427 E_W2 = -0.0748 + 1.0024 * (11/2) V_W2 = 1.0024^2 * (9/12) + 0.0159 print(sprintf(’W1: xbar= %f; s= %f; abs(xbar-5.5)= %f’, E_W1, sqrt(V_W1), abs(E_W1-5.5))) print(sprintf(’W2: xbar= %f; s= %f; abs(xbar-5.5)= %f’, E_W2, sqrt(V_W2), abs(E_W2-5.5)))

Running that code I get [1] "W1: xbar= 5.699500; s= 0.885957; abs(xbar-5.5)= 0.199500" [1] "W2: xbar= 5.438400; s= 0.877214; abs(xbar-5.5)= 0.061600"

As they have approximately the same standard deviation s we would take the estimator that has its mean value closer to 5.5. In this case we see that the second estimator has a mean that is much closer and we would prefer that one.

Question 3.6.19

For a this density we find

Note that

Next we compute E [(Y − μ ) 6 ] = E [(Y − 1) 6 ] = E = ∑ j =0 6

E [Y j ](− 1) 6 − j = ∑ j =0 6

We can evaluate the above using js = 0:6 sum( ( choose( 6, js ) * ( -2 )^js ) / (js + 1) )

(− 1) j .

Running the above we get the value 0.1428571.

Question 3.6.20

Recall that γ 1 = E [(W − μ ) 3 ]. For an exponential random variable like the one here we have that μ = 1 and σ 2 = 1 so that σ = 1. Thus we find

Question 3.6.21

Recall that

For the given random variable we have that μ = and σ 2 =

. This means that

Question 3.6.22

We can expand E [(W − μ ) 3 ] = E [W 3 − 3W 2 μ + 3Wμ 2 − μ 3 ] = E [W 3 ] − 3E [W 2 ]μ + 3μ 3 − μ 3 = E [W 3 ] − 3E [W 2 ]μ + 2μ 3 .

If we are told that E [(W − μ ) 3 ] = 10 and E [W 3 ] = 4 then the above implies that

If μ = 2 this becomes

I see no reason this can’t be true so the value of μ = 2 seems possible.

Question 3.6.23

If E (X ) = μ then E (Y ) = aμ + b and σ Y = | a | σ X = aσ x as a > 0. Then for Y we compute γ1Y =

E [(Y − aμ − b ) 3 ] =

=

=

=γ 1X.

In the same way we have

Question 3.6.24

Part (a): We have E (Y ) = (n + 1)(n + 2) ∫ 0 1 y n +1 (1 − y )dy =

=

E (Y 2 ) = (n + 1)(n + 2) ∫ 0 1 y n +2 (1 − y )dy =

,

when we integrate and simplify. This means that

Part (b): For k ≥ 0 this we would have

when we integrate and simplify.

Question 3.6.25

Part (a): For this we must have

To have this equal one means that c = 5.

Part (b): The k th moment of Y would be computed using

This will exist if and only if k − 6 < − 1 or k < 5. The highest moment that exists is thus k = 4.

Question 3.7.1

We must have ∑ x,y p (x,y ) = 1 which means that

Question 3.7.2

To have our pdf normalized we must have 1 = ∫ 0 1 ∫ 0 1 c (x 2 + y 2 )dxdy = c ∫ 0 1 =c∫01

dy = c

0

1

0

=

1

dy

.

To have this be true means that c = .

Question 3.7.3

To have our pdf normalized we must have 1 = ∫ y =0 1 ∫ x =0 y c (x + y )dxdy = c ∫ y =0 1 = c ∫ y =0 1 To have this be true means that c = 2.

Question 3.7.4

To have our pdf normalized we must have

dy = c ∫ y =0 1 y 2 dy = .

0

y

dy

1 =∫

y =0

1

∫ x =0 y cxydxdy = ∫ y =0 1

= ∫ y =0 1

dy =

0

1

0

y

dy

= .

To have this be true means that c = 8.

Question 3.7.5

This is an example of a multivariate hypergeometric random variable so we have

for 0 ≤ x ≤ 3, 0 ≤ y ≤ 2, and 0 ≤ x + y ≤ 3.

Question 3.7.6

There are four kings and four queens and then 52 − 8 = 44 other cards. Then the probability of x kings and y queens is then given by

for 0 ≤ x ≤ 4, 0 ≤ y ≤ 4, and 0 ≤ x + y ≤ 4.

Question 3.7.7

We would want to consider the event where P (X > Y ). Dividing the numbers in the table by fifty and then adding up the entries where X > Y we get T = matrix( data=c( 11, 6, 4, 9, 10, 3, 5, 0, 2 ), nrow=3, ncol=3, byrow=TRUE ) P = T/sum(T) rownames(P) = c( ’Y0’, ’Y1’, ’Y2’ ) colnames(P) = c( ’X0’, ’X1’, ’X2’ ) print( sum( c( P[1, 2:3], P[2, 3] ) ) )

the above gives the probability of 0.26.

Question 3.7.8

In Question 3.5.18 in Page § we explicitly enumerated the possible sets of heads we could get on three tosses of a fair coin. In that code we represented a head with a one and a tail with a zero.

T = 0:1 outcomes = expand.grid(T, T, T) colnames(outcomes) = c(’T1’, ’T2’, ’T3’) X = outcomes[, 3] # number of heads on the last flip

Y = apply(outcomes, 1, sum) # total number of heads T = table(X, Y) P = T/sum(T) print(P)

This gives X

Y

0 1 2 3 0 0.125 0.250 0.125 0.000 1 0.000 0.125 0.250 0.125

for the joint probability table p X,Y (x,y ).

Question 3.7.9

We can enumerate and count the possible values for the variables X and Y with the following R code D = 1:6 outcomes = expand.grid(D, D) colnames(outcomes) = c(’X1’, ’X2’) X = apply(outcomes, 1, function(x){ sum(x==2) }) Y = apply(outcomes, 1, function(x){ sum(x==3) }) T = table(X, Y) print(T) P = T/sum(T) print(P)

When we run that we get > print(T) Y X 0 1 2 0 16 8 1 1 8 2 0 2 1 0 0 > P = T/sum(T) > print(P) Y X 0 1 2 0 0.44444444 0.22222222 0.02777778 1 0.22222222 0.05555556 0.00000000 2 0.02777778 0.00000000 0.00000000

The later is the joint probability density p X,Y (x,y ). To use p X,Y (x,y ) to find p Z (z ) we simply sum up the probabilities of all of the possible ways to get z where z ∈{ 0, 1, 2} . We find these numbers given by [1] 0.4444444 0.4444444 0.1111111 Question 3.7.10

Part (a): We can find c as the density must be normalized or

Part (b): This would be given by

Question 3.7.11

We can compute this using P (Y < 3X ) = P

= ∫ y =0 ∞ ∫ x =

y

2e − ( x + y ) dxdy

= − 2 ∫ y =0 ∞ e − y (e − y − e − y⁄ 3 )dy = − 2 ∫ y =0 ∞ (e − 2 y − e − 4 y⁄ 3 )dy =−2

= ,

when we integrate and simplify.

Question 3.7.12

This would be

Here as the density is one over the area of the circle πr 2 with r = 2.

Question 3.7.13

We can compute this using P (X < 2Y ) = P = ∫ x =0 1

= ∫ x =0 1 ∫ y =

1

(x + y )dydx

dx =

,

when we integrate and simplify.

Question 3.7.14

We first need to compute p1 =P

= ∫ 0 1 ⁄ 3 2tdt =

p2 =P

= ∫ 1 ⁄ 3 2 ⁄ 3 2tdt =

p3 =P

=1−p1−p2= .

Then the probability that we get X values in the first interval and Y values in the second interval from n points is given by

Evaluating this for x = 1 and y = 2 with n = 5 we get 0.02116886.

Question 3.7.15

If we draw this triangle in the x -y plane we have that the three vertexes should be at (0, 0), (b, 0) and (0,h ). The line connecting the point (0,h ) and (0,b ) is given by

Using these facts computing the desired probability is easy. We have p = ∫ y =0 h⁄ 2 ∫ x =0 − =

∫ y =0 h⁄ 2

y+b

dxdy dy =

.

Here A is the area of the triangle or bh . Using that we get p = .

Question 3.7.16

We would sum out the variable Y in the pdf obtained in that question. Using the result from there and the following R code p_x = c() for( x in 0:3 ){ ys = 0:2 v = sum( ( choose(3, x) * choose(2, ys) * choose(4, 3-x-ys) ) / choose(9, 3) ) p_x = c( p_x, v ) }

we get [1] 0.23809524 0.53571429 0.21428571 0.01190476 Question 3.7.17

We would sum out the variable X to get f Y (y ) and Y to get f X (x ) from the pdf obtained in that question. Using the result from there for these we find > apply(P, 1, sum) # gives f_X(x) 0 1 0.5 0.5

and > apply(P, 2, sum) # gives f_Y(y) 0 1 2 3

0.125 0.375 0.375 0.125 Question 3.7.18

The number in each category is given by a multinomial distribution with equal probabilities p = . This means that

when we evaluate. The marginal probability for the middle is a binomial with p = so

Question 3.7.19

Part (a): For f X (x ) we compute

for 0 ≤ x ≤ 2. For f Y (y ) we compute

for 0 ≤ y ≤ 1.

Part (b): For f X (x ) we compute

for 0 ≤ x ≤ 2. For f Y (y ) we compute

for 0 ≤ y ≤ 1.

Part (c): For f X (x ) we compute

for 0 ≤ x ≤ 1. For f Y (y ) we compute

for 0 ≤ y ≤ 1.

Part (d): First we need to compute c . We need to have 1 = ∫ x =0 1 ∫ y =0 1 c (x + y )dydx = c ∫ x =0 1 = c ∫ x =0 1

dx = c

0

1

=c

Now that we know the value of c for f X (x ) we compute

for 0 ≤ x ≤ 1. For f Y (y ) we compute

for 0 ≤ y ≤ 1.

Part (e): For f X (x ) we compute

for 0 ≤ x ≤ 1. For f Y (y ) we compute

for 0 ≤ y ≤ 1.

Part (f): For f X (x ) we compute

for x ≥ 0. For f Y (y ) we compute

for y ≥ 0 using the fact that the pdf f X,Y (x,y ) is symmetric in x and y .

Part (g): For f X (x ) we compute

y =0

1

dx

=c.

for x ≥ 0. For f Y (y ) we compute

for y ≥ 0.

Question 3.7.20

Part (a): For f X (x ) we compute

for 0 ≤ x ≤ 2. For f Y (y ) we compute

for 0 ≤ y ≤ 2.

Part (b): For f X (x ) we compute

for 0 ≤ x ≤ 1. For f Y (y ) we compute

for 0 < y ≤ 1.

Part (c): For f X (x ) we compute

for 0 ≤ x ≤ 1. For f Y (y ) we compute

for 0 < y ≤ 1. Lets check that these give valid densities. By integrating we can show

and

as they should.

Question 3.7.21

For f X (x ) we compute

for 0 ≤ x ≤ 1.

Question 3.7.22

For f Y (y ) we compute

for y ≥ 0.

Question 3.7.23

The marginal density p X (x ) can be computed as p X (x ) = ∑ y =0 4 − x

x

= =

x

x

y

∑ y =0 4 − x

4−x−y

y

∑ y =0 4 − x

y

4−x−y

4−x−y

.

The sum above can be simplified using the binomial theorem and we have

which is a binomial density.

In the same way the marginal density p Y (y ) can be computed as p Y (y ) = ∑ x =0 4 − y =

x

y

∑ x =0 4 − y

y

4−x−y

x

4−x−y

=

y

=

y

4−y

4−y

,

which is another binomial density.

Question 3.7.24

Part (a): This is the expression for a multinomial random variable of which X is.

Part (b): Following the same logic as in Question 3.7.23 we can show that p X (x ) =

p 1 x (1 − p 1 ) n − x

p Y (y ) =

p 2 y (1 − p 2 ) n − y ,

which are the pdfs of two binomial random variables.

Question 3.7.25

Part (a): We can enumerate the sample space using R as follows C = 0:1 # 0 represents a tail and 1 a head D = 1:6 outcomes = expand.grid(C, D) colnames(outcomes) = c(’X’, ’Y’)

This gives X Y 1 0 1 2 1 1 3 0 2 4 1 2 5 0 3 6 1 3 7 0 4 8 1 4 9 0 5 10 1 5 11 0 6 12 1 6

Part (b): We can compute this using

Question 3.7.26

The probability we draw r red chips, b black chips, and w = n − r − b white chips when we draw n total chips out is given by a multivariate hypergeometric distribution where we have

(8)

for 0 ≤ r ≤ R , 0 ≤ b ≤ B , 0 ≤ w ≤ W , and r + b + w = n . Then in terms of this notation what we want to compute is given by

when we evaluate the above sum using the following R code p = 0 for( w in 0:1 ){ for( r in 0:2 ){ b = 4 - w - r p = p + ( choose(4, r) * choose(3, b) * choose(5, w) ) / choose(12, 4) } } Question 3.7.27

Part (a): For this we have

Part (b): For this we have

Part (c): For this we have

Question 3.7.28

For this question it will help to draw the region where f X,Y (x,y ) is defined in setting up the needed integrations.

Part (a): For this we would have F X,Y (u,v ) = ∫ x =0 u ∫ y = x v f X,Y (x,y )dydx = ∫ x =0 u ∫ y = x v dydx = ∫ x =0 u

dx = u (2v − u ),

for 0 ≤ u ≤ v ≤ 2. Note that F X,Y (2, 2) = 1 as it must.

Part (b): For this we would have F X,Y (u,v ) = ∫ y =0 v ∫ x = y u f X,Y (x,y )dxdy = ∫ y =0 v ∫ x = y u dxdy = ∫ y =0 v (ln(u ) − ln(y ))dy = v (ln(u ) − ln(v ) + 1), for 0 ≤ v ≤ u ≤ 1. Note that F X,Y (1, 1) = 1 as it must.

Part (c): When we draw the region where f X,Y (x,y ) is defined we see that it is a right triangle with its two legs the x and y axis and its hypotenuse connecting the points (0, 1) to (1, 0).

Now if v < 1 − u i.e. the point we want to evaluate F X,Y (u,v ) at is below the line y = 1 − x i.e. the hypotenuse then F X,Y (u,v ) = ∫ x =0 u ∫ y =0 v f X,Y (x,y )dydx = ∫ x =0 u ∫ y =0 v 6xdydx = ∫ x =0 u 6xvdx = 3u 2 v .

Now if v > 1 − u i.e. the point we want to evaluate F X,Y (u,v ) at is above the line y = 1 − x then we will evaluate our integral a bit differently. We have F X,Y (u,v ) = ∫ x =0 v ∫ y =0 v f X,Y (x,y )dydx + ∫ x = v u ∫ y =0 1 − x f X,Y (x,y )dydx = ∫ x =0 v ∫ y =0 v 6xdydx + ∫ x = v u ∫ y =0 1 − x 6xdydx = ∫ x =0 v 6xvdx + ∫ x = v u 6x (1 − x )dx = 3v 3 + (− 2u 3 + 3u 2 + v 2 (2v − 3)) = − 2u 3 + 3u 2 + 5v 3 − 3v 2 .

Note: I’m not sure that above is correct in that it does not seem to have the correct behavior as u and v get larger and larger where it should tend to one. If anyone sees anything wrong with what I have done please contact me.

Question 3.7.29

The pdf f X,Y (x,y ) is the first derivative of the cdf F X,Y (x,y ) so that

Question 3.7.30

The pdf f X,Y (x,y ) is the first derivative of the cdf F X,Y (x,y ) so that

Question 3.7.31

The cdf F X,Y (x,y ) must satisfy F X,Y (1, 1) = 1 so using that we would need to have k = . Then as the pdf f X,Y (x,y ) is the first derivative of the cdf F X,Y (x,y ) we have that

The probability we want to evaluate is then p = ∫ x =0 1 ⁄ 2 ∫ y =1 ⁄ 2 1 k (16xy + 20y 3 )dydx = ∫ x =0 1 ⁄ 2 k

dx =

=

.

Question 3.7.32

This is an identity of double integrals in that we can write the left-hand-side as the double integral

We can then right the right-hand-side as rhs = ∫ −∞ b ∫ −∞ d f X,Y (x,y )dydx − ∫ −∞ a ∫ −∞ d f X,Y (x,y )dydx − ∫ −∞ b ∫ −∞ c f X,Y (x,y )dydx + ∫ −∞ a ∫ −∞ c f X,Y (x,y )dydx .

If we draw these regions in the x -y plane we see that the two integrals are the same.

Question 3.7.33

For a single bulb we have that

For an exponential distribution F X i (x ) = 1 − e − λx so P (X i > 1050) = e − λ (1050) . In this question we have λ = . As each bulb is independent the probability that all four of the bulbs last this long is the above probability raised to the fourth power. When we compute that we get 0.01499558.

Question 3.7.34

These are both described by multidimensional hypergeometric pdfs.

Part (a): There are four aces, kings, and queens and then 52 − 12 = 40 other cards. Thus

where 0 ≤ x ≤ 4, 0 ≤ y ≤ 4, 0 ≤ z ≤ 4, and x + y + z ≤ 6.

Part (b): In these cases there would be 52 − 8 = 44 other cards. Thus

where 0 ≤ x ≤ 4, 0 ≤ y ≤ 4, and x + y ≤ 6. The expression for p X,Z (x,z ) is the same as p X,Y (x,y ) with y replaced by z .

Question 3.7.35

This would be

We write the sum above as

Then the probability we want is given by

Question 3.7.36

Part (a): We have

Part (b): We have

Part (c): We have

once we integrate and simplify.

Question 3.7.37

We have

once we integrate and simplify.

Using this the probability we want is computed as

once we integrate and simplify.

Question 3.7.38

If we consider a 6 × 6 table with the rows the value X of the first die and the columns the value Y of the second die then as each of these combinations is equally likely each pair has a value for its probability of . Next computing the marginal densities from this joint density (or using first principles) we have that P (X = x ) = P (Y = y ) = .

As we then see that

we can conclude that X and Y are independent.

Question 3.7.39

We find the marginal density f X (x ) given by

By symmetry then we would have f Y (y ) = λe − λy . Note that f X,Y (x,y ) = f X (x )f Y (y ) so X and Y are independent.

1 2 3 4 1 (1/4) (2/5) (1/4) (1/5) (1/4) (1/5) (1/4) (1/5) 2 (1/4) (1/5) (1/4) (2/5) (1/4) (1/5) (1/4) (1/5) 3 (1/4) (1/5) (1/4) (1/5) (1/4) (2/5) (1/4) (1/5) 4 (1/4) (1/5) (1/4) (1/5) (1/4) (1/5) (1/4) (2/5) Table 4: The rows represent the possible values for X and the columns represent the possible values for Y . The intersection of the row and column holds the probability of that (X,Y ) combination (when the two fractions are multiplied).

Question 3.7.40

Part (a): Consider Table 4 which shows the probability for each possible value for the pair (X,Y ) for this question (when the fractions given there are multiplied).

Part (b): Given Table 4 we can compute the marginal probabilities of X and Y where we find

Part (c): If these two variables were independent then from the above we would have

which is not true as can be seen from the above table. Thus these two variables are not be independent.

Question 3.7.41

Note that the range in the x -y plane where f X,Y (x,y )≠ 0 is a triangle so the possible values for x depend on what the value of y is. That fact indicates that the variables might not be independent. By drawing the triangle in the x -y plane we see that P (X ≥ 3⁄ 4) ≠ 0 P (Y ≥ 3⁄ 4) ≠ 0.

These two regions combined are the North-East region of the square 0 ≤ X ≤ 1 and 0 ≤ Y ≤ 1 where in fact the joint density f X,Y (x,y ) is zero. Thus we have argued that

Because of this the two random variables cannot be independent.

Question 3.7.42

From the given joint density f X,Y (x,y ) we can compute f X (x ) = ∫ y =0 1 (x + 2y )dy = f Y (y ) = ∫ x =0 1 (x + 2y )dx = (4y + 1).

Note as f X,Y (x,y )≠ f X (x )f Y (y ) we know that X and Y not independent.

Question 3.7.43

We can find P (Y < X ) by integration. We have P (Y < X ) = ∫ x =0 1 ∫ y =0 x f X (x )f Y (y )dydx = ∫ x =0 1 ∫ y =0 x (2x )(3y 2 ) = ∫ x =0 1 2x = ∫ x =0 1 2xx 3 dx =

0

1

= .

y =0

x

dx

Question 3.7.44

In this case the joint cdf would be given by F X,Y (u,v ) = P (X ≤ u,Y ≤ v ) = ∫ 0 u ∫ 0 v = ∫ 0 u ∫ 0 v xydydx = ∫ 0 u x =∫0u

xdx =

0

u

(2y )dydx 0

=

v

dx

.

Question 3.7.45

To evaluate this probability we would compute

Question 3.7.46

To start note that P (a < X < b,c < Y < d ) = ∫ x = a b ∫ y = c d xye − ( x + y ) dydx = = P (a < X < b )P (c < Y < d ).

The above is true because we can show that f X (x ) = ∫ y =0 ∞ xye − ( x + y ) dy = xe − x f Y (y ) = ∫ x =0 ∞ xye − ( x + y ) dx = ye − y .

This in tern means that

and the same for P (c < Y < d ).

Question 3.7.47

Note that we have P (a < X < b,c < Y < d ) = ∫ x = a b ∫ y = c d (2x + y − 2xy )dydx =∫x=ab

dx

=

∫ x = a b (1 − 2x )dx + 2(d − c ) ∫ x = a b xdx

=

+ 2(d − c )

= (d − c )(b − a )

.

(9)

Next we need to compute the two marginal densities where we find f X (x ) = ∫ y =0 1 (2x + y − 2xy )dy = x + f Y (y ) = ∫ x =0 1 (2x + y − 2xy )dx = 1.

This means that P (a < X < b ) = ∫ x = a b

dx = (b 2 − a 2 + b − a ) = (b − a )(b + a + 1)

P (c < Y < d ) = d − c , and the product is then

There are many values of a , b , c , and d that will show these two are not equal. For example if we take a = c = 0, b = and d = 1 then we find then

and

Which are of course not equal.

Question 3.7.48

As we know that P (X,Y ) = P (X )P (Y ) if we let X → g (X ) and Y → h (Y ) then this becomes

or P (U,V ) = P (U )P (V ) showing that U and V are independent.

Question 3.7.49

Yes this can happen and we saw an example of this in the joint density f X,Y (x,y ) = xye − x − y with x > 0 and y > 0 in a previous question.

Question 3.7.50

As all variables are independent this would be the product of n individual pdfs or ∏ i =1 n f X (x i ).

Question 3.7.51

Part (a): This would be

Part (b): The number of X i (from n = 4) that will be less than 1⁄ 2 would be given by a binomial random variable with a p value computed above. Thus for this probability we would get dbinom(1, 4, 1/16) which evaluates to 0.2059937.

Part (c): This would be 4 4 x 1 3 x 2 3 x 3 3 x 4 3 .

Part (d): By independence we know that f X 2 ,X 3 (x 2 ,x 3 ) = 4 2 x 2 3 x 3 3 from which we can evaluate the cdf where we have

Question 3.7.52

The probability that X i is less than (or greater than) 1⁄ 2 is 1⁄ 2 thus this expression will evaluate to

Notes on Section 3.8 I think the second term in the first line in the proof of Theorem 3.8.4 should be

Note that in this expression, the first inequality is reversed from what the book has. Given that expression it next becomes

and then

The rest of the proof follows from here.

Question 3.8.1

Part (a): We have

Here we can take x ≥ 0 and we need to have w − x ≥ 0 or x ≤ w so the above sum becomes

We can evaluate this by writing it as p W (w ) = e − ( λ + μ ) μ w ∑ x =0 w = =

x

∑ x =0 w

x

w

=

.

This is the same type of pdf that X and Y are.

Part (b): We have

Here we can take x ≥ 1 and we need to have w − x ≥ 1 or x ≤ w − 1 so the above sum becomes

for w ≥ 2. This is not the same type of pdf as X and Y .

Question 3.8.2

We have

Here we can take x ≥ 0 and we need to have w − x ≥ 0 or x ≤ w so the above becomes

for w ≥ 0.

Question 3.8.3

We have

Here we can take 0 ≤ x ≤ 1 to have p X (x ) defined but we will need to have both w − x ≥ 0 (or x ≤ w ) and w − x ≤ 1 (or x ≥ w − 1) for p Y (w − x ) to be defined. Thus for a given value of w the above becomes

Following the hint given for this question when we have 0 ≤ w < 1 the above is

and when 1 ≤ w < 2 the above is

Question 3.8.4

Now we can write the joint density of V and W as

But V is independent of X + Y so P (V = v | X = x,Y = w − y ) = P (V = v ) and we have

showing that V and W are independent.

Question 3.8.5

To derive the pdf for W we start by computing the cdf. We have F W (w ) = P (W ≤ w ) = P (Y 2 ≤ w ) = P (− ≤Y≤+ )=∫− Taking the w derivative of the above gives the pdf or f W (w ) =

= f Y(

)

=

(f Y (

) + f Y (−

As Y is nonnegative and −

< 0 we have that f Y (−

f Y (y )dy .

− f Y (−

)

)). ) = 0 and we get

Question 3.8.6

To derive the pdf for W we start by computing the cdf. We have F W (w ) = P (W ≤ w ) = P (Y 2 ≤ w ) = P (− = P (0 ≤ Y ≤

)=∫0

dy =

≤Y≤

)

.

In the above we have used the fact that as Y cannot be negative the probability to observe any negative values is zero. The above means that

Question 3.8.7

To derive the pdf for W we start by computing the cdf. We have F W (w ) = P (W ≤ w ) = P (Y 2 ≤ w ) = P (−

≤Y≤

)

= P (0 ≤ Y ≤ ) = ∫ 0 by (1 − y )dy = 3w − 2w 1 ⁄ 2 .

In the above we have used the fact that as Y cannot be negative the probability to observe any negative values is zero. The above means that

Question 3.8.8

To derive the pdf for W we start by computing the cdf. We have F W (w ) = P (W ≤ w ) = P

=P

=P

=P

=∫0

2

ay 2 e − by dy .

In the above we have used the fact that as Y cannot be negative the probability to observe any negative values is zero. To integrate this we can take v = by 2 so that y dv

= = 2bydy = 2b

dy = 2

dy .

Using these we have

Taking the w derivative of this to get the pdf gives

Question 3.8.9

Part (a): If W = XY then the pdf for W is given by (10) For this problem to have f X (x ) nonzero we need to have 0 ≤ x ≤ 1. To have f Y (w⁄x ) nonzero we need to have

or the requirement that

Using these the expression for f W (w ) becomes

for 0 ≤ w ≤ 1.

Part (b): The limits of integration for the pdf for W in this case are the same as in the previous part of this question. Because of that we find

for 0 ≤ w ≤ 1.

Question 3.8.10

Part (a): If W =

then the pdf for W is given by (11)

For this problem to have f X (x ) nonzero we need to have 0 ≤ x ≤ 1. To have f Y (wx ) nonzero we need to have

or the requirement that

Using these we get

for 0 ≤ w ≤∞ .

Part (b): The limits of integration for the pdf for W in this case are the same as in the previous part of this question. Because of that we find

for 0 ≤ w ≤∞ .

Question 3.8.11

If W =

then

To have f X (x ) nonzero requires that x ≥ 0. To have f Y (y ) nonzero requires that wx ≥ 0 (which will always be true). Thus we have

for 0 ≤ w ≤∞ .

Question 3.9.1

Let X i be the number dawn on the i th draw. Then as we are drawing with replacement we have

The expectation we want to evaluate is given by

Now

Thus we get

Question 3.9.2

For this we would compute E (X + Y ) = ∫ x =0 ∞ ∫ y =0 ∞ (x + y )λ 2 e − λ ( x + y ) dydx = λ 2 ∫ x =0 ∞ = λ ∫ x =0 ∞ xe − λx dx + ∫ x =0 ∞ e − λx dx = λ Question 3.9.3

+ = .

dx

For this we would compute E (X + Y ) = ∫ x =0 1 ∫ y =0 1 (x + y ) = ∫ x =0 1

(x + 2y )dydx dx =

= .

Question 3.9.4

Cathine’s score is S = 4N 1 + 6N 2 where we know the expectations of N 1 and N 2 . Using them we can compute

Question 3.9.5

Computing the expectation on the left-hand-side gives

If this is to equal μ then we need to have ∑ i =1 n a i = 1.

Question 3.9.6

If we let U be the number of days the stock goes up and D the number of days the stock goes down then U + D = n where n is the number of days we watch the stock. Here if the stock can only go up or down we would have q = 1 − p . The total amount the stock will move (in dollars) is then given by

Here U is a binomial random variable with parameters (n,p ). Then we have that the expectation of the above is given by

Note that if p > q = 1 − p then we have that p > so that the above expression is positive.

Question 3.9.7

Here X i = 1 if the i th ball is red.

Part (a): Draw n balls but do not look at any of the colors. Then look at the i th ball’s color. As the probability that it is red must be the same for all i we have that P (X i = 1) = . This means that the expectation of X i is the same for every i .

Part (b): The number of red balls can be written as R = ∑ i =1 n X i so that

Question 3.9.8

Let the product be X 1 X 2 then by independence we have

Question 3.9.9

To be normalized correctly we must have

This means that k =

. Now

This means that E (R ) = k ∫ x =10 20 ∫ y =10 20

(x + y )dydx = k ∫ x =10 20 ∫ y =10 20 xydydx

= k ∫ x =10 20 150xdx = 22500k = 7. 5.

Question 3.9.10

We can do this directly as

As another method we have

using the result from Question 3.8.6.

Question 3.9.11

The area is A = XY thus

Question 3.9.12

We can compute the expectation of the geometric mean as E[

] =E[ = ∏ i =1 n E [

] ]=E[

]n.

To evaluate this we have

Thus

This is to be compared to

By the Arithmetic-Geometric mean inequality we expect that

We can verify that inequality for a few values of n using ns = 1:10 print( ( ns / (ns+1) )^ns )

This gives [1] 0.5000000 0.4444444 0.4218750 0.4096000 0.4018776 0.3965695 0.3926959 [8] 0.3897443 0.3874205 0.3855433

Notice that each number is less than (or equal to) 0.5 (as expected).

Question 3.9.13

For this question we will explicitly enumerate the possible values for X and Y and then compute the required covariance. We can get the joint pdf for X and Y using D = 1:6 outcomes = expand.grid(D, D) colnames(outcomes) = c(’X1’, ’X2’) X = outcomes$X1 Y = apply(outcomes, 1, max) T = table(X, Y) print(T) P = T/sum(T) print(P)

This gives > print(P) Y X 1 2 3 4 5 6 1 0.02777778 0.02777778 0.02777778 0.02777778 0.02777778 0.02777778 2 0.00000000 0.05555556 0.02777778 0.02777778 0.02777778 0.02777778 3 0.00000000 0.00000000 0.08333333 0.02777778 0.02777778 0.02777778 4 0.00000000 0.00000000 0.00000000 0.11111111 0.02777778 0.02777778 5 0.00000000 0.00000000 0.00000000 0.00000000 0.13888889 0.02777778 6 0.00000000 0.00000000 0.00000000 0.00000000 0.00000000 0.16666667

Using this we can compute the needed expectations for the covariance. We find dim(P) = NULL DF = data.frame(X=X, Y=Y, prob=as.double(P)) DF$XY = DF$X * DF$Y EXY = sum(DF$XY * DF$prob) EX = sum(DF$X * DF$prob) EY = sum(DF$Y * DF$prob) print( c( EXY - EX*EY, 105/72 ) )

This gives the value 1.458333.

Question 3.9.14

We can show this with the following manipulations Cov

= E (( aX + b )( cY + d )) − E ( aX + b ) E ( cY + d ) = E ( acXY + adX + bcY + bd ) − ( aE ( X ) + b )( cE ( Y ) + d ) = acE ( XY ) + adE ( X ) + bcE ( Y ) + bd − acE ( X ) E ( Y ) − adE ( X ) − bcE ( Y ) − bd = ac ( E ( XY ) − E ( X ) E ( Y )) = ac Cov

.

Question 3.9.15

Note that X 2 + Y 2 = 1 so that X and Y are functionally dependent. This means that they must be probabilistically dependent (and therefore cannot be independent).

To compute the covariance we have E (X ) = ∫ 0 2 π cos(u )

du =

E (Y ) = ∫ 0 2 π sin(u )

du = −

E (XY ) =

∫ 0 2 π cos(u ) sin(u )du = 0,

when we integrate. Thus we have that Cov Question 3.9.16

= 0 − 0 = 0.

0

2π

0

=0

2π

=0

If we draw the region we will integrate in the x -y plane we see that it is a right triangle in the first and second quadrant formed from the lines y = ± x with its right angle at the origin and its hypotenuse at the level y = 1. It can be drawn with its corners at (0, 0), (1, 1), and (− 1, 1). Then given that region we have E (XY ) = ∫ y =0 1 ∫ x = − y y xydxdy = ∫ y =0 1 y E (X ) = ∫ y =0 1 ∫ x = − y y xdxdy = ∫ y =0 1

−y

−y

y

y

dy = 0

dy = 0.

It does not matter what the value of E (Y ) is because

Note that from the above region where f X,Y (x,y ) is defined we see that P

≠0 P

≠ 0,

but

Thus the above joint probability cannot be equal to the product of the two marginal probabilities and the two random variables must be dependent.

Question 3.9.17

Note

which is an exponential pdf. By symmetry the pdf for Y is another exponential. From the properties of the exponential pdf we know that E (X ) = E (Y ) = Var

= Var

=

.

We need to compute E (XY ) = ∫ x =0 ∞ ∫ y =0 ∞ λ 2 xye − λ ( x + y ) dydx = ∫ x =0 ∞ xe − λx dx = Using this we have that

.

Finally we have

Question 3.9.18

From Question 3.9.3 we know that E (X + Y ) = . Then to compute the variance we need to know E ((X + Y ) 2 ). Doing that E ((X + Y ) 2 ) = ∫ x =0 1 ∫ y =0 1 (x + y ) 2 = ∫ x =0 1 Thus

Question 3.9.19

From the definition of variance we would need to evaluate

Now since Y i is a uniform random variable we have that

Next

Thus we need to compute

This means that

Using everything together this gives

dydx dx =

= .

Question 3.9.20

Using properties of binomial random variables we have E (W ) = 4E (X ) + 6E (Y ) = 4np X + 6mp Y Var

= 4 2 Var

+ 6 2 Var

= 16np X (1 − p X ) + 36mp Y (1 − p Y ).

Question 3.9.21

The total cost C can be written as C = 50U + 60V where U is a Poisson random variable with expectation

calls. This is the product of nine, the number of hours in a normal day, and seven the per-hour call rate during this time. In the same way V is a Poisson random variable with expectation 15 ⋅ 4 = 60.

Using these results and the properties of Poisson random variables we have that E (C ) = 50(63) + 60(60) = 6750 Var = 50 2 (63) + 60 2 (60) = 373500.

Question 3.9.22

Let B i be the width of the i th brick for 1 ≤ i ≤ 50 and M i the width of the i th layer of mortar for 1 ≤ i ≤ 49. Then the total length L of the first row of the wall is then given by

From the information given in this question we then have

and Var

= ∑ i =1 50 Var = 50Var

+ ∑ i =1 49 Var

+ 49Var

= 50

2

+ 49

2

= 0. 2402344.

This means that the standard deviation of L is the square root of the above or 0.4901371 inches. In deriving these results we are assuming independence of B i and M i for all i .

Question 3.9.23

The total resistance is given by R = ∑ i =1 6 R i so that

This means that to satisfy the requirement on the standard deviation of R we must have

This then means that for the individual resistor that σ R 1 ≤ 0. 1632993.

Question 3.9.24

The total the gambler will win T is given by T = ∑ i =1 n W i with W i the win on the i th hand. The random variables W i take the value k with a probability of p and the value zero with a probability of 1 − p . Then E (W i ) = kp + 0(1 − p ) = kp E (W i 2 ) = k 2 p + 0(1 − p ) = k 2 p Var

= k 2 p − k 2 p 2 = k 2 p (1 − p ).

This means that E (T ) = nE (W 1 ) = knp Var = ∑ i =1 n Var assuming independence of the i th hand.

= nk 2 p (1 − p ),

Question 3.10.1

In the notation from this section we want to know P (Y 3 ′ < 5) where Y i ′ is the i th order statistic from n uniform random variables. Here i = 3 and n = 4 so from the formulas given in the book the pdf for Y 3 ′ is given by

For a uniform random variable like considered here we have f Y (y ) = above becomes

Using that we can compute

and F Y (y ) =

so the

when we integrate.

Question 3.10.2

The cdf for this pdf is given by

Using that and the formulas form this section we find that

Thus we can compute that

when we integrate.

Question 3.10.3

The pdf of the maximum of two random variables is given by f Y 2 ′ (y ) = 2F Y (y )f Y (y ). The sixtieth percentile is the number y 0 . 6 such that

The probability we want to evaluate is then

Note that

thus we get

Question 3.10.4

The statement Y 1 ′ < 0. 6 < Y 5 ′ means that the smallest number from our six is less than 0.6 and the largest number is greater than 0.6. The complement of this event is Y 1 ′ > 0. 6 (the smallest number is larger than 0.6) or Y 5 ′ < 0. 6 (the largest number is less than 0.6). Thus we have

We can more easily evaluate the two probability on the right-hand-side. The pdfs we need are given by f Y 1 ′ (y ) = 2ny (1 − y 2 ) n − 1 f Y 5 ′ (y ) = 2ny 2 n − 1 , with n = 5. Using these we have P (Y 1 ′ > 0. 6) = ∫ 0 . 6 1 10y (1 − y 2 ) 4 dy = 0. 107374 P (Y 5 ′ < 0. 6) = ∫ 0 0 . 6 10y 9 dy = 0. 6 10 .

Using these we have

Question 3.10.5

First we will compute P (Y 1 ′ > m ) = ∫ m ∞ n (1 − F Y (y )) n − 1 f Y (y )dy =−∫m∞ =−

(1 − F Y (y )) n − 1 f Y (y )dy m

∞

= (1 − F Y (m )) n .

Next we evaluate

From the definition of m we have that F Y (m ) = 0. 5 so the two expression are P (Y 1 ′ > m ) = 0. 5 n P (Y n ′ > m ) = 1 − 0. 5 n .

If n ≥ 2 the second probability is larger.

As an extension question, using the same type of logic as above we can evaluate

From the definition of m we have that F Y (m ) = 0. 5 thus we see that the two probabilities P (Y 1 ′ > m ) and P (Y n ′ < m ) are equal.

Question 3.10.6

Using the definitions from this section we have that

Thus we want to find n such that

Integrating the left-hand-side we get

Using this we can solve for n and find

Question 3.10.7

From this section of the book the pdf for Y 4 ′ when Y i is from a uniform pdf and n = 6 is given by

Using this the probability that we want is given by

when we integrate.

Figure 3: The plots Question 3.10.8.

Question 3.10.8

For this pdf we have F Y (y ) = y 2 and the pdf’s for Y 1 ′ and Y 5 ′ are given in the text. Specifically we have f Y 2 (y ) = 2y f Y 1 ′ (y ) = 10y (1 − y 2 ) 4 f Y 5 ′ (y ) = 10y 9 .

In the R code for this section we plot these pdfs on the same axis. When we do that we get Figure 3 .

Question 3.10.9

This would be P (Y 1 ′ > 20) = ∫ 20 ∞ f Y ′ (y )dy = ∫ 20 ∞ n (1 − F Y (y )) n − 1 f Y (y )dy 1

∞ n =− 20 = (1 − F Y (20)) .

For the given pdf we can numerically evaluate F Y (20) and find F Y (20) = 0. 5. Thus we get P (Y 1 ′ > 20) = 0. 5 n .

Question 3.10.10

The is equal probability that any one of the samples will be the smallest. Thus this probability is .

Question 3.10.11

From the given graph given in this question we see that f Y (y ) = 2y . Then using samples drawn from a pdf of that form we want to know the values of P (Y 4 ′ > 0. 75) and P (Y 1 ′ > 0. 1). Using formulas from the book the pdf for Y 4 ′ and Y 1 ′ (from n = 5 samples) are given by f Y 4 ′ (y ) = 40y 7 (1 − y 2 ) f Y 1 ′ (y ) = 10y (1 − y 2 ) 4 .

Using these we can integrate to get the requested probabilities. We find P (Y 4 ′ > 0. 75) = ∫ 0 . 75 l f Y ′ (y )dy = 0. 724689 4 P (Y 1 ′ > 0. 1) = ∫ 0 . 1 1 f Y ′ (y )dy = 0. 95088, 1

when we integrate.

Question 3.10.12

The first component will fail at a time Y min = min{ Y 1 ,Y 2 , … ,Y n } where Y i are iid samples from the exponential distribution. For an exponential distribution we have F Y (y ) = 1 − e λy . Using the formula given in this section of the text we have

This is the pdf of another exponential random variable. This time with a parameter nλ . An exponential random variable of this form has a mean given by .

Question 3.10.13

From the suggested theorem we know that

is the pdf of the i th order statistic. If we consider Y to be uniform [0, 1] then the above becomes

As this is a valid pdf it must integrate to one. That statement is equivalent to the statement we were asked to show. Note that this result is a special case of the beta function.

Question 3.10.14

Using results from the previous question we would need to evaluate E [Y i ′ ] = ∫ 0 1 y =

dy ∫ 0 1 y i (1 − y ) n − i dy .

Using the previous question (or by recognizing the above integral to be the beta function) we can evaluate the above expectation to get E (Y i ′ ) = =

∫ 0 1 y ( i +1) − 1 (1 − y ) ( n +1) − ( i +1) dy =

.

Question 3.10.15

We want to evaluate P (Y n ′− Y 1 ′ < 0. 5) when n = 3. From this section of the book the joint pdf for these two order statistics is given by

In the (u,v ) plane we want to integrate over the region in the triangle 0 < u < 1 and u < v < 1 where

If we draw this region in the (u,v ) plane we see that the integral we need to compute to determine this probability can be written as (using n = 3) P = ∫ u =0 1 ⁄ 2 ∫ v = u u +0 . 5 f Y 1 ′ ,Y n ′ (u,v )dvdu + ∫ 1 ⁄ 2 1 ∫ v = u 1 f Y 1 ′ ,Y n ′ (u,v )dvdu = 6 ∫ u =0 1 ⁄ 2 ∫ v = u u +0 . 5 (v − u )dvdu + 6 ∫ 1 ⁄ 2 1 ∫ v = u 1 (v − u )dvdu = 6 ∫ u =0 1 ⁄ 2 du + 6 ∫ 1 ⁄ 2 1 (u − 1) 2 du = + = .

Question 3.10.16

We want to evaluate P (Y n ′− Y 1 ′ < 2) when n = 3. From this section of the book as F Y (y ) = 1 − e −y the joint pdf for these two order statistics is given by

In the (u,v ) plane we want to integrate over the region in the triangle 0 < u < ∞ and u < v < ∞ where

If we draw this region in the (u,v ) plane we see that the integral we need to compute to determine this probability can be written as (using n = 3) P = ∫ u =0 ∞ ∫ v = u u +2 6(e − u − e − v )e − ( u + v ) dvdu = 3(e 2 − 1) 2 ∫ u =0 ∞ e − 3 u − 4 du =

= 0. 747645,

when we integrate.

Question 3.11.1

We first need to compute p X (x ). To do that we note that p X (1) = P { (1, 1), (1, 2)} =

+

=

p X (2) = P { (2, 1), (2, 2)} =

+

=

.

We can combined these two expressions if we note that the numerator in the above can be written as 5x + 3 for the two different x values. This means that p Y | x (y ) = = Question 3.11.2

×

=

.

For this question we will explicitly compute p X,Y (x,y ) in a table form. We can do that with the following R code D = 1:6 outcomes = expand.grid(D, D, D, D, D, D) colnames(outcomes) = c(’D1’, ’D2’, ’D3’, ’D4’, ’D5’, ’D6’) X = apply(outcomes, 1, function(x){ sum(x==4) }) Y = apply(outcomes, 1, function(x){ sum(x[1:2]==4) }) T = table(X, Y) print(T) P = T/sum(T) print(P)

This gives for the probability P (X = x,Y = y ) of > print(P) Y X 0 1 2 0 3.348980e-01 0.000000e+00 0.000000e+00 1 2.679184e-01 1.339592e-01 0.000000e+00 2 8.037551e-02 1.071674e-01 1.339592e-02 3 1.071674e-02 3.215021e-02 1.071674e-02 4 5.358368e-04 4.286694e-03 3.215021e-03 5 0.000000e+00 2.143347e-04 4.286694e-04 6 0.000000e+00 0.000000e+00 2.143347e-05

Here the row label holds X and the column label holds Y . Now that we have P (X,Y ) to compute P Y | x (y ) we will use

To compute P (X ) we sum the columns of the above matrix to get pX = rowSums(P) > print(pX) 0 1 2 3 4 5 3.348980e-01 4.018776e-01 2.009388e-01 5.358368e-02 8.037551e-03 6.430041e-04 6 2.143347e-05

We can use this expression to compute the fraction needed in computing P Y | x (y ). In R we can do this as denom = kronecker(matrix(1, 1, 3), pX) p_Y_given_x = P/denom

Running the above gives > p_Y_given_x Y X 0 1 2 0 1.00000000 0.00000000 0.00000000 1 0.66666667 0.33333333 0.00000000

2 0.40000000 0.53333333 0.06666667 3 0.20000000 0.60000000 0.20000000 4 0.06666667 0.53333333 0.40000000 5 0.00000000 0.33333333 0.66666667 6 0.00000000 0.00000000 1.00000000 Question 3.11.3

The joint distribution of (x,y ) is given by

In the same way p (x ) is given by

Thus using these two we have

Question 3.11.4

From the question statement we have

p(x) =

p ( x,y ) =

Using these two results we have that

If we take X = 2 and Y = 2 the above evaluates to 0.01017576.

.

Question 3.11.5

Part (a): We must have ∑ x∑ yk ( x +y ) =k = 36 k ,

equal to one. This means that k =

.

Part (b): We need to compute p X (x ). We find p X (x ) = ∑ y =1 , 2 , 3 k (x + y ) = k = k (3x + (1 + 2 + 3)) = k (3x + 6).

Thus we have that

Question 3.11.6

Part (a): From the question we have that p X,Y (x,y ) = p (Y = y | X = x )P (X = x ) =

y

x−y

=

x

.

Here we have used the facts that P (X = x ) = and Y | X = x is a binomial random variable.

Part (b): Using the previous expression we would have that

If we evaluate this in R we find pY = rep(NA, 4) xs = 1:3 for( y in 0:3 ){ pY[y+1] = sum( choose(xs, y) * (1/2)^xs )/3 } print(pY)

This gives the values > print(pY) [1] 0.29166667 0.45833333 0.20833333 0.04166667

for y ∈{ 0, 1, 2, 3} .

Question 3.11.7

We first compute p (z ) = ∑ x =1 2 ∑ y =1 2 p (x,y,z ) = ∑ x =1 2 ∑ y =1 2 =

+

+

+

=

.

This means that

Question 3.11.8

While there maybe many ways of working this problem one simple way is to do it numerically. We can do that in the following R code D = 1:2 outcomes = expand.grid(D, D, D) colnames(outcomes) = c(’X’, ’Y’, ’Z’) outcomes$W = apply(outcomes, 1, max) X = outcomes$X Y = outcomes$Y Z = outcomes$Z outcomes$P = ( X * Y + X * Z + Y * Z )/54 P_X_W = matrix( NA, nrow=2, ncol=2 ) for( x in 1:2 ){ for( w in 1:2 ){ mask = (outcomes$X == x) & (outcomes$W == w) P_X_W[x, w] = sum( outcomes$P[mask] ) } } rownames(P_X_W) = c(1, 2) colnames(P_X_W) = c(1, 2) print(P_X_W) p_X = rowSums(P_X_W) print(p_X) p_W_given_x = P_X_W / kronecker(matrix(1, 1, 2), p_X) print(p_W_given_x)

When we run that code we get (along with other output) > print(p_W_given_x) 1 2 1 0.1428571 0.8571429 2 0.0000000 1.0000000

This says that if X = 2 we have that W = 2 with 100% certainty (which is consistent with the definition of W ). If X = 1 then there is approximately a 14% that W = 1 and a 86% chance that W = 2.

Question 3.11.9

Let W = X + Y then from Question 3.8.1 we know that the distribution of W is Poisson with a parameter λ + μ . Then to show this we will use

The numerator of the above fraction is

The denominator of the above fraction is

If we take the ratio of these two and simplify we get

which is the pmf for a binomial random variable with n = w and p =

.

Question 3.11.10

As the number of errors on each of the pages that A prepares is a Poisson random variable with a parameter of two and there are 100 such independent pages the total number of errors that A will make is another Poisson with a parameter of 2(100) = 200. Let X be the random variable denoting the total number of errors that A makes. The same logic holds for B but the total number of errors is now a Poisson random variable with a parameter 3(100) = 300. Let Y be the total number of errors that B makes. We are told that X + Y = 520. From the previous question we know that the distribution of X given W is a binomial with n = 520 and

This means that the probability that fewer than half the errors are from A is

Question 3.11.11

For the left-hand-side we have P (X > s + t | X > t ) = =

= =

= e − s⁄λ .

For the right-hand-side we have

As these two expressions are equal we have the desired expression.

Question 3.11.12

Part (a): To evaluate this we will use

Now we have

In the same way we have

This means that

Part (b): Note that

since if X = 1 then from the domain of definition of the joint pdf we must have Y > 1.

Part (c): For this we have

We need to compute

This means that

for y ≥ x .

Part (d): Using the above expression for f Y | x (y ) we compute

Question 3.11.13

From the definition of the conditional density we have

To use this we need to compute

This means that

Question 3.11.14

From the definition of the conditional density we have

To use this we need to compute

for 0 < x < 1. This

for 0 < y < x . As this is independent of y this is a uniform density.

Question 3.11.15

Using what we are given we have

Using this we find the marginal

for 0 < y < 1.

Question 3.11.16

Part (a): From the joint we can compute the marginal

for 0 ≤ x ≤ 1.

Part (b): Using the marginal we can compute the conditional density

Part (c): Using the conditional density we can compute this probability P

= ∫ y =1 ⁄ 4 3 ⁄ 4 f Y | x =1 ⁄ 2 (y )dy = ∫ y =1 ⁄ 4 3 ⁄ 4

dy

= ∫ y =1 ⁄ 4 3 ⁄ 4 (1 + 3y )dy = .

Part (d): Using the conditional density we have

Question 3.11.17

so we need to compute

for 0 < y < 1. Using that we have

for 0 < y < 1. Using that we can compute

Question 3.11.18

First we need to compute

To do that we need to compute

Using this we have

Using that we have

Question 3.11.19

We will use

To compute that denominator we will use P (X 4 ,X 5 ) = ∫ x =0 1 ∫ x =0 1 ∫ x =0 1 32x 1 x 2 x 3 x 4 x 5 dx 3 dx 2 dx 1 1

= 32

2

0

= 4x 4 x 5 .

Using this we have

Question 3.11.20

Part (a): For this we have

1

3

0

1

0

1

(x 4 x 5 )

Part (b): Using the above we have

Part (c): For this we will use

Thus we need to evaluate

and

Together we can then conclude that

Question 3.12.1

For this we would have

Question 3.12.2

Here if we let C 1 and C 2 be the value of the i th chip we would have

and

This means that

Question 3.12.3

The expression we want to compute is M X (3) when X is a binomial random variable. When X is a binomial random variable then the mgf is given by (12) Using this we have

Question 3.12.4

Using the given pmf we have M X (t ) = E (e tX ) = ∑ k =0 ∞ e kt = ∑ k =0 ∞

k

k

=

=

.

Question 3.12.5

In this question we match these mgf to the ones introduced in this section of the text.

Part (a): This is the mgf for a Gaussian with μ = 0 and

= 6 (or σ 2 = 12).

Part (b): The is the mgf from an exponential random variable with λ = 2.

Part (c): This is the mgf for a binomial random variable with p = and n = 4.

Part (d): This is the mgf for a geometric random variable with p = 0. 3.

Question 3.12.6

Using the given pdf we would have M Y (t ) = E (e Y t ) = ∫ y =0 1 e yt ydy + ∫ y =1 2 e yt (2 − y )dy = Question 3.12.7

+

=

.

Using the given pmf for a Poisson random variable we would have M X (t ) = E (e tX ) = ∑ k =0 ∞ e tk = e − λ ∑ k =0 ∞

t

t

= e − λ e λe = e λ ( e − 1) .

(13)

Question 3.12.8

Using the given pdf we compute the moment generating function. We see that

Using integration by parts we find (14) as we were to show.

Question 3.12.9

Using the given expression for M Y (t ) we find M Y (t ) = e t 2 ⁄ 2 M Y ′ (t ) = te t 2 ⁄ 2 M Y ′′ (t ) = e t 2 ⁄ 2 + t 2 e t 2 ⁄ 2 = (1 + t 2 )e t 2 ⁄ 2 M Y ′′′ (t ) = 2te t 2 ⁄ 2 + (t + t 3 )e t 2 ⁄ 2 = (t 3 + 3t )e t 2 ⁄ 2 .

Using this we see that E (Y 3 ) = M Y ′′′ (0) = 0.

Question 3.12.10

If Y is an exponential random variable then its mgf looks like (15) The n th derivative of this is given by

This means that

Question 3.12.11

From the given mgf

we have the first derivative given by

From this we conclude that

The second derivative of the mgf is given by

This means that

Solving for b 2 this means that

Question 3.12.12

From the given mgf

we see that

This means that

Question 3.12.13

The M Y (t ) we are given is the mgf of a Gaussian random variable with μ = − 1 and 8). Then

= 4 (so σ 2 =

Question 3.12.14

If our mgf looks like (16) then we have that

This means that

Question 3.12.15

In Question 3.12.1 we found M X (t ) = ) = M X ′ (0). We compute

. To find the expectation we will use the fact that E (X

M X ′ (t ) = − =

+ .

If we try to evaluate this at t = 0 (as would be needed to compute E (Y )) we find we have an expression of the form 0⁄ 0 which is indeterminate. To evaluate that expression at t = 0 we will need to use L’Hospital’s rule. Taking the derivative of the numerator and the denominator of the above gives

To this expression we divide the numerator and the denominator by e t to get

This limit (at t = 0) is still of the 0⁄ 0 form so we use L’Hospitals rule again. Taking the derivative of the numerator and the denominator of the above gives

If we take t = 0 in this expression we get

Note that if we take a = 0 and b = n − 1 then this expression is

.

Question 3.12.16

From the given mgf we have

and

From these we have M Y ′ (0) = E (Y ) = 2 M Y ′′ (0) = E (Y 2 ) = 2 + 0 + 2(2) = 6 so Var

= 6 − 2 2 = 2.

Question 3.12.17

From Question 3.12.8 we know that if Y ∼ ye − y then the mgf is given by Equation 14 . Now if we let V = then the pdf for V is λye − λy so using Theorem 3.12.3 (a) with a = we have that

Question 3.12.18

We would have

This is the same mgf as found in Question 3.12.14 Equation 16 when r = 6.

Question 3.12.19

To solve the parts of this question we have to look if the product of the mgf for two terms we add gives another mgf of the same functional form.

Part (a): Using Equation 13 we would have

which is the mgf for another Poisson random variable.

Part (b): Using Equation 15 we would have

which is not the mgf of an exponential random variable.

Part (c): The book gives the form of the mgf for a normal random variable. Using that we would have

which is the mgf for another normal random variable.

Question 3.12.20

The given mgf is for a binomial random variable with n = 5 and p = 3⁄ 4. Thus P (X ≤ 2) can be computed using pbinom(2, 5, 3/4) which gives 0.1035156.

Question 3.12.21

The mgf for the sum n Y = ∑ i =1 n Y i will be the product of the mgf for each Y i . This means that

Then using transformation properties of mgf’s we have that the mgf for the product of and the random variable n Y or (n Y ) would be given by

This is the mgf for a normal random variable with mean μ and variance

.

Question 3.12.22

From the form of the mgf given, the random variable W must be the sum of a Poisson random variable X and a binomial random variable Y . Then P (X + Y ≤ 1) = ∑ k =0 4 P (X + Y ≤ 1| Y = k )P (Y = k ) = ∑ k =0 4 P (X ≤ 1 − k | Y = k )P (Y = k ) = ∑ k =0 1 P (X ≤ 1 − k | Y = k )P (Y = k ).

We can evaluate the above using the following R code

ks = 0:1 print( sum( ppois( 1-ks, 3 ) * dbinom(ks, 4, 1/3) ) )

Running the above gives 0.0590069.

Question 3.12.23

Part (a): Using transformation properties of mgf’s we have that

which is not the mgf for a Poisson random variable.

Part (b): Using transformation properties of mgf’s we have that

Question 3.12.24

Part (a): Using transformation properties of mgf’s we have that

which is the mgf for a normal random variable.

Part (b): Using transformation properties of mgf’s we have that

which is also the mgf for a normal random variable.

Chapter 4 (Special Distributions) Question 4.2.1

When we run the R code for this section we get 0.1577983 for the binomial answer and for the 0.1578431 Poisson answer. These two numbers are so close because n is large and p is small.

Question 4.2.2

When we run the R code for this section we get 0.03083403 for the exact binomial answer and for the 0.03078655 Poisson answer.

Question 4.2.3

For this I get the probability 0.6022812.

Question 4.2.4

Part (a): This probability would be approximately 0.180447.

Part (b): For the given estimate of p and n a table like suggested would look like [1] "P(X >= 1) = 0.864665" [1] "P(X >= 2) = 0.593994" [1] "P(X >= 3) = 0.323324" [1] "P(X >= 4) = 0.142877" [1] "P(X >= 5) = 0.052653" [1] "P(X >= 6) = 0.016564" [1] "P(X >= 7) = 0.004534" [1] "P(X >= 8) = 0.001097" [1] "P(X >= 9) = 0.000237" [1] "P(X >= 10) = 0.000046" [1] "P(X >= 11) = 0.000008"

Thus if we had X ≥ 7 then there is a less than 1% chance of this happening when p is 1⁄ 10000.

Question 4.2.5

For this question I get [1] "binomia1 prob= 0.095618" [1] "poisson prob= 0.095163" Question 4.2.6

For this question I get [1] "the probability of three or more claims in a given year= 0.047423" Question 4.2.7

For this question I get [1] "the probability loosing two or more bags in a given year= 0.121901" Question 4.2.8

For this question I get [1] "the probability of two or more cases (under H0)= 0.000045"

This number is too small to be attributed to chance.

Question 4.2.9

For this question we want to find p such that

Inverting the above expression with n = 100 ⋅ 10 9 gives p > 6. 931472 ⋅ 10 − 12 .

Question 4.2.10

If I table the empirical and Poisson model probabilities with E (X ) = 0. 61 I get 1 2 3 4 5

num_of_fatalities empirical model 0 0.545 0.543350869 1 0.325 0.331444030 2 0.110 0.101090429 3 0.015 0.020555054 4 0.005 0.003134646

The close agreement of the two columns of numbers argues that we can use a Poisson model.

Question 4.2.11

If I table the empirical and Poisson model probabilities with E (X ) = 0. 4353933 I get 1 2 3 4

num_changes empirical model 0 0.66573034 0.647010175 1 0.25280899 0.281703868 2 0.06179775 0.061325983 3 0.01966292 0.008900306

These model coefficients give rise to an expected frequencies of [1] 230.335622 100.286577

21.832050

3.168509

Question 4.2.12

If I table the empirical and Poisson model probabilities with E (X ) = 1. 525 I get 1 2 3 4 5 6

Bags_Lost empirical model 0 0.225 0.21762106 1 0.325 0.33187211 2 0.250 0.25305249 3 0.125 0.12863501 4 0.050 0.04904210 5 0.025 0.01495784

The close agreement of the two right-most columns argues that we can use a Poisson model.

Question 4.2.13

If I table the empirical and Poisson model probabilities with E (X ) = 0. 3628319 I get 1

number_of_countries empirical model 0 0.72566372 0.6957034003

2 3 4 5

1 0.22123894 0.2524233576 2 0.03539823 0.0457936180 3 0.00000000 0.0055384612 4 0.01769912 0.0005023825

The two right-most columns are close in value indicating that a Poisson model might be adequate.

We could also put the groups with the smallest frequencies into a single larger group.

Question 4.2.14

Part (a): If I table the empirical and Poisson model probabilities with E (X ) = 2. 156934 I get 1 2 3 4 5 6 7 8 9 10

number_of_deaths empirical model 0 0.1478102190 0.1156792150 1 0.2436131387 0.2495124674 2 0.2472627737 0.2690910004 3 0.1687956204 0.1934705368 4 0.1012773723 0.1043258095 5 0.0556569343 0.0450047835 6 0.0246350365 0.0161787269 7 0.0072992701 0.0049852073 8 0.0027372263 0.0013440956 9 0.0009124088 0.0003221251

The two right-most columns are close (but perhaps not as close as we would like) in value indicating that a Poisson model could provide a good fit.

Question 4.2.15

If I table the empirical and Poisson model probabilities with E (X ) = 0. 81 I get 1 2 3 4 5 6 7 8

num_of_infestations empirical model 0 0.55 0.44 1 0.20 0.36 2 0.21 0.15 3 0.01 0.04 4 0.01 0.01 5 0.01 0.00 6 0.00 0.00 7 0.01 0.00

The two rightmost columns are not that close in value indicating that a Poisson model is perhaps not the best model. One would expect that if there is clustering of where the mites are located the independence assumption of the Poisson model would not hold.

Question 4.2.16

Here the parameter in the Poisson distribution is 1⁄ 5 or once every five hours. Using that we find [1] "the probability of less than three breakdowns= 0.783358" Question 4.2.17

Here the parameter in the Poisson distribution is 0. 15. Using that we find [1] "the probability of more than two errors= 0.826422" Question 4.2.18

Here we are told that the parameter in the Poisson distribution must satisfy P (X = 0) = e − λ = so λ = log(3). Using that we find [1] "the probability of two or more hits= 0.300463" Question 4.2.19

Here the parameter in the Poisson distribution is 1⁄ 10. Using that we find [1] "the probability of three or more flaws= 0.761897" Question 4.2.20

Here the parameter in the Poisson distribution can be defined in two different ways either a rate in terms of particles per minute or a rate in terms of particles per two minutes. With these we find [1] "Method 1: the probability of three particles in two minutes= 0.028035" [1] "Method 2: the probability of three particles in two minutes= 0.028035" Question 4.2.21

Here the parameter in the Poisson distribution is 0. 1 or the per day injury rate. Using that we find [1] "the probability of two accidents in five days= 0.075816" [1] "the probability of four accidents in two weeks (10 days)= 0.015328"

Notice that the square of the two accident probability is not the four accident probability as the four accident event is not simply two independent two accident events.

Question 4.2.22

From the statement given, we must have a constant λ such that

Solving for λ we get λ = 2. Using the Poisson density we then find that P (X = 4) = 0. 09022352.

Question 4.2.23

Let E be the event that X is even. For this we want to evaluate

The sum on the right we recognize as e − λ times the hyperbolic cosign of λ . Thus we have

Question 4.2.24

We expression we will use from Theorem 3.8.3. to prove this is

To use this we have p W (w ) = ∑ x =0 w

= e − ( λ + μ ) μ w ∑ x =0 w

=

∑ x =0 w

=

.

x

=

w

This is Poisson distribution with a parameter of λ + μ . In the above to evaluate the sum we have used the binomial theorem.

Question 4.2.25

We are told that the conditional density for X 2 is binomial

To compute P (X 2 = x ) recall that P (X 2 = x | X 1 = x 1 ) = 0 if x > x 1 thus we have P (X 2 = x ) = ∑ x = x ∞ P (X 2 = x | X 1 = x )P (X 1 = x 1 ) 1

= ∑ x1= x∞ = ∑ x1= x∞

p x (1 − p ) x 1 − x p x (1 − p ) x 1 − x

=

∑ x1= x∞

=

∑ x 1 =0 ∞

=

∑ x 1 =0 ∞

=

e λ (1 − p ) =

,

This is Poisson distribution with a parameter of λp .

Question 4.2.26

Part (a): I would think yes as the individual crashes are independent events and the number that happen in one interval will not depend on the number that happen in another interval.

Part (b): This would be given by the Poisson density with λ = 2. 5 at X = 2 which is 0.4561869.

Part (c): The time Y between events is an exponential random variable with the same parameter λ .

Here we want to evaluate P (Y < ) which is 0.4647386.

Question 4.2.27

If Y is the time till another death than Y is an exponential random variable with λ = 0. 1. We want to know P (Y > 7) which is 0.4965853.

Question 4.2.28

From Case Study 4.2.4 we have λ = 0. 027 (per month) and we want to compute

when we use the λ above.

Question 4.2.29

The probability that one light burns out in 75 hours is given by p = 0. 561765. The expected number to burn out is then np which in this case is about 28.

Question 4.3.1

Rather than use a table we will use the R function pnorm to evaluate these integrals. In the R code that goes with this question we find [1] 5.782723e-01 8.263912e-01 9.305634e-01 7.801460e-06 Question 4.3.2

Rather than use a table we will use the R function pnorm to evaluate these probabilities. In the R code that goes with this question we find [1] 4.807738e-01 1.951184e-01 8.554277e-01 9.903076e-03 2.013345e-06 Question 4.3.3

Part (a): By the symmetry of the standard normal probabilities density function f Z (z ) these two integrals are equal.

Part (b): The two integrals have an area of overlap i.e. the integral from a to a + is common to both so we need to determine which is larger

If in the first integral we let u = z − 1 (and then change the integration variable back to z ) we can write it as

Now when z > 0 we have that f Z (z ) is a decreasing function so as long as the lower limit of the above integrals i.e. a − > 0 then the second integral I 2 will be larger. By plotting these functions in the case where 0 < a < we see that the second integral I 2 is larger in this case also.

Question 4.3.4

Notice that these are scaled versions of integrals of the standard normal probability density function. Thus we can evaluate them using pnorm . We find [1] 0.9838824 2.3936537 Question 4.3.5

For this question we can use the R function qnorm to determine these values of z .

Part (a): For this we get z = − 0. 4399132.

Part (b): Write this as

which we can use in qnorm to find z = 0. 7600913.

Part (c): Write this as

which as this equals a known number we can write in terms of P (Z ≤ z ) alone. Doing this we get z = 0. 4098861.

Part (d): Using symmetric we can write this as

which as this equals a known number we can write in terms of P (Z ≤− z ) alone. Doing this we get − z = − 1. 281552 which gives z .

Part (e): We can write the left-hand-side as

which as this equals a known number we can write in terms of P (Z ≤ z ) alone. Doing this we get z = 0. 9495193.

Question 4.3.6

Using R we find this to be Q = 1. 34898.

Notes on the Continuity Correction

To apply this to discrete random variables we need to have the probability we want to evaluate in terms of limits with equalities that is if we have a statement like P (a < X < b ) we need to write it as P (a ′≤ X ≤ b ′ ). We can then add and subtract one-half to a ′ and b ′ to apply the continuity correction.

Question 4.3.7

The tax collected will be T = 50X where X is a binomial random variable. We can approximate P (T < 306000) using the normal approximation with the continuity correction. We have

In the last step we have used the continuity correction. The above can be approximated using the normal distribution where we find it to be 0.06520594.

Question 4.3.8

If we let X be the number of usable cabinets that come in the next shipment then X ∼ℬ in (1600, 0.

8). What we are asked to compute is P (1260 ≤ X ≤ 1310) which we will do with the continuity correction and the normal approximation. We find this probability estimated as 0.8716382.

Question 4.3.9

Let X be the number of voters that vote for the incumbent mayor so X ∼ℬ in (400, 0. 55).

Part (a): This is P (X = 200) = 0. 005341676.

Part (b): This is

Question 4.3.10

Let X be the number of free throws the team will make so X ∼ℬ in (100, 0. 7).

Part (a): This is P (75 ≤ X ≤ 80) which we could write as the sum of binomial densities.

Part (b): This approximately given by 0.1520813.

Question 4.3.11

Let X be the number of people who died in the three months following their birthdays so X ∼ℬ in (747, 3⁄ 12). Then we want to compute

when we evaluate this using the normal approximation.

Question 4.3.12

Let X be the number of correct guesses so X ∼ℬ (1500, 1⁄ 5). Then we want to compute

when we evaluate this using the normal approximation. While these results would reject the null hypothesis at 5% they would not reject it at 1%.

Question 4.3.13

From the statements in the book we would want n > and n > to apply the normal approximation. The larger of these two numbers is 21 and so n is “too small”. If we compute the exact probability and then use the normal approximation anyway we get [1] "exact= 0.8401; approx= 0.8418"

Thus the “bad” approximation is still relatively good.

Question 4.3.14

Let X be the number of hot dogs sold so X ∼ℬ (42200, 0. 38). Then we want to find the value of n such that

Using the normal approximation we would find that n > 16119. 92.

Question 4.3.15

We have P (X i = 1) = and n = 200 so that E (X ) = np = 100 and Var = npq = 50. The central limit theorem approximation is given by (here we don’t use the continuity correction)

P (− 5 ≤ X − E (X ) ≤ +5) = P (− 5 ≤ X − E (X ) ≤ +5) =P =P

= 0. 520500.

Question 4.3.16

Let X i be the number of dots showing on the i th toss of the die. Then let S = ∑ i =1 n X i with n = 100. We want to know

To compute this we need E (X i ) =

= 3. 5

E (X i 2 ) =

= 15. 16667

Var = 2. 916667.

Then we can use these to compute E (S ) = 100E (X i ) = 350 and Var Then the probability we want to compute can be computed

= 100Var

= 291. 6667.

using the central limit theorem.

Question 4.3.17

Let X i be the amount won on betting 5$ on the i th play of roulette. Then let S = ∑ i =1 n X i with n = 100. We want to know P (S > − 50). To compute this we need E (X i ) = 5

−5

E (X i 2 ) = 25 Var = 24. 93075.

Then we can use these to compute E (S ) = 100E (X i ) and Var probability we want to compute can be computed P (S ≥− 50) = 1 − P (S < − 50) =1−P using the central limit theorem.

Question 4.3.18

= 100Var

= 0. 6823719,

. Then the

The ratio in Theorem 4.3.2 would be

Question 4.3.19

Let X be the number of orders received during this week. Then we are told that X ∼𝒫 (50). We want to compute

Question 4.3.20

Lets X be the number of leukemia cases then X ∼𝒫 (3) and we want to calculate

Using the exact Poisson distribution and the central limit theorem we find [1] "exact prob= 0 central limit prob= 0.026341"

In either case it is very unlikely that we get this many leukemia cases “by chance”.

Question 4.3.21

Let X be the lifetime of a new set of tires. We want to see if P (X > 25000) = 0. 9. We compute P (X > 25000) = 0. 8413447.

Question 4.3.22

Let X be the IQ or a random student. We would like to determine what percent of the students have P (X < 80) or P (X > 135) which is equal to 1 − P (80 < X < 135). Using the R code we compute this to be 0.1200028. Of the 1400 that are expected this fraction of them will need special assistance.

Question 4.3.23

Let X be the amount of money collected in one day. We want to know P (X > 30000). In the R code I compute this to be 0.02275013.

Question 4.3.24

Let Y be the length of a pregnancy in days then we want to compute P (Y > 305) where 305 is ten months and five days. In the R code I compute this to be 0.007394607. This makes it unlikely that the letter writer is correct.

Question 4.3.25

Part (a): We will classify a non-delinquent as a delinquent if S > 75 when that person is a nondelinquent. Thus we want to compute

In the R code for this problem we compute this to be 0.066807.

The other mistake is computed in the same way. Thus we want to evaluate

Question 4.3.26

Let X be the cross section of a randomly selected plastic tube. Then we want to know 1 − P (12 < X < 13) since that will be the proportion of tubes that are not compliant. In the R code we estimate this probability to be 0.01241933. As we have 1000 tubes in the box we will get 12.41933065 or about 12 tubes that are non compliant.

Question 4.3.27

Let X be the verbal score of a student at State University. Then we want to compute P (X > 660) since that will be the proportion of students that scored better than Marina. In the R code we estimate this probability to be 0.1026373. As we have 4250 students we will get 4250P (X > 660) = 436. 2083203 or about 436 students that do better.

Question 4.3.28

We want the boundaries between the grades to be points that cut the standard normal at the need percentages. For example we would want the boundary between the A’s and B’s to be given by

Then the boundary in terms of the points on the test is

In the R code for this problem we find the boundary locations in z to be given by [1] -1.5547736 -0.9153651

0.1004337

0.8416212

These would translate into boundary locations in terms of the test score of [1] 51.34272 59.01562 71.20520 80.09945 Question 4.3.29

We would write

This can be written as

In R notation we have (17) so in the R code for this chapter we can so1.e for σ and find σ = 29. 65204.

Question 4.3.30

Let X be the weight of an eighteen-year old woman. Then we are told that P (103. 5 < X < 144. 5) = 0. 8. We can transform this into

If the two values 103.5 and 144.5 are equidistant from the average μ we must have

This means that the above probability statement is

Using Equation 17 this means that −

= − 1. 281552. We can solve for σ to get σ = 15. 99623.

Question 4.3.31

In this case we want to compute P (Y < 0. 08) when Y ∼𝒩 (0. 09, 0. 004 2 ). I compute this to be 0.006209665. The person with the 0.075 blood alcohol content should take the test twice and the person with the 0.09 blood alcohol content should take the test once. Taking two measurements and averaging will reduce the variance of the observation by making the true blood Algol content become apparent.

Question 4.3.32

We would “norm” each person and choose the larger of the two. The two normed scores for Michael and then Laura are [1] 1.710526 1.453704

Michael has the larger normed score and should be hired.

Question 4.3.33

Using the distribution of the mean when the standard deviation of the samples is known we compute P ( Y > 103) = 0. 2868877. The probability that P (Y > 103) where Y is an arbitrary measurement is 0.4256343. The probability that exactly three of the nine will exceed 103 is 0.232552.

Question 4.3.34

We can write the given expression in terms of n as

Using Equation 17 we can solve this for n where we find n ≥ 2653. 959.

Question 4.3.35

If C is the combined resistance then we want to compute P (C > 19). In the R code I compute this to be 0.02714591. In the second part we want to know the variance of the individual resistors such that P (C > 19) = 0. 005. If σ is the unknown precision of an individual resistor then we can write this as

In the above I can solve for σ where I find σ = 0. 2241415.

Question 4.3.36

If we let C and P be the diameters of the cylinder and piston respectively then we are told that C ∼𝒩 (41. 5, 0. 4 2 ) and P ∼𝒩 (40. 5, 0. 3 2 ) and we want to determine the probability that P > C or equivalently P − C > 0. As everything is independent the random variable P − C is normal with a known mean and variance and we can calculate this probability. In the R code for this question I find this probability given by 0.02275013.

Question 4.3.37

This is just repeated application of Theorem 4.3.3 to consider sums of more than just two random variables.

Question 4.3.38

From this section of the book we know that Y ∼𝒩 (2, 2 2 ⁄ 9) and Y ∗ ∼𝒩 (1, 1 2 ⁄ 9). We want to know the probability that Y ≥ Y ∗ or equivalently that Y − Y ∗ ≥ 0. As everything is independent this random variable is given by

In the R code for this question I find this probability given by 0.9101438.

Question 4.4.1

Let X be the number of years from now that pass before the first audit. Then X is a geometric random variable and we want to compute

Question 4.4.2

The variable X is a geometric random variable so p X (k ) = p (1 − p ) k − 1 for k ≥ 1. For this type of random variable we know E (X ) = = = 10.

Question 4.4.3

Given this data we will look and see if the empirical probabilities match the theoretical probabilities from the function p X (k ). We compute the empirical probabilities by counting the number of times a value of X = k happen (divided by the number of trials). For example

When we do this for all of the possible values of k and compare these numbers with the theoretical pmf we get k empirical model 1 1 0.08 0.19 2 2 0.26 0.14 3 3 0.20 0.11 4 4 0.14 0.08 5 5 0.10 0.06 6 6 0.08 0.04 7 7 0.06 0.03 8 8 0.06 0.03 9 9 0.02 0.02

Note that many of these numbers are not very close indicating that a geometric pdf is not a good match.

Question 4.4.4

If X is the random variable indicating the birth at which the first girl appears. Then according to the rules that they follow if X = 1 they stop and end with a family of size one. If X = 2 they stop and end with a family of size two. In general if X = x they will have a family of size x . Thus the expected family size is E (X ) = = 2.

Question 4.4.5

From the definition we find F X (t ) = P (X ≤ t ) = ∑ k =1 ⌊ t ⌋ p X (k ) = ∑ k =1 ⌊ t ⌋ p (1 − p ) k − 1 = p ∑ k =1 ⌊ t ⌋ (1 − p ) k − 1 = p ∑ k =0 ⌊ t ⌋− 1 (1 − p ) k =p

= 1 − (1 − p ) ⌊ t ⌋ .

Question 4.4.6

To evaluate the given probability note that as X is an integer we have

To use this result we need to know p the probability that we get the sum of four when we toss three die. The only way to get a sum of four from three die is to get two ones and one two. To get any specific three numbers say (2, 1, 1) will have a probability

As there are = = 3 ways to permute the two ones and the one two and have the sum stay the same the probability that we get a four is then 3 3 = . Using this we have the probability we want given by 0.058260.

Question 4.4.7

We can compute this as P (n ≤ Y ≤ n + 1) = ∫ n n +1 λe − λy dy = e − λn (1 − e − λ ), when we integrate and simplify.

Question 4.4.8

If we let X be the number of trials preceding the first success and Y be the trial at which this first success occurs (so that Y has a pmf given by the geometric distribution as given in the text) then X = Y − 1 and the pmf of X is given by

Then the moment generating function for X (by direct evaluation) is given by E (e tX ) = ∑ k =0 ∞ e tk p X (k ) = ∑ k =0 ∞ e tk p (1 − p ) k =p

.

To use Theorem 3.12.3 since X = Y − 1 we would have a = 1 and b = − 1 so

which simplifies to the same expression as we had before.

Question 4.4.9

The moment generating function for a geometric random variable is given by (18) Taking the first derivative we get

Evaluating this at t = 0 we get

Taking the second derivative we get

Evaluating this at t = 0 we get (19) Using these two we can compute the variance

Question 4.4.10

The probability distribution of the sum will have a moment generation function that is the product of the two moment generation function. Thus we have

This is not of the form

for any value of p . Thus the sum is not a geometric random variable.

Question 4.4.11

We want to compute E (t X ) when X is a geometric random variable. We find E (t X ) = ∑ k =1 ∞ t k p X (k ) = ∑ k =1 ∞ t k (1 − p ) k − 1 p =

∑ k =1 ∞ (t (1 − p )) k =

=

.

If we take the t derivative of this we find E (t X ) = =

+

.

If we evaluate this at t = 1 we get

Taking the second derivative of E (t X ) we get

If we evaluate this at t = 1 we get t =1

= E (X (X − 1)) =

+

+

=

+

.

To get E (X 2 ) from this expression we need to add E (X ) = to the above expression. When we do this get

when we simplify. Note that this is the same results we found in Equation 19 . As we have agreement between E (X ) and E (X 2 ) the variance will also be the same.

Question 4.5.1

If X is the random variable representing the number of houses that are entered and visited in getting his quota of five in-home visits then X is a negative binomial random variable with r = 5 and p = 0.

3 and we want to compute P (X < 8) = P (X ≤ 7). In the R code for this section we find this to be 0.0287955.

Question 4.5.2

If X is the number of missile attacks needed before the third hit. Then X is a negative binomial random variable and we want P (X = 7). In the R code for this section we find this to be 0.0972405.

Question 4.5.3

From the problem description, X is a negative binomial random variable with r = 2 and p = 0. 5.

Then in that case we can compare the empirical probability of getting X = x using the data in the question with the theoretical probability given by the negative binomial distribution. In the R code for this question we get k empirical model 1 2 0.24 0.25 2 3 0.26 0.25 3 4 0.19 0.19 4 5 0.13 0.12 5 6 0.08 0.08 6 7 0.05 0.05 7 8 0.03 0.03 8 9 0.01 0.02 9 10 0.01 0.01

Notice that these are so close to each other indicating that perhaps the student cheated in generating these numbers.

Question 4.5.4

From the problem statement X is a negative binomial random variable with r = 3 and p = 0. 15 and the expressions we want to compute are P (X ≥ 5) and E (X ). These are computed in the R code for this problem where we get [1] 0.9880187 20.0000000 Question 4.5.5

In the expression for the expectation E (X ) given in the question notice that we can write

Using this we have

If we decrement the index in the sum by one we get E (X ) = ∑ ∞ k = r +1 = ∑ k = r +1 ∞

p r +1 (1 − p ) k − 1 − r p r +1 (1 − p ) k − ( r +1) = .

In this derivation we have used the normalization condition of the negative binomial pmf with parameters r + 1 and p (i.e that summing the probability over all possible outcomes gives one).

Question 4.5.6

If Y is a negative binomial random variable as given in the text i.e. it represents the number of trials needed to obtain the r th success then X as defined in this question is X = Y − r . As Y has a pmf of

using y = x + r we have

Using the identity

=

we can write the coefficient above as

to get the result quoted in the book.

Question 4.5.7

From Question 4.5.6 we have that X = Y − r where Y is a negative binomial random variable as given in the book. Thus

and

Next using Theorem 3.12.3 in the book we conclude that

Question 4.5.8

From this pmf we see that X i is a negative binomial random variable with r = 3 and p = . Using the solution from Question 4.5.10 in this book we have that X is a negative binomial random variable with r = 3 + 3 + 3 = 9 and the same value of p . Thus we can compute P (10 ≤ X ≤ 12) = F X (12) − F X (9) = 0. 6603512. See the R code for this section.

Question 4.5.9

The moment generating function of a negative binomial random variable is given by

The derivative with respect to t of this is given by

While we could simplify this expression using algebra, as we only need to evaluate it at t = 0 to compute E (X ) we will simply evaluate it “as is”. We find

when we simplify.

Question 4.5.10

As the X i are independent the moment generating function for X is the product of the moment generating function of the X k . Thus M X (t ) = =

r1 r 1+ r 2+

r2 +rk

rk

.

If we define r s ≡ ∑ i =1 k r i (here s stands for sum) then we see that X has the moment generating function of a negative binomial random variable with parameters r s and p . This means that we can evaluate everything about X . Namely we have p X (k )

=

p r s (1 − p ) k − r s for k = r s ,r s + 1,r s + 2, …

E (X ) = Var

=

.

Question 4.6.1

From the question, the total length of time until the last gauge wears out Y is given by a gamma distribution with r = 3 and λ = .

Question 4.6.2

The average time between calls to a technician is (months) and if we assume that the waiting time between calls is an exponential distribution we will need to have its parameter given by λ = 3 (per month). Then if we let Y be the total time for twenty four of these calls to take place then Y is given by a gamma distribution with r = 24 and λ = 3. Thus

This distribution will have an average of

in months.

Question 4.6.3

As Y is a gamma random variable, using the given numerical values in this question and the formulas E (Y ) = and Var = we can compute r and λ for the underlying gamma distribution.

We can do this by dividing the expression for E (Y ) by that for Var . When I do this I find r = 3 and λ = 2. Then for an individual i we can compute p = P (1 < Y i < 2. 5) = 0. 5520244. Then the number of the Y i that would fall in this interval is a binomial random variable with n = 100 and p from above. This will have an expectation np = 55. 2024397.

Question 4.6.4

Let the random variable X be given by X = λY then the transformation of variables formula gives that

As y (x ) = and we know the functional form for p Y (y ) is a gamma pdf we have that

Notice that this is another gamma pdf but with parameters r and λ = 1.

Question 4.6.5

If Y has a gamma pdf the first derivative of f Y (y ) with respect to y is given by

If we set this equal to zero and solve for y we find y = . To show that this is a maximum we can either plot the gamma pdf and observe that this location is clearly a maximum of the graph or take the second derivative of f Y (y ), evaluate it at and show that it is negative.

Question 4.6.6

Using the hint we start with

In the above integral let v = z 2 ⁄ 2 so that

Using this substitution the expression above becomes

As Z is a standard normal random variable it has a mean of zero and a variance of one so that we know that the expression E (Z 2 ) = 1. Because of this we can solve for the integral above to find

Next note that Γ

= Γ

and using the definition of the gamma function we have

As this last integral is the same one we computed from E (Z 2 ) we can equate the two and we get

If we solve for Γ

we get the desired expression

Question 4.6.7

To do this we will use the fact that Γ(r ) = (r − 1)Γ(r − 1) (several times) so that Γ

=

Γ

=

=

Γ Γ

=

,

which is the result we wanted to show. We have also used the fact that Γ an earlier question).

=

Question 4.6.8

From the definition of the pdf for a gamma random variable we have E (Y m ) = ∫ 0 ∞ y m =

dy

∫ 0 ∞ y m + r − 1 e − λy dy .

To evaluate that integral we let v = λy so that dv = λdy . Using this we have E (Y m ) = =

∫ v =0 ∞

e−v

(m + r − 1)! =

=

∫ 0 ∞ v m + r − 1 e − v dv .

Here r is assumed to be an integer so that we can use Γ(r ) = (r − 1)!.

Question 4.6.9

As the moment generating function for a gamma random variable is

The derivative with respect to t of this is given by

Using this we have

The second derivative of M Y (t ) with respect to t is given by

(this is proved in

= (− r − 1) =

−r−2

−r−2

.

Using this we have

Thus

Question 4.6.10

Using the results from the previous problem recall that the second derivative of M Y (t ) was given by

Following the derivation of this result we see that the m th derivative of M Y is given by

Thus evaluating this expression at t = 0 we get = = =

,

the expression we were trying to show.

Chapter 5 (Estimation) Question 5.2.1

As this is the same pdf as in Example 5.1.1 in this section the maximum likelihood estimate of 𝜃 was ∑ i =1 n k i which in this case is

Question 5.2.2

Let D i be the random variable denoting the color of the chip on the i th draw (possible outcomes can be R for red and W for white). Then the likelihood of the draw we do observe is given by

If the proportion of reds is then this becomes

If the proportion of reds is then this becomes

As the first of these two expressions for L is larger we have that the maximum likelihood estimate of p is .

Question 5.2.3

Forming the likelihood for a sample of size n we have

Taking the logarithm of L we get

Taking the derivative of this with respect to λ and setting that equal to zero gives

Solving this for λ we find

Using the data given in this question we find λ e = 0. 1219512.

Question 5.2.4

Forming the likelihood for a sample of size n we have

Taking the logarithm of L we get

Taking the derivative of this with respect to 𝜃 and setting that equal to zero gives

Solving this for 𝜃 we find the maximum likelihood estimator for 𝜃 is

Question 5.2.5

Forming the likelihood for a sample of size n we have

Taking the logarithm of L we get

Taking the derivative of this with respect to 𝜃 and setting that equal to zero gives

Solving this for 𝜃 we find the maximum likelihood estimator for 𝜃 is

Using the numbers in the book we find 𝜃 e = 0. 7333333.

Question 5.2.6

Forming the likelihood for a sample of size n we have

Taking the logarithm of L we get

Taking the derivative of this with respect to 𝜃 and setting that equal to zero gives

Solving this for 𝜃 we find the maximum likelihood estimator for 𝜃 is

Using the numbers in the book we find 𝜃 e = 0. 4562949.

Question 5.2.7

Forming the likelihood for a sample of size n we have

Taking the logarithm of L we get

Taking the derivative of this with respect to 𝜃 and setting that equal to zero gives

Solving this for 𝜃 we find the maximum likelihood estimator for 𝜃 is

Using the numbers in the book we find 𝜃 e = 7. 996476.

Question 5.2.8

Forming the likelihood for a sample of size n we have

Taking the logarithm of L we get

Taking the derivative of this with respect to p and setting that equal to zero gives

Solving this for p we find the maximum likelihood estimator for p is

Using the numbers in the book we find p e = 0. 6582031. We can then use this probability to compute the expected frequencies where we find [1] 665.443359 227.446461

77.740490

26.571456

9.082041

3.104213

Question 5.2.9

Part (a): From Example 5.2.1 when the model is Poisson we have λ e = k = ∑ i =1 n k i . From the data given we have

Using this I find the expected frequencies given by [1]

8.0 16.0 16.0 10.6

8.4

Part (b): The agreement is reasonably good.

Question 5.2.10

Part (a): The maximum likelihood estimate of 𝜃 for a uniform random variable involves forming the the likelihood for a sample of size n we have

Notice that L (𝜃 ) gets larger as 𝜃 gets smaller. Thus to maximize L we would like to take 𝜃 as small as possible. The smallest it can be is the largest sample observed y max . Thus from the numbers in the book we find 𝜃 e = 14. 2.

Part (b): The maximum likelihood estimate of (𝜃 1 ,𝜃 2 ) for a uniform random variable involves forming the the likelihood for a sample of size n we have

Notice that L gets larger as 𝜃 1 and 𝜃 2 get closer together. Thus to maximize L we would like to take 𝜃 2 − 𝜃 1 as small as possible. The smallest 𝜃 2 can be is the largest observed sample or y max so 2 = y max . The largest 𝜃 1 can be is the smallest observed sample or y min so 1 = y min . From the numbers in the book we find 1 = 1. 8 and 2 = 14. 2.

Question 5.2.11

Forming the likelihood for a sample of size n we have (20) Taking the logarithm of L we get

Taking the derivative of this with respect to 𝜃 and setting that equal to zero gives

This would give 𝜃 = 0 as the solution which is not a possible value for 𝜃 given the samples observed. Looking at Equation 20 we see that L gets larger as 𝜃 → 1. From the definition of the pdf 𝜃 must be smaller than all of the samples y i . Thus the largest 𝜃 can be is the value y min and that is the maximum likelihood estimate. Using the numbers in the book we find 𝜃 e = 0. 21.

Question 5.2.12

Forming the likelihood for a sample of size n we have (21) Taking the logarithm of L we get

Taking the derivative of this with respect to 𝜃 and setting that equal to zero gives

which has no solution. Looking at Equation 21 we see that L gets larger as 𝜃 → 0. From the definition of the pdf 𝜃 must be larger than all of the samples y i . Thus the smallest 𝜃 can be is the value y max and that is the maximum likelihood estimate.

Question 5.2.13

Forming the likelihood for a sample of size n we have

Taking the logarithm of L we get

Taking the derivative of this with respect to 𝜃 and setting that equal to zero gives

Solving this for 𝜃 we find the maximum likelihood estimator for 𝜃 given by (22) If we have collected data on 25 individuals then n = 25.

Question 5.2.14

Part (a): Forming the likelihood for a sample of size n we have

Taking the logarithm of L we get

Taking the derivative of this with respect to 𝜃 and setting that equal to zero gives (23) Solving this for α we find the maximum likelihood estimator for α is

Part (b): If β is also unknown then we need to solve Equation 23 with the equation

In deriving the above we have used the derivative identity

Question 5.2.15

Using the results from Example 5.2.4 in this section where the book computed

This expression would need to be set equal to zero to get

and solve the resulting equation for σ 2 to get

Question 5.2.16

From the given pdf the expectation can be computed as E (Y ) = ∫ 0 𝜃 y =

∫ 0 𝜃 y 2 dy =

dy =

(𝜃 3 − 0) =

0

𝜃

.

Thus setting this equal to the empirical mean E (Y ) = ∑ i =1 n y i = y we can derive an expression for 𝜃 . Doing this we have

When we put in the given numbers we get 𝜃 e = 75. From the previous question the maximum likelihood estimate of 𝜃 is given by y max which for the sample given is 92.

Question 5.2.17

From the given pdf the expectation can be computed as E (Y ) = ∫ 0 1 yf Y (y ; 𝜃 )dy = ∫ 0 1 (𝜃 2 + 𝜃 )y 𝜃 (1 − y )dy = (𝜃 2 + 𝜃 ) = (𝜃 2 + 𝜃 ) = 𝜃 (𝜃 + 1)

=

.

The method of moments would then set this equal to the empirical mean y and solve for 𝜃 . When we do that we get

Question 5.2.18

For the Poisson distribution we can recall that E (Y ) = λ so

this is the same estimate we would use if λ were estimated using the method of maximum likelihood.

Question 5.2.19

From the given pdf the expectation can be computed as

Using the method of moments we would set this equal to the empirical mean y to get the equation = y . Solving for λ we find λ e = .

Question 5.2.20

In working this problem lets first check that the given pdf is normalized correctly. To that end we compute

as it should be. Next lets compute E (Y ). We find E (Y ) = ∫ 𝜃 1 − 𝜃 2 𝜃 1 + 𝜃 2 yf Y (y ; 𝜃 1 ,𝜃 2 )dy =

∫ 𝜃 1 − 𝜃 2 𝜃 1 + 𝜃 2 ydy =

𝜃 1− 𝜃 2

𝜃 1+ 𝜃 2

((𝜃 1 + 𝜃 2 ) 2 − (𝜃 1 − 𝜃 2 ) 2 ) =

=

(4𝜃 1 𝜃 2 ) = 𝜃 1 .

(24)

Next lets compute E (Y 2 ). We find E (Y 2 ) = ∫ 𝜃 1 − 𝜃 2 𝜃 1 + 𝜃 2 y 2 f Y (y ; 𝜃 1 ,𝜃 2 )dy = =

𝜃 1− 𝜃 2

𝜃 1+ 𝜃 2

=

= (3𝜃 1 2 + 𝜃 2 2 ).

(25)

To use the method of moments we will set Equation 24 equal to the empirical mean ∑ i =1 n y i and set Equation 25 equal to the empirical mean ∑ i =1 n y i 2 and solve for 𝜃 1 and 𝜃 2 . Doing this in Equation 24 is simple and we get

The method of moments formulation using Equation 25 for E (Y 2 ) gives

Using what we know about the estimate for 𝜃 1 e we can solve the above for 𝜃 2 and find

Question 5.2.21

For this given pdf we need to evaluate E (Y ) = ∫ k ∞ yf Y (y ; 𝜃 )dy = ∫ k ∞ 𝜃k 𝜃 y − 𝜃 dy = 𝜃k 𝜃 =

(0 − k − 𝜃 +1 ) =

k

∞

.

The method of moments would sett this equal to the empirical mean or y and we would then solve for 𝜃 . Doing this we find

This is different than the maximum likelihood estimate given by Equation 22 .

Question 5.2.22

From the given pdf we have

The method of moments estimate would set this equal to the empirical mean or y . This then gives an estimate of 𝜃 . Using the numbers in the book we get

Question 5.2.23

If we have n samples y i ∼ N (μ,σ 2 ) then as we know that E (Y ) = μ the first method of moments equation is to set μ equal to the empirical mean or μ = y . This is the same as the maximum likelihood estimate. Now recalling that

Using what we know for an estimate of μ we can solve the above for σ 2 to get that the second method of moments equation for σ is

The maximum likelihood estimate for σ 2 is given by σ 2 = ∑ i =1 n (y i − y ) 2 = =

= ∑ i =1 n y i 2 − y 2 ,

(26)

which is the same expression as above.

Question 5.2.24

Using Theorem 4.5.1 for a negative binomial pdf we have

and

The method of moments would set E (Y ) = ∑ i =1 n y i and E (Y 2 ) = ∑ i =1 n y i 2 and then solve these two equations for r and p . We thus need to solve

∑ i =1 n y i = ∑ i =1 n y i 2 =

+

,

for r and p . Doing this we get p = r =

.

Where for notational simplicity I have used E (Y ) ≡ ∑ i =1 n y i and E (Y 2 ) ≡ ∑ i =1 n y i 2 .

Question 5.2.25

From a geometric random variable we know that E (Y ) = . The method of moments estimate of p is then the inverse of the empirical mean. From the given data we have that p e = 0. 4789272. We can then compute the expected frequencies assuming a geometric random variable with this value of p . When we do that we get [1] 119.73

62.39

32.51

16.94

8.83

4.60

2.40

1.25

These numbers are not a great a match to the actually observed frequencies.

Question 5.2.26

For a two parameter pdf the method of moments equation from Definition 5.2.3 are the equations ∫ −∞ ∞ yf Y (y ; 𝜃 1 ,𝜃 2 )dy = ∑ i =1 n y i ∫ −∞ ∞ y 2 f Y (y ; 𝜃 1 ,𝜃 2 )dy = ∑ i =1 n y i 2 .

This first equation is the same as the equation E (Y ) = y . Next if we square the first equation and subtract it from the second equation we get

Note that the left-hand-side is the expression for Var while the right-hand-side can be shown to equal ∑ i =1 n (y i − y ) 2 (see the derivation of Equation 26 ). Thus we have started with the two equation from Definition 5.2.3 and ended with the equivalent set of equations E (Y ) = y and Var = 2.

Interval Estimation Question 5.3.1

Running the R code associated with this section of the text we get

[1] 103.7423 112.0577

for the 95% confidence interval.

Question 5.3.2

In the R code we compute [1] 0.7258476 0.8067840

for the confidence interval. Notice that this interval does include the normal value of 0.8 indicating that it is believable that the exposure has no effect on the FEV ratio.

Question 5.3.3

A 95% confidence interval of the mean (for the women tested) is given by [1] 64.43212 77.23455

As this does not include the value of eighty it is not believable that males and females metabolize methylmercury at the same rate (women seem to metabolize it faster).

Question 5.3.4

A 95% confidence interval of the mean is given by [1] 175.4595 201.3405

As the national average is contained in this interval this indicates that the diet does in fact not have any effect.

Question 5.3.5

The length of the 95% confidence interval for the mean is given by

To make this less than 3.06 we need to take n ≥ 335. 5718 as n must be an integer we would take n ≥ 336.

Question 5.3.6

The “confidence” of intervals like the type given would be the area under the standard normal between the two limits specified. Thus for the given intervals we get [1] 0.94 1.00 0.45

rounded to two digits.

Question 5.3.7

The given intervals are 68.69% confidence intervals for the true mean μ . This means that the probability that a given interval contains the mean is 0.6869. The probability that at least four intervals contain μ can be obtained from the binomial distribution where we find that value to be 0.5014409.

Question 5.3.8

That interval is centered symmetrically on the sample mean and can thus provide information about hypothesis testing i.e. could the value μ = μ 0 be the mean (i.e. is the value μ 0 in the computed interval).

Question 5.3.9

While notionally the interval given is correctly calculated, a histogram of the data indicate that it has quite a large skew to the right and thus is not well approximated by a normal distribution. If the underlying distribution is not normal there is no guarantee that the interval calculated using the techniques from this section will truly be a 95% confidence interval.

Question 5.3.10

Using the formula in the book I compute [1] 0.3151819 0.3959292 Question 5.3.11

If we let p be the unknown probability that a person will watch less than a quarter of the super bowl advertisements. Using the formula in the book I compute a 90% confidence interval [1] 0.2537464 0.2999482 Question 5.3.12

If we let p be the unknown probability that a person will prefer Budweiser beer. then Using the formula in the book I compute a 95% confidence interval on p of [1] 0.4423159 0.6376841

Budweiser would have stated that p > 0. 5 while Schlitz would argue that the fact that 0.5 is in the above interval indicates that there is no statistical consequence to the above results and that the p observed is just due to random fluctuations.

Question 5.3.13

Using the formula in the book I compute a confidence interval on p of [1] 0.610064 0.649936

Notice that the old measurement of 0.54 is not in this interval.

Question 5.3.14

The value of = will be the center of the interval or the value 0.6. The width of a confidence interval of level α for p is given by

For this level of α I compute z α⁄ 2 = 0. 6744898. Using the known interval width we compute n ≈ 121.

3164. As this is not an integer we would round n to 121.

Question 5.3.15

The largest the width of the α confidence interval on p can be is

If we make this less than 0.02 and take α = 0. 01 then we find n ≥ 16587. 24. As n must be an integer we round this to n ≥ 16588.

Question 5.3.16

Foley would first have to get 2174 votes to get “even” with Nethercutt. From the 14000 − 2174 = 11826 votes that remain he would have to win at least one half of these (or 5913) to win. That means that he had to win 2174 + 5913 = 8087 votes from the 14000. Let p be unknown then the number of votes F that Foley gets is a binomial random variable with parameters n = 14000 and p . We want to know the smallest value of p such that

The left-hand-side is the probability that Foley gets enough votes under the assumed binomial model. In the R code for this section I plot the left-hand-side as a function of p . I find that the left-hand-side becomes greater than 0.2 when p is larger than about 0.5741884.

Question 5.3.17

The first interval covers p in approximately 0.4971422 of the intervals of this type while the second interval covers p in approximately 0.5 of the intervals of this type. Thus they are both equally likely to contain p .

Question 5.3.18

We find

and

We have g ′ (p ) = 0 when p = and as the second derivative is negative this is a maximum of g . The largest value of g is then when p = and takes the value .

Question 5.3.19

In the R code for this problem we get [1] "d= 0.031021" [1] 0.5594859 0.6205141

for the 95% confidence interval.

Question 5.3.20

In the R code for this problem we get [1] "p_hat= 0.425743; d= 0.068953"

Here d is the margin of error. If we add the margin of error to the estimate we don’t get a number over 0.5 indicating that it is not believable that the true value of p is as larger as 0.5.

Question 5.3.21

If we look at the derivation of Definition 5.3.1 what we need to change when n is not negligible relative to N is to now use

Then the difference between the two endpoints of the confidence interval is the same but with the added fraction and d is now given by

Question 5.3.22

Part (a-b): In the R code for this problem we find [1] "ci Part (a): (0.317798, 0.402202)" [1] "ci Part (b): (0.320329, 0.399671)"

Notice that the confidence interval with the finite correction factor is smaller than the one that does not include it.

Question 5.3.23

The margin of error decrease like so to decrease it by two we need to increase n by a factor of four.

From the given original margin of error we find n = 267 so it would go to 1068.

Question 5.3.24

This conclusion is not true. An estimate in the difference between the probability that candidate A wins vs. that candidate B wins would be 0. 52 − 0. 48 = 0. 04. The given margin of error states that the true probability difference can be between

indicating that either of the two candidates can still be leading. A situation like this is discussed in more detail in the text to this section.

Question 5.3.25

We can use the formula given in the book of (27) to determine the needed sample size n in both cases. We find [1] 421.7885 478.8909

for the required two values of n when (α,d ) = (0. 04, 0. 05) and (α,d ) = (0. 08, 0. 04) respectively. Thus the second condition requires a larger value of n .

Question 5.3.26

By following the proof of the expression for the sample size n in terms of d and α for this problem we have d = 0. 05 and α = 0. 01 but as we are given an upper bound on p of 0. 4 we can bound

This allows us to find an upper bound on n which using the numbers given in this problem we find to be 636.9501 thus we would take n = 637.

Question 5.3.27

Using the formula for n in Equation 27 I find n ≥ 1026. 484. As n must be an integer we will take n = 1027.

Question 5.3.28

Part (a): Using Equation 27 I compute n ≥ 575. 6252. To make n an integer we take n = 576.

Part (b): If we know that p < 0. 1 then the largest that the product p (1 − p ) can be is 0. 1(0. 9) = 0. 09. We can use this to obtain a tighter bound on the needed sample size n . Following the procedure derived in the text we find n ≥ 207 which is a reduction in the sampling requirements by more than a half.

Properties of Estimators Question 5.4.1

Here our estimator is = (X 1 + X 2 ) where X i is the number drawn on the the i th draw. We want to evaluate P (| − 3| > 1. 0) which we can do by writing it as

P (| − 3| > 1. 0) = 1 − P (− 1 < − 3 < +1) = 1 − P (2 < < 4) = 1 − P (4 < X 1 + X 2 < 8).

Now to be valid the sum of the two terms X 1 + X 2 can be the numbers five, six, or seven. We find P (X 1 + X 2 = 5) = P (X 2 = 4| X 1 = 1)P (X 1 = 1) + P (X 2 = 3| X 1 = 2)P (X 1 = 2) + P (X 2 = 2| X 1 = 3)P (X 1 = 3) + P (X 2 = 1| X 1 = 4)P (X 1 = 4) =4

= .

In the same way we have P (X 1 + X 2 = 6) = ∑ k =1 2 P (X 2 = 6 − k | X 1 = k )P (X 1 = k ) + ∑ k =4 5 P (X 2 = 6 − k | X 1 = k )P (X 1 = k ) =2

+2

= .

Again we have P (X 1 + X 2 = 7) = ∑ k =2 5 P (X 2 = 7 − k | X 1 = k )P (X 1 = k ) = (5 − 2 + 1)

= .

This means that

Warning: This is a different result than in the back of the book. If anyone sees anything wrong with what I have done please contact me. In the R code for this problem I do a Monte-Carlo simulation which argues that the above is in fact correct.

Question 5.4.2

Part (a-b): For this part we want to calculate P (| − 3| < 0. 2) = P (− 0. 2 < − 3 < 0. 2) = P (2. 8 < < 3. 2) = P (2. 8 < Y max < 3. 2).

To use the pdf of the maximum of n uniform random variables given by Equation 32 recall that F Y (y ) = so that pdf of Y max is given by (28) This means that P (2. 8 < Y max < 3. 2) =

∫ 2 . 8 3 . 2 y n − 1 dy

=

(3. 2 n − 2. 8 n ).

If 𝜃 = 3 and n = 6 the above is equal to 0.8118677. If n = 3 this is 0.4005926.

Question 5.4.3

We want to evaluate P (X < 250) which we do by converting to a standard normal Z and get the probability of 0.1853552. This is done in the following R code (without the continuity correction)

p = 0.52 n = 500 m = n*p s = sqrt(n*p*(1-p)) pnorm( (250-m)/s ) Question 5.4.4

If transform the probability P (19. 0 < Y < 21. 0) as

When we evaluate this we get 0.3108435.

Question 5.4.5

Now as E [X i ] = λ for all i we would have E [ ] = E [ X ] = ∑ i =1 n E [X i ] = λ .

Question 5.4.6

If we take = cY min then recalling the pdf for the minimum given by Equation 31 when specialized to the uniform distribution gives (29) Using this we find that

Thus if we take c = n + 1 we will have an unbiased estimator for 𝜃 .

Question 5.4.7

For Y with the given density we find that the cdf of Y looks like

Using this with the pdf for the minimum given by Equation 31 when specialized to this distribution gives

Using this pdf we find that

when we integrate. Based on this we see that

and we have an unbiased estimator.

Question 5.4.8

Recall that Y 3 ′ is the third order statistic. From the book the pdf of the i th order statistics is given by (30) When Y is a uniform random variable f Y (y ) = and F Y (y ) = thus the above when i = 3 and n = 4 looks like

Thus using that we have

when we integrate. Thus an unbiased estimator of 𝜃 would be Y 3 ′ . For the data given here this estimate would be

This estimate could give a value that we know to be wrong if for example it provides a value for that is smaller than the largest element in the set for example if Y 4 ′ > Y 3 ′ . For example if our sample was 14, 10, 18, 45 then Y 4 ′ = 45 > = 30. From the sample itself we know that 𝜃 ≥ 45 so the estimated = 30 cannot be correct.

Question 5.4.9

Note that for both i values

Thus

If we take c = the expected value above equals and our estimator is unbiased.

Question 5.4.10

Following the hint lets compute

Thus Y 2 is not an unbiased estimator of 𝜃 2 but = 3Y 2 is.

Question 5.4.11

We are told that E [W ] = 𝜃 and want to know if E [W 2 ] = 𝜃 2 . Using

we see that E [W 2 ] will equal 𝜃 2 only if Var

= 0 or W is a constant.

Question 5.4.12

The maximum likelihood estimate of σ 2 is given by

We can take the expectation of this as E[

2

] = ∑ i =1 n E [Y i 2 − 2μY i + μ 2 ] = ∑ i =1 n = ∑ i =1 n σ 2 = σ 2 ,

showing that it is an unbiased estimator.

Question 5.4.13

Using Equation 32 with the uniform distribution gives the pdf given by Equation 28 . This question asks is what the median of the function Y max is. Lets first find what the median of Y max is. We need to find the ∗ value y such that

To evaluate this using the pdf in Equation 28 we have

If we set this equal to and solve for y ∗ we get

Thus the median of

Y max is

which is not equal to 𝜃 unless n = 1. Thus this estimator is biased (unless n = 1).

Question 5.4.14

For = nY min we need to recall that when we have n samples Y i such that Y i ∼ f Y (y ) (and a distribution function F Y (y )) then the pdf for the minimum from the sample of n or Y min is given by (31) In the same way the pdf of the maximum of n variables is given by (32) Now when Y is an exponential random variable we have

Using this we find (33) and (34) Note that the pdf of Y min is another exponential distribution with a parameter and thus it has an expectation given by . Using this we have that

Thus is unbiased.

For the other note that as E [Y i ] = 𝜃 for every i we have that

Question 5.4.15

To evaluate E ( W 2 ) note that

Because we have Var

=

, we get

Note that lim n →∞ E ( W 2 ) = μ 2 so W 2 is an asymptotically unbiased estimate for μ 2 .

Question 5.4.16

The maximum likelihood estimate of σ 2 is that

2

= ∑ i =1 n (Y i − Y ) 2 . In this section of the text it was shown

As n →∞ the right-hand-side this limits to σ 2 so our maximum likelihood estimator is an asymptotically unbiased estimate of σ 2 .

Question 5.4.17

Part (a): For each of these two estimators we compute E[

1

] = E [X 1 ] = p (1) + (1 − p )(0) = p

E[

2

] = E [X ] = ∑ i =1 n E [X i ] = ∑ i =1 n p = p ,

showing that each estimator is unbiased.

Part (b): To compute the variances we have

Now

Thus

Next we have

As X ∼ℬ (n,p ) we have that Var

which is the value of Var

= np (1 − p ) so that

.

Question 5.4.18

Taking the variance of each estimator we have Var

=

Var

Var

= 36Var

.

As f Y is symmetric we would expect that the variance of Y min and Y max would be equal. This is proven using arguments around Equation 37 below. This would indicate that 1 has the smaller variance.

Question 5.4.19

Part (a): For

1

we have

Now to evaluate the above let v = so that dv =

and the above becomes

E[ 1] =𝜃 For

2

we have

Finally, for

3

=𝜃

= 𝜃 (0 + 1) = 𝜃 .

in Question 5.4.14 on Page § we show that

3

is unbiased.

Part (b): To derive the results in this section we will use the fact that (35) Then for

1

we have

Now we need to compute

Using this we have that

Next for

2

Finally, for

we have

3

we will need to compute

We now use Equation 33 to find

Using that we find (36) Warning: This is different than the result in the back of the book. If anyone sees anything incorrect with what I’ve done pleas contact me.

Part (c): These are simply ratios of the variances computed above.

Question 5.4.20

For these two estimators we have

Thus the relative efficiency of

1

Var

= Var

=λ

Var

= Var

=

to

2

= .

is given by

Question 5.4.21

Using the pdfs for a minimum and a maximum of n random variables given by Equations 31 and 32 , when f Y (y ) is a uniform distribution, we get Equations 29 and 28 respectively. Using these we can compute E [Y min 2 ] = ∫ 0 𝜃 y 2

dy =

E [Y max 2 ] = E [Y min ] = E [Y max ] =

.

Using these we can compute the variances of Y min and Y max and find (37) With these we find expressions for the variances given by (38) and

(39) Using these expressions I get that

as we were asked to show.

Question 5.4.22

Expanding we get

as both random variables are independent. The above equals

If we call this expression F (c ) then F (c ) will be smallest when F ′ (c ) = 0. Taking that derivative, setting the result equal to zero, and solving for c we get

so that

and our estimator for

is

Question 5.5.1

For this pdf we will evaluate the Cramer-Rao lower bound (CRLB). We find ln(f Y (Y ; 𝜃 )) = − ln(𝜃 ) − ln(f Y (Y ; 𝜃 ) = − + ln(f Y (Y ; 𝜃 ) = Taking the expectation of the above gives

Now for an exponential random variable we have

−

.

Thus

From this the CRLB is given by

As we know that Var is also equal to the above expression we have that is an efficient estimator i.e. can be said to be the best.

Question 5.5.2

For this pdf we have ln(p X (X ; λ )) = − λ + X ln(λ ) − ln(X !) ln(p X (X ; λ )) = − 1 + ln(p X (X ; λ )) = −

.

From these we see that

Thus the CRLB is given by . Now

Thus is an efficient estimator for λ i.e. its the variance equals the CRLB. Question 5.5.3

Using the normal distribution (40) we will assume that σ is known and compute the CRLB. To do that we need to evaluate ln(f Y (y ; μ )) = − ln(2π ) − ln(σ ) − ln(f Y (y ; μ )) = ln(f Y (y ; μ )) = −

.

Using this we see that the CRLB is given by estimator.

. Note that this is the variance of Y so Y is an efficient

Question 5.5.4

From the given pdf we find ln(f Y (y ; 𝜃 )) = − ln(𝜃 ) ln(f Y (y ; 𝜃 )) = − .

This is a constant so that the expectation of this squared is

and we find the CRLB to be

Now Equation 38 give the variance of where we find

Note that this expression is less than the CRLB computed above. This is possible and not a violation of the statement given in the book because for this pdf the domain of f Y where f Y ≠ 0 depends on 𝜃 . Another example like this is discussed in this section of the text.

Question 5.5.5

To start we will compute the CRLB for the given pdf. We find ln(f X (x ; 𝜃 )) = (x − 1) ln(𝜃 − 1) − x ln(𝜃 ) ln(f X (x ; 𝜃 )) =

−

ln(f X (x ; 𝜃 )) = −

+

.

As we are told that E [X ] = 𝜃 , the expectation of the last expression above is given by

From this we see that the CRLB is As the variance of X is given by

. =

the statistic X is efficient.

Question 5.5.6

Part (a): To start, note that f Y is the pdf of a gamma random variable with parameters r and 𝜃 . Thus for any of the samples Y i we have E [Y i ] = r𝜃 Var Because of that we find

= r𝜃 2 .

Part (b): For this we will calculate the CRLB for f Y (y ; 𝜃 ). To do that we compute ln(f Y (y ; 𝜃 )) = (r − 1) ln(y ) − − ln((r − 1)!) − r ln(𝜃 ) ln(f Y (y ; 𝜃 )) = ln(f Y (y ; 𝜃 )) = −

− +

.

From this we see that

Thus the CRLB is

Now the variance of is given by

which is the same as the CRLB. Thus is a minimum variance estimator for 𝜃 .

Question 5.5.7

To work this problem we start by taking the 𝜃 derivative of the expression

to get (41) Using the fact that (42) We can write Equation 41 as

Taking another derivative of this with respect to 𝜃 we get

Using Equation 42 on the second term of the above we get

In terms of the expectation operator this relationship can be written (43) If we multiply by n and take the inverse of both sides we have shown the equivalence of the Cramer-Rao inequality.

Notes on Section 5.6

For many of the problems below we will show that an estimator h (k 1 ,k 2 , … ,k n ) is sufficient by writing the likelihood in the “product” form (44) Here h [k 1 ,k 2 , … ,k n ; 𝜃 ] is the function that maps the observed data k 1 ,k 2 , … ,k n to a summary statistic that is normally related to an estimate we might use to extract information about the unknown parameter value 𝜃 . This is basically Theorem 5.6.1 discussed in the book.

Question 5.6.1

Here h (k 1 ,k 2 , … ,k n ) = ∑ i =1 n k i and we will write the likelihood as

Note that we can write this in the form required by Equation 44 if we take g

n

= p n (1 − p ) ∑ i =1 k i

b (k 1 ,k 2 , … ,k n ) = 1, and

is sufficient for p .

Question 5.6.2

This estimator is not sufficient because we can find two samples of the same size (here n = 3) that have the same value for but do not give the same information about p . Consider the two samples (x 1 ,x 2 ,x 3 ) = (1, 1, 0) (x 1 ,x 2 ,x 3 ) = (0, 0, 1).

Then = 3 in both cases. Given that in the first sample we had two heads while in the second sample we only had one head, we expect p to be larger in the first sample than in the second. Thus not being different in this case indicates that is not sufficient.

Question 5.6.3

If we know that is significant for 𝜃 then we have a representation of the likelihood given by Equation 44 .

Then using an invertable transformation ℱ (⋅ ) on as ′ = ℱ ( ) (or using the inverse = ℱ − 1 ( ′ )) we can write our likelihood L (𝜃 ) as

where we have defined the function to be the above composition. Thus we see that we have a representation for the likelihood in terms of the variable ′ as in Equation 44 and it is also sufficient for 𝜃 .

Question 5.6.4

We will write the likelihood of σ 2 as

This is written in the form of Equation 44 if we take h (Y 1 ,Y 2 , … ,Y n ) = ∑ i =1 n Y i 2 g [h ; σ 2 ] = b = Thus

2

exp .

is sufficient for σ 2 .

Question 5.6.5

Following the results from Question 5.5.6 where we had = Y = ∑ i =1 n y i we get the idea that we should try to see if ∑ i =1 n Y i is a sufficient statistic. Towards that end we will write the likelihood as

This can be written as

which is of the form needed for Equation 44 and ∑ i =1 n Y i is a sufficient statistic for 𝜃 .

Question 5.6.6

To start this question lets find the maximum likelihood estimate for 𝜃 . Forming the likelihood we have (45)

Taking the natural logarithm of both sides we get

Taking the 𝜃 derivative of this and setting it equal to zero gives

Solving for 𝜃 we get

Note that 𝜃 is a function of W = ∏ i =1 n Y i .

Now from Equation 45 we see that L (𝜃 ) can be written in the form of Equation 44 if we take h (Y 1 ,Y 2 , … ,Y n ) = W = ∏ i =1 n Y i g [h ; 𝜃 ] = 𝜃 n Thus W is a sufficient statistic for 𝜃 .

𝜃−1

b = 1.

Question 5.6.7

Part (a): Forming the likelihood for 𝜃 we have (46) The product of the indicator functions can be simplified as

and thus we can write the likelihood L (𝜃 ) in the form

This expression is of the form needed for Equation 44 and Y min is sufficient for 𝜃 .

Part (b): That Y max is not sufficient for 𝜃 can be shown by considering two sets of data that have the same value for Y max but that contain different “information” about 𝜃 . Two such sets could be

In the first set we know that 𝜃 ≤ 1 while in the second we only know 𝜃 ≤ 2. Thus Y max (while taking the same value in both sets) does not convey the same information on 𝜃 .

Question 5.6.8

For this pdf, using the same trick as in the previous question, we write the likelihood as

We can write this in the form Equation 44 if we take h (Y 1 ,Y 2 , … ,Y n ) = Y max I [y max ≤ 𝜃 ]

g [h ; 𝜃 ] = b = 1.

Thus Y max is a sufficient statistic for 𝜃 .

Question 5.6.9

The likelihood of a pdf in exponential form would be given by L (𝜃 ) = ∏ i =1 n g (w i ; 𝜃 ) = ∏ i =1 n e K ( w i ) p ( 𝜃 )+ S ( w i )+ q ( 𝜃 ) n

n

= e nq ( 𝜃 ) e ∑ i =1 S ( w i ) e ∑ i =1 K ( w i ) p ( 𝜃 ) .

If we write this as

which is of the form needed for Equation 44 so that = ∑ i =1 n K (w i ) is sufficient for 𝜃 .

Question 5.6.10

To write the given pdf in exponential form we have

This is in exponential form with p (λ ) = − λ K (y ) = y , n and thus a sufficient statistic is ∑ i =1 K (y i ) = ∑ i =1 n y i .

Question 5.6.11

We can write f Y (y ; 𝜃 ) in exponential from as f Y (y ; 𝜃 ) = 𝜃 (1 + y ) − 𝜃 − 1 = e ln ( 𝜃 ) e ln ((1+ y ) − 𝜃 − 1 )

= e ln ( 𝜃 ) e − ( 𝜃 +1) ln (1+ y ) = e − ln (1+ y )( 𝜃 +1)+ ln ( 𝜃 ) .

Now this is in exponential form with p (𝜃 ) = − (𝜃 + 1) K (y ) = ln(1 + y ) q (𝜃 ) = ln(𝜃 ).

Because of this a sufficient statistic is ∑ i =1 n K (y i ) = ∑ i =1 n ln(1 + y i ).

Consistency Question 5.7.1

Because we know that the sample mean is distributed as Y n ∼𝒩 (μ, requirement as

) we can write the given probability

which we can manipulate as

or

or

Here Φ(⋅ ) is the cdf of the standard normal. We can take the “inverse” of both sides. In R notation this would be

Solving for n gives n ≈ 16. 90965 so to make n an integer we would take n = 17.

Question 5.7.2

Note that this estimator is unbiased as

as our normal random variable has μ = 0. Now by independence of the sample Y i we have

Thus to evaluate this we need to evaluate Var . As Y i ∼𝒩 (0,σ 2 ) let us write Y i = σX where X ∼𝒩 (0, 1). Thus Y i 2 = σ 2 X 2 and X 2 is a chi-squared random variable with one degree of freedom. Thus

where we have used the fact that Var

= 2(1) = 2. Using this we have that

Then from Chebyshev’s inequality with W = S n 2 we have

If we take the limit n →∞ we see that P (| S n 2 − σ 2 |≥ 𝜖 ) = 0 so P (| S n 2 − σ 2 | < 𝜖 ) = 1 thus S n 2 is consistent estimate.

Question 5.7.3

Part (a): For

n

= Y 1 consider P (| Y 1 − λ | < 𝜖 ) = P (λ − 𝜖 < Y 1 < λ + 𝜖 ) = ∫ λ − 𝜖 λ + 𝜖 λe − λy dy 2

= e − λ (e λ𝜖 − e − λ𝜖 ), when we integrate. Note that this is a fixed constant (a number) independent of n and cannot be made equal to zero as would be required to have a consistent estimate.

Part (b): For this expression for

n

we again consider

When we add n exponential random variables with rate λ we get another exponential random variable with rate so we know the pdf of n . Specifically p n (y ) = nλe − nλy and the above probability can be computed as

If we take n →∞ we have that both e − nλ ( λ − 𝜖 ) → 0 and e − nλ ( λ + 𝜖 ) → 0. Thus we have shown that

This is a contradiction to the statement that

n

is consistent as the above would need to go to one.

Question 5.7.4

Part (a): The statement of an estimator

n

being squared-error consistent is that (47)

Now fix n and let b = E [ n ] and consider E [( n − 𝜃 ) 2 ] = E [( n − b + b − 𝜃 ) 2 ] = E [( n − b ) 2 + 2( = Var

n

− b )(b − 𝜃 ) + (b − 𝜃 ) 2 ]

+ 2E [( n − b )](b − 𝜃 ) + (b − 𝜃 ) 2 .

Now E [ n − b ] = 0 by the definition of b so the above is

As both terms on the right-hand-side are nonnegative if the left-hand-side goes to zero as n →∞ both terms on the right-hand-side must also. As (b − 𝜃 ) 2 → 0 we have that b → 𝜃 that is E [ n ] → 𝜃 as n →∞ and our estimator is asymptotically unbiased.

Question 5.7.5

For this question we need to consider the limit of

as n →∞ . Using Equation 34 we know the pdf of Y max . Using that to evaluate the right-hand-side of the above we need to compute

Note that this expression tends to zero as n →∞ showing that Y max is a squared-error consistent estimate for 𝜃.

Question 5.7.6

Note that for a symmetric distribution the median of the 2n + 1 order statistics is unbiased. Thus we have

Using Chebyshev inequality for W = Y n +1 ′ and the variance given in the question we have

This goes to zero as n →∞ thus Y n +1 ′ is consistent for μ .

Bayesian Estimation Question 5.8.1

We are told that p X (k | 𝜃 ) = (1 − 𝜃 ) k − 1 𝜃 and

𝜃 r − 1 (1 − 𝜃 ) s − 1 ,

p (𝜃 ) =

Then using Bayes rule we have p (𝜃 | X = k ) ∝ p X (k | 𝜃 )p (𝜃 )

∝ (1 − 𝜃 ) k − 1 𝜃 (𝜃 r − 1 (1 − 𝜃 ) s − 1 ) = 𝜃 r (1 − 𝜃 ) k + s − 2 .

This is a function of 𝜃 that is in the form a beta distribution with parameters r + 1 and k + s − 1.

Question 5.8.2

If the loss is a quadratic loss then the estimator for 𝜃 we should use the mean of the posterior distribution. In example 5.8.2 in the book we compute the posterior distribution g Θ (𝜃 | X = k ) and find it to be a beta distribution with parameters k + r and n − k + s . Using the formula for the mean of a beta distribution, to minimize the quadratic loss the estimate of 𝜃 would be

We want to write this in terms of the prior expectation of 𝜃 (which was likelihood estimate of 𝜃 (which is ). We can do that by writing = =

+ +

Question 5.8.3

Part (a): From the discussion in the book when

with a prior on 𝜃 given by a beta distribution or

.

) and the maximum

We find for the posterior distribution of 𝜃

If r = s = 135 then r + s = 270 so the posterior distribution is explicitly given by

Part (b): Under squared-error loss the Bayes estimate for 𝜃 is given by the mean of g Θ (𝜃 | X = k ) which is a beta distribution with parameters k + 135 and n − k + 135. This means that our estimator is given by

Note that we can write this as =

+

=

+

=

+

.

This is a weighted average of the maximum likelihood estimate and the mean of the prior pdf .

Question 5.8.4

As the uniform prior can be considered a beta pdf with r = s = 1 by following Example 5.8.2 the posterior distribution of 𝜃 will be a beta pdf with parameters k + r = k + 1 and n − k + s = n − k + 1. This means that our Bayes estimate for 𝜃 is given by

Question 5.8.5

The Bayes estimate is which is a linear combination of the unbiased maximum likelihood estimate and an “offset” thus the estimator is biased. As the offset gets smaller as n →∞ it is asymptotically unbiased.

Question 5.8.6

We are told that p Y (y | 𝜃 ) =

y r − 1 e − 𝜃y

p (𝜃 ) =

𝜃 s − 1 e − μ𝜃 ,

so the posterior distribution of 𝜃 is p Θ (𝜃 | Y = y ) ∝ p (Y = y | 𝜃 )p (𝜃 ) =

𝜃r+s−1e −(μ+y)𝜃.

From the form of this expression when viewed as a function of 𝜃 this is a gamma pdf with =

parameters r + s and μ + y .

Question 5.8.7

When we have n samples that we are adding together where Y i ∼ Gamma(r,𝜃 ) we have that

Using the result from Question 5.8.6 above we find that 𝜃 will have a gamma posterior distribution with parameters nr + s and μ + w = μ + ∑ i =1 n y i . This is because in the previous problem we would make the replacements r → nr y →w.

Question 5.8.8

This would be the mean of the above gamma distribution which taking w = ∑ i =1 n y i is = =

+ +

.

Now is the mean of the prior distribution on 𝜃 and is the maximum likelihood estimate of 𝜃 from one sample W = w where W ∼ Gamma(nr,𝜃 ).

Question 5.8.9

In the setting for Example 5.8.2 the marginal pdf of X would be given by integrating

with respect to 𝜃 . To do that we will recall that

Assuming that every integer expression is positive (so that gamma functions are really factorial functions) for p X (k ) we find p X (k ) = =

=

,

which was the desired expression.

Chapter 6 (Hypothesis Testing) Question 6.2.1

Part (a): We would compute the z score of the observed data z = and compare it to the value of − z α for the given value of α . For the numbers given here I compute [1] "z= -1.611111; -z_alpha= -1.405072; p_value= 0.053578"

As the value of z is less than the value of − z α we can reject the null hypothesis.

Part (b): We would compute the z score of the observed data z = and see if it satisfies − z α⁄ 2 < z < +z α⁄ 2 for the given value of α . For the numbers given here I compute [1] "z= 2.750000; z_alpha/2= 2.575829; p_value= 0.005960"

As the value of z is larger than the value of z α⁄ 2 we can reject the null hypothesis.

Part (c): We would compute the z score of the observed data z = and compare it to the value of z α for the given value of α . For the numbers given here I compute [1] "z= 1.170732; z_alpha= 1.126391; p_value= 0.120853"

As the value of z is larger than the value of z α we can reject the null hypothesis.

Question 6.2.2

For acceptance we would require P (| Z |≤ z α⁄ 2 ) = 1 − α = 0. 94. We can manipulate the inner inequality to be

Computing each of these two values we will accept H 0 if y is between [1] 88.9852 101.0148

and reject otherwise.

Question 6.2.3

Part (a): No. The result could be significant at the α = 0. 05 level but not the α = 0. 01 level.

Part (b): Yes. The α = 0. 01 rejection is a more stringent condition.

Question 6.2.4

We would test H 0 : μ = 32500 vs. H a : μ > 32500. In the R code for this question we find [1] "z= 1.258720; z_alpha= 1.644854; p_value= 0.104066"

Thus we cannot reject H 0 in favor of H a .

Question 6.2.5

No. One way to see this is to note that the P-value in the case where we reject in favor of H 1 : μ > μ 0 must satisfy

Here z is the observed z value which we assume is positive. This might happen if say P (Z ≥ z ) = + 𝜖 . The P-value of the test where the alternative is H 1 : μ ≠ μ 0 is given by

In the hypothetical case above we would have P (Z ≥ z ) > α and we would not reject the null hypothesis.

Question 6.2.6

The critical region is the region where we reject the null hypothesis which for this question is an interval very close to (surrounding) the value μ 0 = 30. The level of significance is the probability that the test statistic is in the critical region when H 0 is true. When H 0 is true μ = μ 0 = 30 and we expect some samples to fall in the critical region above (especially as n increases). This is a reason that this interval is a poor choice. For the numbers given here we get the level of significance for the given interval and [1] "level of significance (alpha)= 0.053153" [1] "reject H_0 if y < 27.099485 or y > 32.900515" Question 6.2.7

Part (a): Using the R code for this question I find [1] "z= 2.145247; z_alpha/2= 1.959964; p_value= 0.031933"

As the observed z value is larger than z α⁄ 2 we can reject the null hypothesis and argue that the machine should be readjusted.

Part (b): We need to have the samples come from a normal distribution.

Question 6.2.8

Using the R code for this question this is reported in the R output above. The fact that the Pvalue is smaller than the requested levels of significance α supports the conclusions drawn above.

Question 6.2.9

Using the R code for this question I find [1] "z= 0.920000; p_value= 0.357573"

We would reject H 0 if α was specified to be any number larger than the P-value above.

Question 6.2.10

The test she should conduct is H 0 : μ = 120 vs. H a : μ > 120. Using the R code for this question I find [1] "z= 3.064129; p_value= 0.001092"

We would reject H 0 for most levels of α since the P-value in the above is so small.

Question 6.2.11

The test she should conduct is H 0 : μ = 145. 75 vs. H a : μ ≠ 145. 75. Using the R code for this question I find [1] "z= 2.105263; z_alpha/2= 1.959964; p_value= 0.035268"

As the value of z is larger than z α⁄ 2 we would reject H 0 and conclude that there is a statistically significant difference.

Question 6.3.1

For this question we can use a large sample test. In the R code for this question we find [1] "z= 0.905822; z_alpha= 1.644854; P-value= 0.182515"

Because the value of z is not larger than z α we cannot reject the null hypothesis and must conclude that the production of predator sounds is not an effective technique. The P-value above indicates that we would reject the null hypothesis for any value of α larger than about 0.18.

Question 6.3.2

For this question we can use a large sample test. In the R code for this question we find [1] "z= -1.959152; z_alpha/2= 1.959964; P-value= 0.050095"

Because the value of z is not larger than z α⁄ 2 in magnitude (barley) we cannot strictly reject the null hypothesis. The P-value indicates that we could reject the null hypothesis at α = 0. 06 or α = 0. 1 however.

Question 6.3.3

For this question we can use a large sample test. As the empirical proportion of supporters is now = 0. 6 and has decreased relative to the previous estimate of 0.65 the hypothesis test we will perform is H 0 : p = 0. 65 vs. H a : p < 0. 65. In the R code for this question we find, [1] "z= -1.148339; -z_alpha= -1.644854; P-value= 0.125414"

Because the value of z is not less than − z α we cannot reject the null hypothesis.

Question 6.3.4

In the R code for this question we assume a number of successes k and then compute the Pvalue for this one sided test. We then look and see what is the first value of k where that Pvalue is less than 0.14. We find that value to be k = 98.

Question 6.3.5

As the median of the exponential distribution given in Example 5.3.2 is = − ln(0. 5). If we let p be the probability that a given sample will be less than then the hypothesis test we will consider is H 0 : p = 1⁄ 2 vs. H a : p ≠ 1⁄ 2. If I let x be the number of samples less than the theoretical median above then in the R code associated with this question I’m getting [1] "x= 26; n= 60; p0= 0.500000; p_hat= 0.433333" [1] "z= -1.032796; P-value= 0.301700"

As the P-value is so large we have no reason to reject the null hypothesis.

Question 6.3.6

The null hypothesis would state that the number of people dying in the month before their birth month would be = 0. 08333333. The proportion observed in this data is 0.04597701.

Because of this I’ll perform the following hypothesis test H 0 : p = 1⁄ 12 vs. H a : p < 1⁄ 12. In the R code for this question I find [1] "x= 16; n= 348; p0= 0.083333; p_hat= 0.045977" [1] "z= -2.521382; P-value= 0.011689"

With a P-value this small we can reject the null hypothesis at the 5% level easily.

Question 6.3.7

If we suspect that p > 0. 5 then we would anticipate getting more than np = 7 × (1⁄ 2) = 3. 5 events in our sample. Thus we can consider k ∗ taking values from four to n . For each of these values of k ∗ the α probability would be that we got a sample this large (or larger) from a process with p = 1⁄ 2. We can evaluate these to find [1] 0.5000000 0.2265625 0.0625000 0.0078125

to be the α values for k ∗ given by 4, 5, 6, 7.

Question 6.3.8

Now from the given table in the book if we set our threshold at k ∗ = 8 then we find that

while if we set our threshold at k ∗ = 9 we have α = 0. 010078. Neither of these is equal to 0.05 exactly. To make a randomized decision rule that has the probability of exactly α = 0. 05 we can let p 8 be the probability we will decide to reject or accept with the threshold k ∗ = 8 (as in the previous question) so that 1 − p 8 is the probability we will decide to reject or accept with k ∗ = 9. Then to decide on the value for p 8 note that we should impose P (Reject H 0 | H 0 is true) = P (X ∈{ 8, 9}| H 0 is true)p 8 + P (X ∈{ 9}| H 0 is true)(1 − p 8 ) = 0. 070544p 8 + 0. 010078(1 − p 8 ) = 0. 05.

Solving for p 8 we get p 8 = 0. 495175. At the beginning of the experiment we will flip a biased coin to decide what our threshold for rejection should be.

Question 6.3.9

For this we compute 0.07055664. The plot is presented when running the R code for this question.

Question 6.4.1

For this question we would compute β = P (Type II Error| μ = 500) = P (− z α⁄ 2 ≤ Z ≤ +z α⁄ 2 | μ = 500) =P =P =P

.

This expression we can evaluate using functions in R as the expression (when μ = 500) is a normal random variable. Using the R code for this question we can evaluate the above to get 1 − β = 0. 073369 (which is the power).

Question 6.4.2

The decision rule changed to take α = 0. 1 rather than α = 0. 05. In that case we would reject H 0 if

when we evaluate using the numbers in this question.

Question 6.4.3

The decision rule from that question was to reject H 0 if y was not contained in the limits [1] 88.9852 101.0148

Then in this case we have β = P (Type II Error| μ = 90) = P (y l ≤ Y ≤ y u | μ = 90) =P

.

We can evaluate the above where we find 1 − β = 0. 3757861.

Figure 4: The power plot for Question 6.4.4.

Question 6.4.4

Following the steps in Question 6.4.1 we find that (48)

The power curve is a plot of 1 − β as a function of μ . When the R code for this section is run we get the plot given in Figure 4 .

Question 6.4.5

We are asked to find β the probability of a Type II error. This is β = P (Type II Error| μ = 220) =P =P

.

In the R code for this problem we find this given by 0.627919.

Question 6.4.6

Part (a): This is a bit of a “weird” test, in that we will reject the hypothesis that H 0 : μ = 60 if in fact y is close to the value of 60. To work the problem however, we are told that the value of y ∗ is set to enforce the probability of a Type I error or the condition

or

For this question this is

The above is equivalent to

As μ 0 = 60 the above becomes

Let v =

then using the cumulative normal function Φ we can write this as

As Φ(− v ) = 1 − Φ(v ) the above is

This means that

Using the numbers given in this questions we can evaluate this to get y ∗ = 0. 1171265.

Part (b): For this we want to compute 1 − β = P (Reject H 0 | H 1 ) = P (60 − y ∗ ≤ y ≤ 60 + y ∗ | H 1 ).

Converting this into a standard normal random variable we can evaluate it to find the power given by 0.02279142.

Part (c): The correct way would be to reject H 0 if y does not satisfy

These are the limits [1] 57.58412 62.41588

The power of this test is given by 0.6219825 a significant improvement over the previous number.

Question 6.4.7

This is similar to the example done in the text. The value of α sets the critical value for the test. This means that α = P (Reject H 0 | H 0 ) =P

,

and so the critical threshold on y is (49) which depends on n . The condition on the power is that 1 − β = P (Reject H 0 | H 1 ) = P ( Y < y ∗ | H 1 )

=P

,

which means that (50) Putting Equation 49 into Equation 50 we can solve for n to get

Using the numbers for this question this gives n = 95. 652441 rounding this up to an integer given n = 96.

Question 6.4.8

In the R code for this section using Equation 48 with n = 45 I compute [1] "n= 45; beta= 0.081638"

As the value of β computed above is less than the requirement of 0.2 the value of n is large enough.

Question 6.4.9

For this one-sided test we have that

and β =P =P

.

This means that

From the above we can solve fore z α and then α . In the R code for this question we find [1] "z_beta= -1.036433; z_alpha= 0.741344; alpha= 0.647496"

Note that this does not match the answer in the back of the book. If anyone sees anything wrong with what I have done please contact me.

Figure 5: The plot for Question 6.4.10.

Question 6.4.10

Part (a): This would be

Part (b): This would be

Part (c): When the R code for this section is run we get the plot given in Figure 5 . The gray area (to the right) represents α while the blue area (to the left) represents β .

Question 6.4.11

For this table we should keep Figure 6.4.1 in mind. Using that table we find [1] "alpha= 0.064286; beta= 0.107143"

A Type I error convicting an innocent person would seem worse than not catching a guilty person (the Type II error).

Question 6.4.12

Let w be the number of white ball in the urn and r the number or red balls in the urn. Let X be the random variable denoting the number of white balls drawn when we draw three balls.

Then

In the same way if we have 60% white then w = 6 and r = 4 so that

Evaluating this in the two cases gives [1] "power(w=6,r=4)= 0.666667" [1] "power(w=7,r=3)= 0.816667" Question 6.4.13

The pdf of the maximum of n samples from a uniform distribution is

We want to pick k such that

is the given value of α = 0. 05. If we solve the above for k we find k = 1. 979588.

Question 6.4.14

For this we want to compute

Question 6.4.15

For this we would have

If we want this to be equal to 0.95 we must have p = 0. 95 1 ⁄n .

Question 6.4.16

For this question we would calculate α using

For the terms in the sum we have P (X = 5| H 0 ) = P (X 1 = 1)P (X 2 = 4) + P (X 1 = 2)P (X 2 = 3) = 0. 09375 P (X = 6| H 0 ) = P (X 1 = 2)P (X 2 = 4) = 0. 015625. summing these two numbers means that α = 0. 109375.

Question 6.4.17

We might want to compute α for this test. We have

Here C is the critical region. To derive an expression for β in terms of 𝜃 recall that 1 − β = P (Reject H 0 | H 1 ) = ∫ 0 1 ⁄ 2 (1 + 𝜃 )y 𝜃 dy = when we integrate.

Question 6.4.18

Part (a): We would have

Part (b): We would have

which gives β = 0. 7618967.

,

Question 6.4.19

We would have

for p = . Evaluating the above we get β = 0. 875.

Question 6.4.20

We would have

If we integrate and solve for β we find

Question 6.4.21

The pdf of a sum of two uniform random variables is given by taking the convolution between two uniform pdfs. When we carry out this algebra we get

We now want to find the value of k such that

If we assume that k < 𝜃 = 2 (if this is not true we can keep working) then we would have

Integrating and solving for k we find k = two we have found an answer.

= 0. 6324555. As this value is less than

Question 6.4.22

This is just like the previous question but now we have to find the pdf of the product of two uniform random variables. Following the hint in the book we let W = Y 1 Y 2 then from the theorem stated the pdf for W is given by

where we have used the fact that the uniform pdf f is zero outside of [0,𝜃 ]. Now as x takes values over the interval [0,𝜃 ] the fraction will take values from +∞ to . The expression f (w⁄x ) will be zero for any argument where > 𝜃 which happens if x < . Our integral for f W (w ) then becomes

when we integrate. Lets check that this is a valid pdf. We have that f W (w ) ≥ 0 for all w and that

when we perform the integral. Finally to answer the question we need to compute k ∗ such that

Note that the integral of f W (w ) is

Using this, if we take 𝜃 = 2 (i.e. H 0 is true) and set the above expression equal to 0.05 we have a single nonlinear equation for x . Solving for x we find x = k ∗ = 0. 034843.

Question 6.5.1

The generalized likelihood ratio is defined to be (51) Here in the numerator the space of ω over which we optimize the parameters of is “smaller” and represents constraints on the parameters of the likelihood (notably a hypothesis like H 0 ) and in the denominator the space Ω is “larger” and represents the possible unconstrained parameters (like H a ).

In this case the likelihood is given by

The maximum needed in the numerator of λ when p = p 0 is just L (p 0 ). The maximum likelihood estimate needed to compute max Ω L (𝜃 1 , … ,𝜃 k ) is to take the derivative of L (p ) with respect to p , set the result equal to zero, and solve for p . Taking this derivative and setting the result equal to zero we get

n

If we divide this by p n − 1 (1 − p ) ∑ i =1 k i − n − 1 we get

Solving this for p gives

This is the “unconstrained” solution representing H 1 : p ≠ p 0 . Using these for λ we get

Question 6.5.2

When we have an exponential pdf for y it takes the form of f (y ) = λe − λy and we have a likelihood given by

In viewing Equation 51 the maximum needed in the numerator of λ when λ = λ 0 is just L (λ 0 ).

The maximum likelihood estimate needed to compute max Ω L (𝜃 1 , … ,𝜃 k ) is to take the derivative of L (λ ) with respect to λ , set the result equal to zero, and solve for λ . Once we have done this we find this estimate given by

With this the GLRT value is then

This expression is actually a random variable as the y i are random realizations y ∼ f Y (y ).

Now to design a test that will reject at the level of α we have to find a threshold λ ∗ such that

If we introduce the random variable W ≡ λ 0 ∑ i =1 n Y i (note that this is a multiple of the sum of n exponential random variables) then the above is equal to

or

Let w ∗ = n − n e n λ ∗ be a constant then the above expression is the integral (52) Here we are told that for the numerical value of α to take α = 0. 05.

Thus to evaluate the left-hand-side of the above we have to determine/derive p W (w ) which is related to the density of the sum of n exponential random variables. In a previous chapter we were told that the sum of n exponential random variables is a gamma random variable (if they are independent). Thus let S ≡ ∑ i =1 n Y i then given H 0 is true so λ = λ 0 we have

This means that using the pdf transformation properties of random variables W =λ0S∼

gS

= =

e−w.

Now that we have the pdf of W we can go back and express Equation 52 . That expression becomes

We would integrate the left-hand-side and then solve for w ∗ as a function of α . Once we have w ∗ we would then use w ∗ = n − n e n λ ∗ to determine λ ∗ and the threshold for our GLRT test.

Question 6.5.3

For this question the “small” parameter space is

while the “large” parameter space is

The likelihood under any hypothesis is given by

Under ω the expression for L (ω ) is the above evaluated at μ = μ 0 .

Under the Ω the maximum likelihood estimate of μ is y = ∑ i =1 n y i .

Using the two results above the GLRT is then λ =

=

= exp Lets expand the two quadratics in the exponential where we get

If we sum this over i we get

This means that

Because of this we would base our test on the statistic

.

Question 6.5.4

The likelihood ratio in this case would be

If we expand the quadratics in the argument of the exponential (without the 1⁄ 2 for now) we get

If we sum this over i we get

Using this (and putting back the 1⁄ 2) we have that

Our test would then be based on the value of

This measures where the sample mean y falls relative to the midpoint between μ 0 and μ 1 (the proposed means).

Question 6.5.5

Part (a): The likelihood under the two hypothesis spaces ω and Ω are given by L (ω ) = p k (1 − p ) n − k with p = L (Ω) = p k (1 − p ) n − k with p ≠ .

Under the unconstrained set Ω the maximum likelihood estimate of p is given by = . Thus we have L (ω e ) = L (Ω e )

k

n−k

=

n

=

k

n−k

.

This means that the generalized likelihood ratio (GLRT) is given by λ =

=

.

The generalized likelihood ratio test will reject H 0 if 0 < λ < λ ∗ . This means that

Taking the logarithm of both sides we get

or simplifying we can write this as

If we take the negative of both sides we get

which is what we were to show.

Figure 6: A plot of the data and the model for Question 6.5.5.

Part (b): Consider the left-hand-side of the above expression and define f (k ) as

Note that f (k ) = f (n − k ) and f (k ) has a maximum when

Solving for k we get k = . If we draw a plot of f (k ) we get Figure 6 . Notice that the critical region above can be expressed by an expression of the form

for some constant c . This can be converted to the given expression by dividing by n as k = .

Question 6.5.6

A sufficient statistic for 𝜃 means that for our estimator = h (X 1 ,X 2 , … ,X n ) that we can factor our likelihood as L (𝜃 ) = p (𝜃 e ; 𝜃 )b (k 1 ,k 2 , … ,k n ). Recall that for our generalized likelihood ratio test we need to form

Using representation above we can write λ for our critical region (is region C where we reject H 0 ) as

Notice that this expression depends on the sufficient statistic .

Chapter 7 (Inferences Based on the Normal Distribution) Notes on the computation of F p,m,n In R we can compute values of F p,m,n using the function qf(p, m, n) . For example to verify the value of F 0 . 95 , 3 , 5 (a numerical value found in the text) we can evaluate

qf(0.95, 3, 5) where we get the value 5.409451.

Question 7.3.1

Note that f U (u ) is nonnegative for u ≥ 0. Next we need to compute

If we let v = then dv =

and I becomes

I =∫0∞ =

2 m⁄ 2 − 1 v m⁄ 2 − 1 e − v 2dv ∫0∞v

−1

e − v dv = 1,

showing that our density is normalized (as it must be to be a density).

Question 7.3.2

The moment generating function is defined as

In this case this is M U (t ) = =

∫ 0 ∞ e tu u

−1

∫0∞u

e−

−1

e − du u

du .

To evaluate this we use v =

u

dv =

du ,

to get M U (t ) = =

∫0∞

e−v ∫0∞v

−1

e − v dv =

.

(53)

To compute the expectation we will use E (χ m 2 ) = M U (1) (0) where we see that we need to compute the first derivative of M U (t ). We find

so that evaluating this at t = 0 we get

Next we will compute the expectation of the square of a chi-squared random variable using E ((χ m 2 ) 2 ) = M U (2) (0). To do that we need to compute the second derivative of M U (t ). We find

so that evaluating this at t = 0 we get

Using these two results we can compute the variance of a chi-squared random variable using Var

= E ((χ m 2 ) 2 ) − E (χ m 2 ) 2 = m (m + 2) − m 2 = 2m .

Question 7.3.3

Recall that the sum of three independent standard normal random variables squared is a chi squared random variable with three degrees of freedom. Thus we can compute the value of this random variable and compare it with the quantiles of the chi squared distribution. If z 2 is the value of the sum of the three random variables squared then we find

As the right-hand-side of the above is not that small it is believable that the numbers came from the given distribution.

Question 7.3.4

We start with

which follows from the fact that the random variable is a χ n − 1 2 random variable and the variance of such a random variable. Now the left-hand-side of the above is

Using this in the first expression and solving for Var

and we get

Question 7.3.5

To be a consistent estimator means that (54) As we know from the above question that

so that Var

→ 0 as n →∞ and Equation 54 must hold true.

As another way to see this consider Chebyshev’s inequality on the random variable S 2 . As E (S 2 ) = σ 2 this is

as n →∞ .

Question 7.3.6

If we take n = 200 then using the fact that

we can evaluate the 40% of Y by finding the 40% of Z and then solving for Y in the above.

When I do that I get the value 194.9331.

Question 7.3.7

Using the R code for this section we find [1] 0.9833387 0.1321459 9.0000000 Question 7.3.8

As the given fraction has a F distribution with 7 and 9 degrees of freedom we can evaluate the probability that a variable of that type is between the two given limits. We find these two probabilities to be [1] 0.04944707 0.02516584

so the first situation is more likely.

Question 7.3.9

Part (a): We want to find x such that

The left-hand-side is given by

In R notation the second term on the left-hand-side is the same as

so to evaluate the above we can use

Part (c): For this part we want to find x such that

Using the above observations and the R code for this section we find [1]

6.2393753

0.6508271

9.0019577 14.9771317

2.2797631

Question 7.3.10

Assume that we have a sample of size m in forming S 1 2 and a sample of size n in forming S 2 2 . Then from the statement in the book the ratio

is the needed ratio for the F distribution and the above fraction has a F m − 1 ,n − 1 distribution.

Simplifying the left-hand-side we have that

As this ratio limits to one as m and n approach infinity we have the desired limit on F m − 1 ,n − 1 or equivalently on F m,n .

Question 7.3.11

This fact follows from the fact that an F random variable (with m and n degrees of freedom) is one of the form for V and U chi squared random variables with m and n degrees of freedom. This means that 1⁄F is a random variable of the form and is thus an F random variable but this time with n and m degrees of freedom.

Question 7.3.12

From the discussion in the previous question we would have

if and only if

Running the R code for this problem we get the values [1] 0.05162358 0.05162358 [1] 4.45897 4.45897

which demonstrate the above relationship.

Question 7.3.13

Recall the density for the t -distribution

To start this question we will use the identity

on the factor

as follows. The inverse of this term looks like =

− 2

→et ⁄2, as n →∞ when we use the identity above and the fact that

−1⁄2

→ 1.

We now have to evaluate the limit of the normalizing constant in f T n (t ) as n →∞ . Recall that Stirling’s formula in terms of the Γ(z ) function is z

Γ(z + 1) ∼ Γ

so

= Γ

=

−1

.

Using these we want to evaluate the limit of

as n →∞ . To do that note that using Stirling’s approximations we can write C n as Cn ≈

= =

1⁄2

= =

1⁄2

1⁄2

1⁄2

.

Now as n →∞ we can write the expression inside of the brackets as

as n →∞ . To further evaluate this, as n →∞ , note that we have a limit of the type 1 ∞ . To evaluate this we will write it as

which is now a limit of the type ∞× ln(1) = ∞× 0. We can evaluate this using L’Hospital’s rule. It is simpler, however, to use the first term in the Taylor series of ln(1 + x ) ≈ x . Then we see that

Using this we see that

as n →∞ . This means that we have found the limit of C n to be given by

Combining this with the other limit and we have the desired result.

Question 7.3.14

If we take n = 1 then

As Γ(2) = 1 and Γ(1⁄ 2) =

we have that

As this density must be normalized we see that the given integral must integrate to one.

Question 7.3.15

Consider that

with the last step by symmetry of the integrand about y = 0. As y gets large the integrand looks more and more like

An integral (to infinity) of an integrand of this form will converge if the power of y is larger than one which is equivalent to

Question 7.4.1

Using the R code for this section we find [1] 0.1500304 0.7999350 0.8500353 0.8399316 Question 7.4.2

Part (a): Using R notation to simplify the expression on the left-hand-side we have ∫ − x x f T 22 (ξ )dξ = pt(x, 22) − pt(− x, 22) = pt(x, 22) − (1 − pt(x, 22)) = 2pt(x, 22) − 1 = 0. 98, so that pt(x, 22) = 0. 99. We can solve this for x using the R function qt .

Part (b): We want to evaluate

or

We can solve this for x using the R function qt .

Using the above observations and the R code for this section we find [1]

2.508325 -1.079469

Question 7.4.3

1.705618

4.302653

From the definition of t α,n and the fact that the t distribution has long tails we expect that the expression that reaches out further into the tails (i.e. t 0 . 05 ,n − t 0 . 1 ,n ) will have a larger value. In the R code we demonstrate this with n = 10 where the two expressions evaluate to [1] 0.4402775 0.2791256

showing the claimed expression.

Question 7.4.4

The given expression is a random variable following a t distribution with n − 1 = 8 degrees of freedom. Thus the two values of a are given by [1] 1.396815 1.859548 Question 7.4.5

From the text the given expression is a random variable following a t distribution with n − 1 = 10 degrees of freedom. Thus we find k = 2. 228139.

Question 7.4.6

For this we would use the fact that

This would give that

Question 7.4.7

Running the R code for this problem we get [1] 0.8687496 1.1532504

for the confidence interval.

Question 7.4.8

For this question we will use the R function t.test . Running this on the given data we get the output (among others)

95 percent confidence interval: 1319.364 1712.636 Question 7.4.9

Part (a): For this question we will use the R function t.test . Running this on the given data we get the output (among others) 95 percent confidence interval: 30.82328 40.01005

Part (b): A plot of this type is generated in the R code for this question. When that R code is run we see that there does not appear to be any trend to this data.

Question 7.4.10

Running the R code for this problem we get [1] 103.3925 108.0829

for the confidence interval.

Question 7.4.11

Running the R code for this problem we get [1] 176.0111 211.0722

for the confidence interval. Notice that this interval is inside the normal range.

Question 7.4.12

The the confidence intervals of the mean is given in the book. Using that we compute the values of y and s to be [1] 47.300000 4.879309 Question 7.4.13

Variability between the two samples can make the 95% confidence interval smaller if the computed s is sufficiently smaller.

Question 7.4.14

A 95% confidence interval of the mean revenue is given by

[1] 54266.94 64813.06

This would be a range where the company would expect to find the average revenue.

Question 7.4.15

We want to know what percentage of the corresponding t distribution (when scaled correctly) the given intervals cover. From the expressions given we can determine the value of n and then the percentage. In the R code for this problem we compute [1] 0.9499976 0.7999904 0.9449981 0.9499983 Question 7.4.16

Part (a): For this I compute [1] 3.741285 4.548001

for the confidence interval.

Part (b): Notice that there are quite a few samples in the 11 − 12 bin at the end. This makes me think that the distribution of Y might be skewed. This would say that the distribution for Y is not very close to normal and the conditions needed for the stated theorem would not hold.

Question 7.4.17

We want to test H 0 : μ = 0. 8 vs. H a : μ ≠ 0. 8. In the R code for this problem we compute [1] "t= -1.709116; p_value= 0.104614"

Thus we do not have enough evidence to reject H 0 .

Question 7.4.18

In the R code for this problem we compute [1] "t= 17.413788; p_value= 0.000000"

Thus we have strong evidence to reject H 0 .

Question 7.4.19

For this problem we will use the R function t.test . When we run that code we get the following (partial) output

t = -2.25, df = 14, p-value = 0.04106 alternative hypothesis: true mean is not equal to 40 95 percent confidence interval: 34.27047 39.86287

Given that the confidence interval does not contain the value of 40 we can reject the null hypothesis H 0 at this level of significance.

Question 7.4.20

For this problem we will use the R function t.test . When we run that code we get the following (partial) output t = 0.79567, df = 33, p-value = 0.4319 alternative hypothesis: true mean is not equal to 0.618 95 percent confidence interval: 0.5879594 0.6866288

Given the large P-value we cannot reject the null hypothesis.

Question 7.4.21

Here our hypothesis test is H 0 : μ = 0. 0042 vs. H a : μ < 0. 0042. We will work this problem using the R function t.test . When we run that code we get the following (partial) output t = -2.4772, df = 9, p-value = 0.01758 alternative hypothesis: true mean is less than 0.0042 95 percent confidence interval: -Inf 0.004122001

Given the small value for the P-value we can conclude that the plastic coating seems to be beneficial i.e. makes the maximum pit depth smaller.

Question 7.4.23

Due to the shape of the exponential density (with its largest values close to the origin) on any given draw it is more likely to draw samples with a small value of y . This would give the empirical mean y a smaller value than the true mean μ and any statistic constructed from y − μ will be skewed to the left.

Question 7.4.24

The first distribution is uniform while the second is peaked more near the value y = 1. This means that for the second distribution a small sample is more likely to have its empirical mean y closer to one (rather than the true value of ). This means that y − μ will be skewed to

the positive. We would expect the t test to work worse with this pdf than with the uniform one.

Question 7.4.25

As n increases the pdf f T n − 1 (t ) is converging to the normal pdf.

Question 7.4.26

The data set in (c) seems to be very skewed in that it has samples separated from whatever centering location one would choose to specify. This would indicate that the t test might perform poorly with small samples of data from this form.

Question 7.5.1

Using the R function qchisq we find these three numbers to be [1] 23.684791

4.605170

2.700389

Question 7.5.2

Using the R function pchisq we find these numbers to be [1] 0.9499933 0.9000124 0.9499992 0.9899983 Question 7.5.3

Using R we find these numbers to be [1]

2.087901

7.260944 13.763391 17.538795

Question 7.5.4

Part (a): This is equivalent to

We can set up a small root finding routine to find the value of n where this is true.

Part (b): This is equivalent to

We can set up a small root finding routine to find the value of n where this is true.

The other parts of this question can be done as the previous two parts above. When we implement these ideas in the R code for this chapter we get the following for the values of n [1] 14.23483 45.73260 31.00056 20.39664 Question 7.5.5

In the R code for this problem we find this to be 233.9927.

Question 7.5.6

The given statement is equivalent to

As we know that

∼ χ n − 1 2 we note that we are looking for the value of n such that

Using the R code for this problem we find n = 9 is the first value of n where that is true.

Question 7.5.7

From the statements in the book we know that (55) In this expression we will solve for σ 2 . As a first step note that the above is equivalent to

or solving for σ 2 we get (56) The two limits on σ 2 give a 100(1 − α )% confidence confidence interval on σ 2 . Taking the square root gives a 100(1 − α )% confidence interval for σ .

Question 7.5.8

For this question we will use Equation 55 where we solve for S 2 to get

Then matching the above to a and b from the question statement in the R code we find a and b given by [1]

5.487164 21.017586

Question 7.5.9

Part (a): To do this we use Equation 56 (the square root of the limits). When we compute these two numbers on this data I find limits of [1] 20.12728 42.16950

Part (b): The two one-sided intervals are given by (57) and (58) Computing these two intervals on this data gives (21. 10699, ∞ ) and (0, 39. 16185).

Question 7.5.10

For this we use Equation 56 (taking the square root of the limits) where in the R code we finds the confidence interval for σ to be given by [1] 0.3014431 0.8000719

The value of 0.262 quoted for the one year is smaller than the left-most limit of the above range indicating that the 5-year CD rates are indeed more variable than the 1-year.

Question 7.5.11

The confidence interval for σ would be used more often as it is in the same units as the data and would be easier to understand in the context of the experiments being studied.

Question 7.5.12

Part (a): As we know that is given by a chi-squared random variable with n − 1 degrees of freedom. From what the mean and variance of a chi-squared random variable we can standardize the above random variable using the expression

Here I have defined the random variable Z . As we are told that Z is asymptotically normal we can form a large sample confidence interval by using the above in

If we solve the above for σ 2 we get

The limits around σ 2 give a 100(1 − α )% confidence interval for σ 2 . Taking the square root gives a large sample confidence interval for σ .

Part (b): Using the above formula on the data gives [1] 21.06209 45.99563

for the confidence interval for σ . This is somewhat different than the one found in Case Study 7.5.1 of (20. 5, 40. 0) but rather close. Given that n = 19 we don’t really seem to have a large sample size so we don’t expect the results to match exactly.

Question 7.5.13

To do this we use Equation 56 . Then with what we are told for this question we have = 54. 47

(59)

= 261. 90, with α = 0. 1. We don’t know n or the value of s 2 but we have two equations and so can determine them using these two relationships. Taking the ratio of the two equations above gives

The value of n that gives the closest match to the right-hand-side is n = 11. Using this value into Equation 59 and we get s 2 = 111. 5719.

Question 7.5.14

Part (a): Introduce the random variable S n defined as

Now from Theorem 3.12.3 the moment generating function (MGF) for S n is ∏ i =1 n M Y i (t ).

Then using the result from Example 3.12.3 with λ = the MGF for Y i is

Using these two results the MGF for S n = n Y is

Now using Theorem 3.12.3 the MGF for (n Y ) is (60) Recall that the MGF for a χ n 2 random variable is given in Question 7.3.2. where we found it to be

Thus taken together this means that Equation 60 is the MGF for a χ 2 random variable with 2n degrees of freedom.

Part (b): A confidence interval for 𝜃 can be derived using

Solving for 𝜃 we get

The confidence interval is given by the two expressions on either side of 𝜃 .

Question 7.5.15

We want to test H 0 : σ 2 = 30. 4 2 H a : σ 2 < 30. 4 2 .

To do this we will compute χ 2 = find

with σ 0 = 30. 4. In the R code for this problem we

[1] "chi^2= 14.285177; p_value= 0.289679"

The P-value is to large for us to reject H 0 .

Question 7.5.16

We want to test H0 : σ2=12 Ha : σ2>12.

To do this we will compute χ 2 =

= (n − 1)s 2 . In the R code for this problem we find

[1] "s^2= 0.424724; chi^2= 12.316987; p_value= 0.997128"

Notice that the value of s 2 computed on this data is in fact smaller than the hypothesized value of σ 0 2 = 1. The P-value is to large for us to reject H 0 .

Question 7.5.17

Part (a): We want to test H 0 : μ = μ 0 = 0. 101 H a : μ > μ 0 = 0. 101.

To do this we will compute t =

. In the R code for this problem we find

[1] "t= 0.674392; p_value= 0.253392"

for the P-value of a one-sided test. As this P-value is large we cannot reject the null hypotheses.

Part (b): For this part we want to test H 0 : σ 2 = (0. 1567) 2

H a : σ 2 < (0. 15671) 2 .

To do this we will compute χ 2 =

. In the R code for this problem we find

[1] "chi^2= 9.687947; p_value= 0.006954"

With a P-value this small we should reject the null hypothesis and conclude that the variance of the portfolio is smaller.

Chapter 8 (Types of Data: A Brief Overview) One way to help in determining if the data give is categorical or not is to ask yourself the question “can I replace the given numbers with fractional numbers and get a reasonable table?”. For some studies where you are counting the number people who agree/disagree you can’t do that. For example you can’t have 10.5 people who agree and 11.5 who disagree.

Data that must be natural numbers is almost always categorical.

Question 8.2.1

The data is quantitative, the units are dissimilar, so we have regression data.

Question 8.2.2

The data is quantitative, the units are similar, there are two factors involved, and the samples are independent, so we have two-sample data.

Question 8.2.3

The data is quantitative, the units are similar, there is only one factor, so we have one-sample data.

Question 8.2.4

The data is quantitative, the units are similar, there are three factors, the samples are dependent (as each group was subjected to three densities), so we have randomized block design.

Question 8.2.5

The data is quantitative, the units are dissimilar, so we have regression data.

Question 8.2.6

The data is quantitative, the units are similar, there are two factors involved, and the samples are independent (different lambs), so we have two-sample data. If we viewed the samples as dependent (each lamb had the same ewe) we might say we had paired data.

Question 8.2.7

The data is quantitative, the units are similar, there are three factors, the samples are independent, so we have k -sample data.

Question 8.2.8

The data is quantitative, the units are dissimilar, so we have regression data. We would probably be interested in learning how age affects the percent of speed related fatalities.

Question 8.2.9

The data is quantitative, the units are similar, there is only one factor, so we have one-sample data.

Question 8.2.10

The data is quantitative, the units are dissimilar, so we have regression data.

Question 8.2.11

The data is quantitative, the units are dissimilar, so we have regression data.

Question 8.2.12

The data is quantitative, the units are similar, there is only one factor, so we have one-sample data.

Question 8.2.13

The data is quantitative, the units are similar, there are two factors involved, and the samples are independent, so we have two-sample data.

Question 8.2.14

The data is quantitative, the units are dissimilar, so we have regression data.

Question 8.2.15

The data is quantitative, the units are similar, there are three factors, the samples are independent, so we have k -sample data.

Question 8.2.16

The data is quantitative, the units are similar, there are more than two factors, the samples are dependent (in the same city), so we have a randomized block design experiment.

Question 8.2.17

The data is qualitative (is the student working, taking classes, or playing) so we have categorical data.

Question 8.2.18

The data is quantitative, the units are similar, there are more than two factors (the city is one), the samples are independent, so we have k -sample data.

Question 8.2.19

The data is quantitative, the units are similar, there are two factors involved, and the samples are independent, so we have two-sample data.

Question 8.2.20

The data is qualitative (complete overhaul vs. minor revisions).

Question 8.2.21

The data is quantitative, the units are similar, there are two factors involved, and the samples are dependent, so we have paired data.

Question 8.2.22

The data is quantitative, the units are similar, there are more than two factors, the samples are independent, so we have k -sample data.

Question 8.2.23

The data is qualitative, who is responsible for giving each drug type.

Question 8.2.24

The data is quantitative, the units are similar, there are more than two factors, the samples are independent, so we have k -sample data.

Question 8.2.25

The data is qualitative, survive or not survive.

Question 8.2.26

The data is quantitative, the units are similar, there are two factors involved (southeast and northeast), and the samples are independent, so we have two-sample data.

Question 8.2.27

The data is qualitative, we are breaking up people into discrete categories.

Question 8.2.28

The data is quantitative, the units are similar, there are two factors involved (male or female), and the samples are independent, so we have two-sample data.

Question 8.2.29

The data is quantitative, the units are similar, there are two factors involved, and the samples are dependent, so we have paired data.

Question 8.2.30

The data is quantitative, the units are similar, there are more than two factors (freshmen, sophomores, and juniors), the samples are independent, so we have k -sample data.

Question 8.2.31

The data is quantitative, the units are similar, there are more than two factors, the samples are dependent (in the same year or company), so we have a randomized block design experiment.

Question 8.2.32

The data is quantitative, the units are similar, there are two factors involved (how fractions are taught), and the samples are independent, so we have two-sample data.

Chapter 9 (Two-Sample Inferences) Question 9.2.1

Running the R code for this problem we get

[1] "t= 1.750214; p_value= 0.048107; t_crit= 2.539483"

we accept H 0 .

Question 9.2.2

Running the R code for this problem we get [1] "t= -3.096576; p-value= 0.001163; t_crit= -1.654808"

thus we would reject H 0 .

Question 9.2.3

Running the R code for this problem we get [1] "t= 5.686948; p_value= 0.000000; t_crit= 1.646284"

thus we would reject H 0 .

Question 9.2.4

Running the R code for this problem we get [1] "t= -1.669141; p_value= 0.047570; t_crit= -1.645101"

thus we would reject H 0 at the 5% level.

Question 9.2.5

Running the R code for this problem we get [1] "t= -0.531655; p_value= 0.595956; t_crit= 2.617776"

thus we would accept H 0 .

Question 9.2.6

Running the R code for this problem we get [1] "t= -1.679668; p_value= 0.054695; t_crit= -1.729133"

thus we would not reject H 0 at the 5% level (but we would reject at the 10% level).

Question 9.2.7

Running the R code for this problem we get

[1] "t= 0.685531; p_value= 0.518604; t_crit= 2.446912"

thus we would not reject H 0 .

Question 9.2.8

Running the R code for this problem we get [1] "t= -1.426852; p_value= 0.177197; t_crit= 3.012276"

thus we would accept the null hypothesis.

Question 9.2.9

Running the R code for this problem we get [1] "t= 2.155780; p_value= 0.033894; t_crit= 1.987934"

thus we would reject the null hypothesis at the 5% level.

Question 9.2.10

Running the R code for this problem we get [1] "t= 4.743416; p_value= 0.000081; t_crit= 1.734064"

thus we would reject the null hypothesis.

Question 9.2.11

Part (a-b): Running the R code for this problem we get [1] "Part (a): |xbar-ybar|_min= [1] "Part (b): |xbar-ybar|_min=

22.881; t_crit= 166.977; t_crit=

2.947" 1.729"

Question 9.2.12

Running the R code for this problem we get [1] "z= 0.224783; p_value= 0.822148" Question 9.2.13

Part (a): As we must drive twice per day (there and back) we want to know if the sum of two times driving through town will be quicker than the sum of two times driving by interstate. That is what is the probability that

Here X i is the time to drive to work via the interstate and Y i is the time to drive to work through town. Thus we want to evaluate

Given that we know the mean and variance of the random variable D ≡ ∑ i =1 2 Y i − ∑ i =1 2 X i we can subtract both sides by the mean and divide by the variance to get

and use properties of the standard normal to evaluate this. In the R code for this problem we find this to be 0.358623.

Part (b): In this case we want to know the probability that

As we know the distributions of the expressions on the right and left-hand-side of the above we can evaluate the probability that the above event happens. Using the R code for this problem we find this to be 0.209034.

Question 9.2.14

Recall that X ∼N Y ∼N

.

As everything is independent we have

From this we see that the ratio referred to in this problem is Z ∼ N (0, 1).

Question 9.2.15

From the definition of s P 2 given by

(61) and the fact that E [s X 2 ] = E [s Y 2 ] = σ 2 we have that

Question 9.2.16

A variant of this question is worked in Appendix 9.A.1 where there it is assumed that σ 2 is not known. To work this question one can follow the arguments made there (but assuming that σ 2 is known). What changes is that we no longer have to take derivatives of the likelihood with respect to σ 2 . Below we sketch that procedure.

In using the generalized likelihood test we need the “restricted” space ω (for H 0 ) which in this case is

This is the statement that the two means of the two normals are restricted to have the same value.

We also need the “unrestricted” space Ω (for H 1 ) which in this case is

This is the statement that the means are unconstrained.

In forming the likelihood under ω (or L (ω )) we would get the same expression as Eq. 9.A.1.1 in the book. To maximize L (ω ) we take derivatives with respect to the common mean μ , set the result equal to zero, and solve for μ . When we do that we find that

Using this expression we see that ∑ i =1 n (x i − μ ω e ) 2 = ∑ i =1 n (x i − x + x − μ ω e ) 2 = ∑ i =1 n = ∑ i =1 n (x i − x ) 2 + n ( x − μ ω e ) 2 = (n − 1)

X

2

+ n ( x − μ ω e) 2 ,

where we have used the fact that ∑ i =1 n (x i − x ) = 0 on the middle (second) term. In the same way we would have

Using these, we can express L (ω ) evaluated at the maximum likelihood solution as L (ω e ) =

n+m

exp

.

In forming the likelihood under Ω (or L (Ω)) we would get the same expression as Eq. 9.A.1.4 in the book. To maximize L (Ω) we take derivatives with respect to μ X and μ Y , set the results equal to zero, and solve for each. When we do that we find that μ X e = x and μ Y e = y . Using these, we can express L (Ω) evaluated at the maximum likelihood solution as

Using these two expressions the ratio needed in the generalized likelihood criterion is given by

To simplify this some recall that μ ω e =

and thus

x −μ ωe = y −μ ωe =

.

Using these we see that λ = exp = exp

.

Using the formulation of the generalized likelihood ratio we reject H 0 : μ X = μ Y when 0 < λ ≤ λ ∗ . From the above equation for λ this means that we reject H 0 when ( x − y ) 2 > λ ∗∗ . This is the same form as the t -test of two means with a known common variance which is to be expected.

Question 9.2.17

In this case we will not assume that the variances of the two samples are equal. Using the test that is appropriate in this case we find [1] "w= 4.743416; p_value= 0.000157; t_crit= 1.761310"

Given such a small P-value we would reject the null hypothesis.

Question 9.2.18

Having a smaller degrees of freedom as mentioned in this question (i.e. ν < n + m − 2) is a disadvantage in the case where σ X 2 = σ Y 2 as the t distribution t ν is “wider” than that for t n + m − 2

. This means that for a given true difference in μ X − μ Y the P-value when using t ν will be smaller than the P-value when using t n + m − 2 . This means that we will be more likely to not reject H 0 when if the difference is significant and we should. Another way to say this is that the test that assumes that σ X 2 = σ Y 2 will have more power in this case.

Question 9.2.19

Part (a): When I compute statistics on the two groups I find [1] "n= 24; xbar= 6.958; sx= 3.149; m= 24; ybar= 8.125; sy= 3.279"

As the values of S x ≈ S y we can use the test that assumes the variances are equal.

Part (b): As this data represents a quantity that cannot be negative it is hard to justify the normality assumption needed to apply the tests from this section. This data also has long positive tails as there are samples that are almost three standard deviations from the mean (even though we only have n = m = 24 samples).

Question 9.2.20

For this problem I’ll use the test that does not need to assume equal variances. When I run the R code for this question I get [1] "xbar= 29.433; sx= 9.647; ybar= 35.729; sy= 15.328" [1] "w= 4.743416; p_value= 0.000394; t_crit= 2.763769"

The small P-value indicates that we should reject H 0 and conclude that the mean of the samples from Y (here republicans) is larger than the mean of the samples from X (here democrats).

Question 9.3.1

Running the R code for this problem we get [1] "alpha= 0.05; f_crit_L=0.287878; f_crit_R= 3.473699; f= 0.307707"

As our f value is between the two limits we can accept H 0 that the variances are equal.

Question 9.3.2

Running the R code for this problem we get [1] "alpha= 0.10; f_crit_L= 0.164090; f_crit_R= 4.120312; f=2.625000"

As our f value is between the two limits we can accept H 0 that the variances are equal.

Question 9.3.3

Running the R code for the first part of the problem we get [1] "alpha= 0.05; f_crit_L= 0.395812; f_crit_R= 2.526451; f=0.684093"

As our f value is between the two limits we can accept H 0 that the variances are equal. Next testing H 0 : μ X = μ Y we get [1] "t= 2.661960; p_value= 0.011324; t_crit= 2.024394"

This indicates that we can reject H 0 at the 5% level.

Question 9.3.4

Running the R code for this problem we get [1] "alpha= 0.05; f_crit_L= 0.248386; f_crit_R= 4.025994; f=3.188640"

As our f value is between the two limits we can accept H 0 that the variances are equal.

Question 9.3.5

Running the R code for this problem we get [1] "alpha= 0.05; f_crit_L= 0.248386; f_crit_R= 4.025994; f= 0.288960"

As our f value is between the two limits we can accept H 0 that the variances are equal.

Question 9.3.6

Running the R code for this problem we get [1] "alpha= 0.05; f_crit_L= 0.312745; f_crit_R= 3.391728; f= 1.452551"

As our f value is between the two limits we can accept H 0 that the variances are equal.

Question 9.3.7

Once we load the data we can test H 0 : σ X 2 = σ Y 2 against the alternative (that they are not equal).

In the R code for this chapter we find that [1] "alpha= 0.05; f_crit_L= 0.207586; f_crit_R= 6.757172; f=0.286477"

Given this we will not reject H 0 (even though f is “close” to the lower limit) and thus we can use the theorem that assumes equal variances to test assess whether H 0 : μ X = μ Y .

Question 9.3.8

Running the R code for this problem we get [1] "alpha= 0.05; f_crit_L= 0.225568; f_crit_R= 4.433260; f= 1.169516"

As our f value is between the two limits we can accept H 0 that the variances are equal. Thus we can use the theorem that assumes equal variances to test assess whether H 0 : μ X = μ Y .

Question 9.3.9

To work this question we start by following Appendix 9.A.1. In using the generalized likelihood test we need the “restricted” space ω (for H 0 ) which in this case is

This is the statement that the two variances of the two normals are restricted to have the same value. We also need the “unrestricted” space Ω (for H 1 ) which in this case is

This is the statement that the two variances are unconstrained.

In forming the likelihood under ω (or L (ω )) we would get the same expression as Eq. 9.A.1.4 in the book.

L (ω ) =

n+m

exp

.

Given this, we take derivatives with respect to μ X , μ Y , and σ 2 , set the results equal to zero and solve for these parameters. When we do this we get μ X e = x , μ Y e = y , and

These are the same results as found in Appendix 9.A.1.

Thus using these we find the likelihood actually given by L( ω e) =

exp

=

e−

(n+m)

=

,

which is the same result found in Appendix 9.A.1.

Next for L (Ω) we would have

Given this expression to maximize its value, we take derivatives with respect to μ X , μ Y , and σ X 2 , σ Y 2 , set the results equal to zero and solve for these parameters. When we do this we get μ X e = x , μ Y e = y , and σ X 2 = ∑ i =1 n (x i − x ) 2 σ Y2 =

∑ j =1 m (y i − y ) 2 .

Then putting these expressions into L (Ω) we find L (Ω e ) =

n⁄ 2

=

m⁄ 2

exp

exp

n⁄ 2

m⁄ 2

e−

.

From these two expressions for L (ω e ) and L (Ω e ) we have λ =

=

n⁄ 2

m⁄ 2

= =

n⁄ 2

m⁄ 2

,

which is the expression we were trying to show.

Question 9.4.1

Running the R code for this problem we get [1] "z= 1.762420; p_value= 0.077998; z_crit= 1.959964"

Thus we cannot reject H 0 and must conclude that the two proportions are the same.

Question 9.4.2

Running the R code for this problem we get [1] "z= -2.376140; p_value= 0.017495; z_crit= 1.959964"

Here we can reject H 0 at the 5% level.

Question 9.4.3

Running the R code for this problem we get [1] "z= -0.170293; p_value= 0.864780; z_crit= 1.959964"

Here we cannot reject H 0 .

Question 9.4.4

Running the R code for this problem we get [1] "z= -0.924701; p_value= 0.355121; z_crit= 2.575829"

Here we cannot reject H 0 .

Question 9.4.5

Running the R code for this problem we get [1] "z=

4.246855; p_value= 0.000011; z_crit= 2.326348"

Thus we can easily reject H 0 and conclude that there must be a difference in proportions from before and after the new policies.

Question 9.4.6

Running the R code for this problem we get [1] "z= 1.702513; p_value= 0.088659" Question 9.4.7

From the total number of students (8605) minus the number of students who live on campus (6001) we get the number of students who live off campus to be 2604. From the total number of women enrolled (4134) minus the 2915 who live on campus we get 1219 women who live off campus. This means that our statistical test will have x = 2915 n = 6001 y = 1219 m = 2604.

Using these numbers we find [1] "z= 1.503433; p_value= 0.132727; z_crit= 1.959964"

Thus we cannot reject H 0 .

Question 9.4.8

Running the R code for this problem we get [1] "z= -7.929770; p_value= 0.000000; z_crit= 1.959964"

Thus we can easily reject H 0 and conclude that there must be a difference in the divorce rates.

Question 9.4.9

Running the R code for this problem we get [1] "z= 0.251006; p_value= 0.801809; z_crit= 1.959964"

Thus we would accept H 0 and the player is correct.

Question 9.4.10

Using the nightmare data I compute − 2 ln(λ ) = 0. 3869864. In this section of the book the statement is made that we reject H 0 if this expression is larger than 3.84 (at the 5% level) and thus we don’t reject H 0 at that level. As the 1% level would be more difficult to reject we would not reject H 0 either. This conclusion is equivalent to that reached in the case study example in the book.

Question 9.5.1

Running the R code for this problem we get [1] 0.709072 1.551245

for our confidence interval in μ X − μ Y . As this interval does not contain zero we can reject the hypothesis that μ X = μ Y .

Question 9.5.2

Running the R code for this problem we get [1] 0.6173921 1.6429254

for our confidence interval in μ X − μ Y . As this interval does not contain zero we can reject the hypothesis that μ X = μ Y .

Question 9.5.3

Running the R code for this problem we get

[1] "Equal variances: ci= (-13.358858, 6.692191)" [1] "Unequal variances: ci= (-13.641526, 6.974859)"

As expected the confidence intervals under the unequal variances assumption is larger than the one under the equal variance assumption.

Question 9.5.4

From the book’s Eq. 9.5.1 we will solve for μ X − μ Y . To start, we multiply “both sides” by the expression S p

to get

Next subtract X − Y from “both sides” to get

Taking the negative of everything (and flipping the direction of the inequalities) then gives

Finally writing this in the more standard notation we get

Question 9.5.5

As we know that

Following similar steps to the previous problem in this case we would have

Question 9.5.6

Running the R code for this problem we get [1]

0.5414515 10.9607479

for our confidence interval in σ X 2 ⁄σ Y 2 . As this interval contains the value one we cannot reject the hypothesis that σ X 2 = σ Y 2 .

Question 9.5.7

Running the R code for this problem we get [1] 0.05890421 2.13351793

for our confidence interval in σ X 2 ⁄σ Y 2 . As this interval contains the value one we cannot reject the hypothesis that σ X 2 = σ Y 2 .

Question 9.5.8

We are told that

This means that

If we solve for

we get

which is the expression we wanted to show.

Question 9.5.9

Running the R code for this problem we get [1] -0.02133250

0.05070715

for our confidence interval in p X − p Y . As this interval contains the value zero we cannot reject the hypothesis that p X = p Y .

Question 9.5.10

Running the R code for this problem we get [1] -0.03319675

0.09569675

for our confidence interval in p X − p Y . As this interval contains the value zero we cannot reject the hypothesis that p X = p Y .

Question 9.5.11

From the statement given for this question we have

If we solve for p X − p Y we get the statement in Theorem 9.5.3. This is the same “procedure” as we did for Questions 9.5.4 and 9.5.5.

Question 9.5.12

Running the R code for this problem we get [1] -0.14109398 -0.02479991

for our confidence interval in p X − p Y . As this interval does not contain the value zero we can conclude that the suicide rate for women is larger than that for men.

Chapter 10 (Goodness-Of-Fit Tests) Question 10.2.1

This would be given by a multinomial probability with n = 3 + 2 + 1 = 6 and p i as given in the table. Using the R code dmultinom in the stats library I get this probability to be 0.000885691.

Question 10.2.2

Again using dmultinom I compute this to be 0.02966309.

Question 10.2.3

We first need to compute the probability that we are a member of each group. In the R code I find these to be [1] 0.878573062 0.117129194 0.004297744

The multinomial density would assign the sample (6, 3, 1) the probability 0.00266798.

Question 10.2.4

We first need to compute the probability that we are a member of each group. In the R code I find these to be [1] 0.878573062 0.117129194 0.004297744

The multinomial density would assign the sample (2, 4, 1) the probability 0.09481.

Question 10.2.5

In this case the probability breaks up into four distinct groups depending on where y falls. We have for y ∈ (− 60, − 20) or to the left of the pipeline (with no damage caused), y ∈ (− 20, 0) or to the left of the pipeline with damage, y ∈ (0, 20) to the right of the pipeline with damage, and finally y ∈ (20, 60) to the right of the pipeline and no damage. Given the expression for f Y (y ) we can integrate to find the probability of each group. That is we need to compute p 1 = ∫ − 60 − 20

dy =

p 2 = ∫ − 20 0

dy =

p 3 = ∫ 0 20

dy =

p 4 = ∫ 20 60

dy = .

Once we have these we can use the multinomial probability to compute the probability requested to find 0.006890905.

Question 10.2.6

This is an application of the multinomial probability and we find it to have a value of 0.01111803.

Question 10.2.7

Part (a): For this question we first need to compute the probability that y is in each of the given ranges. These give us “group” probabilities p i for i = 1, 2, 3, 4. To compute these we have p 1 = ∫ 0 1 ⁄ 4 3y 2 dy = p 2 = ∫ 1 ⁄ 4 2 ⁄ 4 3y 2 dy = p 3 = ∫ 2 ⁄ 4 3 ⁄ 4 3y 2 dy = p 4 = ∫ 3 ⁄ 4 1 3y 2 dy =

.

Then the probability we have X i occurrences of y falling in a group is given by the multinomial probability density function with n = 50 and the probabilities computed.

Part (b): As an individual X i is a binomial random variable it has Var = 3 and using p 3 computed above this number is 10.43701.

= np i (1 − p i ). When i

Question 10.2.8

We find the moment generating function given by M X 1 ,X 2 ,X 3 (t ) = E (e t 1 X 1 + t 2 X 2 + t 3 X 3 ) = ∑ k 1 ,k 2 ,k 3 e t 1 k 1 + t 2 k 2 + t 3 k 3 = ∑ k 1 ,k 2 ,k 3

p 1 k 1p 2 k 2p 3 k 3

(p 1 e t 1 ) k 1 (p 2 e t 2 ) k 2 (p 3 e t 3 ) k 3

= (p 1 e t 1 + p 2 e t 2 + p 3 e t 3 ) n , using the multinomial theorem.

Question 10.2.9

Using the previous problem any of the marginal generating functions can be obtained by setting all but one of the t i s equal to zero. For example, using the above expression setting all but t 1 equal to zero we would get

When we notice that this is the moment generating function for a binomial random variable we have a proof for Theorem 10.2.2.

Question 10.2.10

The likelihood for a given sample where X 1 = k 1 ,X 2 = k 2 , … ,X t = k t is given by

where C is the multinomial coefficient. Taking the logarithm of this we get

To find the maximum likelihood estimates for p i we want to maximize ℒ subject to the constraint that ∑ i =1 t p i = 1. To do this we form the Lagrangian as

To maximize this subject to the constraint we need to compute derivatives, set them equal to zero, and solve for p 1 ,p 2 , … ,p t and λ . These derivatives are Lagrangian =

−1=0

(

Lagrangian = ∑ i =1 t p i − 1 = 0.

(

If we solve Equation 62 for p i we get (64) If we put this expression into Equation 63 we get

As ∑ i =1 t k i = n the above can be solved for λ to get λ = − n . Using this in Equation 64 we get that

as we were to show.

Question 10.3.1

We can make this transformation by squaring and simplifying. We have D = ∑ i =1 t = ∑ i =1 t = ∑ i =1 t

− 2 ∑ i =1 t X i + n ∑ i =1 t p i

= ∑ i =1 t

− 2n + n = ∑ i =1 t

−n,

as we were to show.

Question 10.3.2

As we know the probabilities associated in each class this question is a direct application of the goodness-of-fit tests. In the R code for this section I compute [1] "d= 0.968750; d_critical= 4.605170; p_value= 0.616082"

Indicating that we cannot reject the null hypothesis and conclude that a hypergeometric random variable is reasonable.

Question 10.3.3

If we sample with replacement the number of white balls drawn in each sample is now a binomial random variable with n = 2 and p = . In this case running the R code for this section we get [1] "d= 3.298611; d_critical= 4.605170; p_value= 0.192183"

This indicates that drawing with replacement is also a reasonable model.

Question 10.3.4

If the hour of birth was uniform we would expect to have the fraction 4⁄ 24 of the n = 2650 be born between midnight and 4 AM. We can test if the actual number 494 is “close” to the expected number using goodness-of-fit tests. In the R code for this problem we find [1] "d= 7.441208; d_critical= 2.705543; p_value= 0.006375"

This indicates that we can reject the null hypothesis of uniform probabilities.

Question 10.3.5

To use the techniques from chapter 6 we apply the fact that a Z-transformed expected number of success is approximately normally distributed and we can compute approximate P-values using the cumulative normal distribution. Using this fact in the R code for this problem we get [1] "z= 2.727858; z_critical= 1.644854; p_value= 0.003187"

This gives the same conclusion as before, i.e. to reject the null hypothesis. In this case it the null hypothesis that we are rejecting is the statement: that X 1 , the number of babies born between midnight and 4 AM, is a binomial random variable with n = 2650 and p 0 = . The statistic in Chapter 6 is the square root of the statistic considered in this chapter.

Question 10.3.6

In the R code for this section I find [1] "d= 10.018097; d_critical= 3.841459; p_value= 0.001550"

This indicates that we can reject the null hypothesis and conclude that there is evidence to suggest that a higher proportion of schizophrenics are born in the early months of the year.

Question 10.3.7

In the R code for this section I find [1] "d= 12.226151; d_critical= 11.070498; p_value= 0.031817"

This indicates that we can reject the null hypothesis at the 5% level but not the 1% level. This gives some evidence to the claim that the distribution of colors is not as is stated.

Question 10.3.8

In my version of the book the table given didn’t look exactly like it coded the information in the question statement. To work this problem I’ll take the first row to be the statement that the American League (AL) won four games of a nine year stretch meaning that the National League (NL) won the other five years as the World Series is played every year. Thus all rows combined we have n = 50 examples during which we had k = 22 National League wins. This is to be compared with the expected number of = 25 using goodness-of- fit tests. In the R code for this problem we find [1] "d= 0.720000; d_critical= 3.841459; p_value= 0.396144"

This large of a P-value means we can’t reject the null hypothesis and conclude that there is no evidence that the probability p that the American League wins p ≠ .

Question 10.3.9

If the winner of a race was independent of the starting position we would expect the probability a horse wins from any position to be uniform. As there are eight starting positions we expect 1⁄ 8 of the winners to come from any given position. As there are a total of 144 races this would be an expected number of 18 winners from each spot. We can use goodness-of-fit tests to see if the actual number of winners from each spot differs significantly from this number. In the R code for this question we find [1] "d= 18.222222; d_critical= 14.067140; p_value= 0.011007"

Thus we can reject the null hypothesis at the 5% level and conclude that it looks like starting position is a factor in which horse wins a given race.

Question 10.3.10

Given the assumed normal distribution for the random variable y we first need to compute the model probability of the classes. These are the numbers

and etc for the other p i values for the other suggested ranges of y values. In computing the last bin we will want to satisfy the np i ≥ 5 requirement. This means that we might need to suggest a binning, compute the expected count in each bin, and combine any bins that don’t have the above count requirements. At first I tried bins with the following “edges” [1] 220 230 240 250 260 270 280 290 300

These gave the expected frequencies of [1] 0.714 2.790 7.460 13.662 17.141 14.735 8.679 3.501

We see that some of these expected frequencies are “too small”. If we don’t adjust our bins to enforce np i ≥ 5 (i.e. just keep these bins) a direct application of the χ 2 goodness-of-fit test gives [1] "d= 12.783606; d_critical= 12.017037; p_value= 0.077560"

This indicates that we can not reject the null hypothesis at the 5% level.

Question 10.3.11

Here n = 50 and we need to compute the probabilities of y to be in the expected intervals.

Computing the probabilities for each category p i we find p 1 = ∫ 0 1 y 2 dy = p 2 = ∫ 1 2 y 2 dy = p 3 = ∫ 2 3 y 2 dy =

.

We can compare the expected frequencies against the observed frequencies using a goodness-offit test. In the R code for this problem we find [1] "d= 23.521203; d_critica1= 5.991465; p_value= 0.000008"

which indicates that we can easily reject the null hypothesis.

Question 10.4.1

If the responses are independent then in each home questioned we expect the number of “Yes” to be binomially distributed with some unknown probability p . We can take all of the responses and estimate p . Doing this we find p ≈ 0. 54166. We can then use this estimate to compute the expected number of responses (from the 200) to be of any of the four categories and then use a goodness-of-fit test to validate if the binomial model is a good one or not. In the R code that works this problem we find [1] "d= 11.603748; d_critical= 5.991465; p_value= 0.003022"

We would reject the null hypothesis.

Question 10.4.2

If this data can be described with a Poisson pdf. Then we would expect

We can estimate λ using the data given where we find = 0. 5. Using this estimate and the adjusted goodness-of-fit tests we compute [1] "d= 0.759981; d_critical= 11.344867; p_value= 0.859013"

Indicating that the data does appear to be generated from a Poisson distribution.

Question 10.4.3

As in the previous question we first estimate λ and find = 2. 52963. Using this value of λ we can compute the expected Poisson probabilities and then the expected counts in each. This gives us some bins with less than five samples. Thus we need to combine bins. Combining the last eight bins into one we get the following expected counts [1] 21.52 54.43 68.84 58.05 36.71 18.57 11.89

We can then use goodness-of-fit statistics (now with t = 7) where we find [1] "d= 46.753982; d_critical= 11.070498; p_value= 0.000000"

With a P-value this small we should reject the null hypothesis. The Poisson pdf requires an independence assumption between plots. This might not be true as neighboring plots of plants might contaminate each other.

Question 10.4.4

Our hypothesis is that the number of horse kick fatalities in a given year is a Poisson random variable. We estimate the mean number of horse kicks in a year to be = 0. 61. Applying a goodness-of-fit test with his value of λ I get [1] "d= 0.705366; d_critical= 11.344867; p_value= 0.871941"

This result indicates that there is no reason to doubt the validity of a Poisson model.

Question 10.4.5

We first need to estimate the parameter λ in the exponential distribution. In the R code for this question we find this to be λ = 0. 8226121. Using this value we can compute the probabilities for each of the bins and the expected counts. For the expected counts we get [1] 118.31 51.97 22.83 10.03 4.41 1.94 0.85 0.37

Note that some of these bins have less than the desired count of five or more samples. Using goodness-of-fit tests on this (as is) we find [1] "d= 7.064702; d_critical= 12.591587; p_value= 0.314905"

Thus we cannot reject the null hypothesis and an exponential model looks to be a good fit. We would get the same qualitative conclusion (fail to reject) if we had combined bins to ensure that np i ≥ 5 as is done in the book and several of the questions.

Question 10.4.6

For this question we first need to propose “breaks” (in the nomenclature of the R hist command) such that the expected number of points that fall in each bin is np i ≥ 5. For this problem I took them to be [1]

830

880

930

980 1030 1080

Using the normal assumption I then compute the expected number of points that should fall in these bins. I get [1]

5.850 11.893 14.502 10.607

4.652

As each of these numbers is larger than five (or very close to five) I did not choose to combine bins to make them larger. Since we have estimated two parameters (the mean and the standard deviation) we find [1] "d= 10.824457; d_critical= 4.605170; p_value= 0.004462"

These results indicate that the normal model should be rejected.

Question 10.4.7 (the number of boys)

For this problem we estimate the probability of a boys being born to be p = 0. 5333333. We then get an expected number of families with t ∈{ 0, 1, 2} boys born (from our n = 120) to be [1] 26.13333 59.73333 34.13333

Using these we compute [1] "d= 0.612245; d_critical= 3.841459; p_value= 0.433944"

These results indicate that we cannot reject the binomial model.

Question 10.4.8

We will created breaks in the data given and see if the observed number in each bin from the n = 100 we started with is close enough to the expected number. As I am assuming that the data came from a uniform distribution there are no parameters to estimate and this question can be worked with the material from the previous section. I find [1] "d= 2.400000; d_critical= 16.918978; p_value= 0.983453"

The large value found for the P-value means that we cannot reject the null hypothesis.

Question 10.4.9

Looking back at that case study we see that we estimated the value of λ to be 3.87. We can then run the R code associated with this question where we find [1] "d= 12.974499; d_critical= 18.307038; p_value= 0.225101"

With a P-value this large we cannot reject the null hypothesis.

Question 10.4.10

If we think this data comes from a Poisson model we first have to estimate what λ would be under such a model. We find = 0. 3628319. Then using the R code associated with this section of the book we have [1] "d= 0.584396; d_critical= 3.841459; p_value= 0.444594"

This large a P-value indicates that we cannot reject the Poisson hypothesis.

Question 10.4.11

We first have to estimate p where we find p = 0. 2617801. Then we can compare expected frequencies with observed frequencies where we find [1] "d= 9.276668; d_critical= 7.814728; p_value= 0.025830"

Here I have had to combine a large number of bins to get approximately five or more expected samples. This small of a P-value indicates that we should reject the null hypothesis.

Question 10.4.12

We would expect that if the numbers were indeed random that we would have equally likely chance of getting any of the numbers between one and n . As we don’t know n we must estimate it, decide on a binning, count the number of samples in each bin, and determine if these numbers are “close enough” to the actual observed frequencies. As the largest winner has a number of 115 we might take that number as our estimate of n . We then introduce “breaks” at the locations [1]

0

20

40

60

80 100 120

Then we would get expected frequencies for the n = 50 tickers of [1] 8.333333 8.333333 8.333333 8.333333 8.333333 8.333333

this is to be compared with the actual frequencies in these bins for the data which are given by [1]

9

5

7 11

8 10

Using this we can compute

[1] "d= 2.773913; d_critical= 9.487729; p_value= 0.596344"

This means that we can’t reject the null hypothesis of random data.

Question 10.5.1

For this problem we will use the R function expected _count _under _null developed when working through the book [1 ] . Using this function will make working these exercise easier.

When we run the R code for this question we get [1] "d2= 2.766916; df= 1; p_value= 0.096231"

Based on this we can not reject the null hypothesis and must conclude that these to groups are independent.

Question 10.5.2

In the R code for this question we get [1] "d2= 1.537446; df= 1; p_value= 0.214998"

Based on this we cannot reject H 0 and must conclude that the two variables are independent.

Question 10.5.3

In the R code for this question we get [1] "d2= 42.244521; df= 3; p_value= 0.000000"

Based on how small the P-value is we should reject H 0 and must conclude that the two variables are not independent.

Question 10.5.4

In the R code for this question we get [1] "d2= 50.340434; df= 1; p_value= 0.000000"

Based on how small the P-value is we should reject H 0 and must conclude that the two variables are not independent. This means that the risk of an abnormal birth is affected by when a rubella infection is contracted.

Question 10.5.5

In the R code for this question we get [1] "d2= 4.793779; df= 1; p_value= 0.028563"

Based on the P-value we can conclude that aspirin use and breast cancer are related.

Question 10.5.6

In the R code for this question we get [1] "d2= 3.814363; df= 4; p_value= 0.431712"

Based on the P-value is we cannot reject H 0 and conclude that the two variables are independent.

Question 10.5.7

In the R code for this question we get [1] "d2= 12.609028; df= 1; p_value= 0.000384"

Based on how small the P-value is we should reject H 0 and must conclude that the two variables are not independent.

Question 10.5.8

In the R code for this question we get [1] "d2= 10.291451; df= 3; p_value= 0.016244"

This means that at the 5% level we can reject H 0 and must conclude that the two variables are not independent.

Question 10.5.9

In the R code for this question we get [1] "d2= 2.197166; df= 1; p_value= 0.138265"

Based on this we cannot reject H 0 and must conclude that the two variables are independent.

Question 10.5.10

Part (a): In the R code for this question we get z = − 0. 2736586 Part (b): The fact that z 2 = 0. 08 and that was the value of d 2 found in the case study is because the statistic in the chi-squared test is the square of the “difference in means statistic” used in computing z .

Chapter 11 (Regression)

Question 11.2.1

When we run the R code for this section we get the following estimate of the coefficients in the linear model (Intercept) 25.232

x 3.291

The prediction from this model when x = 18 is 84.47201.

Figure 7: Left: A scatter plot of the data and least squares linear model for Question 11.2.2. Right: A residual plot of the data for Question 11.2.3.

Question 11.2.2

When we run the R code for this section we get the plot given in Figure 7 (left).

Question 11.2.3

When we run the R code for this section we get the residual plot given in Figure 7 (right). The residual plot looks strange in that for large regions of the Temperature variable the data points appear to be increasing as temperature increases. This might be an indication that adding nonlinear terms would help the regression.

Question 11.2.4

For the left plot, notice that all of the points are above the line y = 0. Thus the residuals are all positive which is strange. This implies that there is a bias to this data. Removing this bias would give a better fit to the data.

For the right plot, notice that the residuals alternative above and below the line y = 0 in a pattern (if a point is above y = 0 the next point will be below the line y = 0 and vice versa). This implies that there is an autocorrelation to this data. Removing this autocorrelation would give a better model fit.

Question 11.2.5

The value of x = 12 is very far from the range of the x values used to produce the model. This is extrapolation which is known to perform poorly in many cases.

Question 11.2.6

This type of a residual plot could be observed if the data comes from two “clusters” around different regions in the x domain. One might worry if the model was not expected to be the same (i.e. linear) in both of these regions.

Figure 8: Left: A scatter plot of the data and least squares linear model for Question 11.2.7. Right: A residual plot of the data (and model) for Question 11.2.8.

Question 11.2.7

When we run the R code for this section we get the plot given in Figure 8 (left). Notice that there are two points in the lower left corner that are very far from the linear fit line. These two schools don’t seem to fit the pattern of the others. In fact there are several schools that seem to fall far from the linear fit. It would seem very likely that the reason they are such poor fits has nothing to do with the level of spending. We find the coefficients of the linear model given by (Intercept) 81.088

Spending_Per_Pupil 0.412

From these coefficients, it looks like increasing the spending per pupil will increase the graduation rate percentage (as the coefficient b > 0).

Question 11.2.8

Part (a): For the model

by using the R code in this section I compute Coefficients: (Intercept) 0.2546

Plant_Diversity 1.7026

Part (b): A residual plot is given in Figure 8 (right). The residual plot has no pattern indicating that a linear fit is appropriate.

Figure 9: Left: A residual plot of the data (and model) for Question 11.2.9. Right: A residual plot of the data (and model) for Question 11.2.10. See the text for more details on characteristics of this plot.

Question 11.2.9

The linear model for this problem has coefficients (Intercept) 114.716

Index_of_Exposure 9.231

and the residual plot for this model is given in Figure 9 (left). Based on this residual plot the linear model seems to be plausible.

Question 11.2.10

The linear model for this problem has coefficients (Intercept) 6.106

x 4.997

and the residual plot for this model is given in Figure 9 (right). Based on this plot a linear fit looks reasonable. One thing that is “strange” about this data is that the points alternative above and below the least squares fit line. This can be seen in the residual plot by noticing that the data points sequentially fall above and below the line y = 0. This implies that there is an autocorrelation to this data. Removing this autocorrelation would give a better model fit.

Question 11.2.11

The linear model

for this problem has coefficients (Intercept) 0.6078

Plumage_Index 0.8407

and the residual plot for this model is given in R code associated with this problem. Based on this plot a linear fit looks reasonable.

Question 11.2.12

In the derivation in the book of the least squares estimates of a and b (under the “proof” section) we get the following linear equations na +

b = ∑ i =1 n y i

a+

b = ∑ i =1 n x i y i .

If we write this as a linear system we get the expression given.

Question 11.2.13

If we use the expression for a given by a = y − b x we get a prediction ŷ (x ) given by

In this expression if we let x = x we get ŷ ( x ) = y . Thus the least squares line passes though the point ( x , y ).

Question 11.2.14 (regression through the origin)

The least square estimator for β 1 is obtained by finding the value of

1

such that the residual sum

of squares RSS(β 1 ) is minimized. Taking the derivative of the given expression for RSS( with respect to 1 and setting the resulting expression equal to zero we find

1

)

or

Solving this expression for

1

we find (65)

To study the bias introduced by this estimator of β 1 we compute

showing that this estimator is unbiased. To study the variance of this estimator we compute

the requested expression. An estimate of is given by the usual

which has n − 1 degrees of freedom.

Question 11.2.15

Using the formula from the previous question we estimate = 0. 03543993.

Question 11.2.16

Part (a): For this model we would minimize the following residual sum of squares

Taking the derivative of this expression with respect to a we get

If we set this equal to zero and solve for a we get (67) Part (b): For this model we would minimize the following residual sum of squares

Taking the derivative of this expression with respect to b we get

If we set this equal to zero and solve for b we get (68) Question 11.2.17

For this problem we use Equation 68 with a ∗ = 100. Using this formula we get b = − 5. 190269.

Question 11.2.18

Part (a-b): In the model y = bx we would estimate b = 0. 06116. Using this model we would predict y = b (120) = 11. 0088.

Question 11.2.19

The residual sum of squares (that we would want to minimize) for this problem would be

Taking the derivatives of the RSS with respect to a , b , and c and then setting the resulting expressions equal to zero we get = − 2 ∑ i =1 n (y i − a − bx i − c sin(x i )) = 0 = − 2 ∑ i =1 n (y i − a − bx i − c sin(x i ))x i = 0 = − 2 ∑ i =1 n (y i − a − bx i − c sin(x i )) sin(x i ) = 0.

These become the linear equations ∑ i =1 n y i − na − b ∑ i =1 n x i − c ∑ i =1 n sin(x i ) = 0 ∑ i =1 n x i y i − a ∑ i =1 n x i − b ∑ i =1 n x i 2 − c ∑ i =1 n x i sin(x i ) = 0 ∑ i =1 n y i sin(x i ) − a ∑ i =1 n sin(x i ) − b ∑ i =1 n x i sin(x i ) − c ∑ i =1 n sin(x i ) 2 = 0.

These simplify to na + b ∑ i =1 n x i + c ∑ i =1 n sin(x i ) = ∑ i =1 n y i a ∑ i =1 n x i + b ∑ i =1 n x i 2 + c ∑ i =1 n x i sin(x i ) = ∑ i =1 n x i y i a ∑ i =1 n sin(x i ) + b ∑ i =1 n x i sin(x i ) + c ∑ i =1 n sin(x i ) 2 = ∑ i =1 n y i sin(x i ).

Question 11.2.20

Part (a): We want to fit the model y = ae bx to this data. Taking the logarithm of both sides we can fit a linear model to determine log(a ) and b . We can then extract a from log(a ), In the R code for this problem I compute [1] "Nonlinear model parameters: a= 104.239332, b=

Part (b): For this I compute [1] "halflife=

4.737895"

-0.146299"

Figure 10: Left: A residual plot for Question 11.2.21. Right: A plot of the data and the model for Question 11.2.22.

Question 11.2.21

Part (a): As in the previous question we will take the logarithm of y and fit a linear model. We can then transform the coefficients back to get a and b in the model y = ae bx . We find [1] "Nonlinear model parameters: a=

4.677850, b=

0.048440"

Part (b): The model predicted debt in 2007 is 8.365494 trillion which is close to the projected 9 trillion Part (c): A residual plot for this data is given in Figure 10 (left). Notice that the residual steadily decrease to the point x = 6 and then increase from that point onwards. As the profile of the residual plot is quadratic looking (as a function of x ) this means that a better model would be given by

That is, we expect the residual from the model y − ae bx to be a quadratic when viewed as a function of x .

Question 11.2.22

Part (a): If we fit a model of the suggested form we find [1] "Nonlinear model parameters: a= 15474.512035, b=

-0.101949"

A plot of the data and the exponential model fit are given in Figure 10 (right).

Part (b): The model predicted price would be 5041.76.

Part (c): The difference in price between the the new car and the model price of a car one year old x = 1 is 16200 − ae b = 2225. 345. This is a substantial fraction (13%) of the initial car price.

Question 11.2.23

I find the model parameters for y = ae bx given by [1] "Nonlinear model parameters: a= 819.620452, b=

0.128467"

Question 11.2.24

Part (a): The solution of this differential equation is of the given form for y . Taking the derivative of the expression for y will also give the differential equation.

Part (b): Taking the logarithm of both sides of y = ae bx would give

so the points (x i , log(y i )) are linear. The points (x i ,y i ) would be linear on logarithmic (in y ) paper.

Question 11.2.25

Part (a): For the model y = ax b we find [1] "Nonlinear model parameters: a=

13.486560, b=

10.537981"

11.996195, b=

0.876444"

Question 11.2.26

Part (a): For the model y = ax b we find [1] "Nonlinear model parameters: a=

Part (b): Using the model just derived when x = 2500 we predict ŷ = 11406. 33.

Figure 11: Left: A plot of the data and the model for Question 11.2.27. Center: A plot of the data and the model for Question 11.2.30. Right: A plot of the data and the model for Question 11.2.31.

Question 11.2.27

For the model y = ax b we find [1] "Nonlinear model parameters: a=

0.074161, b=

1.436862"

A plot of the data an the model is given in Figure 11 (left).

Question 11.2.28

For the model y = a + we find y = 1. 536 + ŷ = 80. 844.

. If x = then using the model just found we find

Question 11.2.29

These are all transformations that if one apples to the proposed nonlinear (x,y ) relationship will give a linear relationship for (a,b ) that can be fit with least squares.

Question 11.2.30

Using the transformation proposed in the book we can get a linear relationship for a and b .

Fitting this transformed expression we can estimate these two parameters We find [1] "Nonlinear model parameters: a=

5.384099, b=

-1.293225"

A plot of the data and the nonlinear model is given in Figure 11 (center).

Question 11.2.31

Running the R code for this problem I find [1] "Nonlinear model parameters; a=

5.558722, b=

-0.133141"

A plot of the data and the nonlinear model is given in Figure 11 (right).

Question 11.3.1

When we fit a linear model to this data we get coefficients given by (Intercept) 13.8

age -1.5

All of the information needed to perform a hypothesis test that β 1 ≠ 0 is given in the summary command when applied to the linear model. That command even provides a P-value which for this data I find to be 0.2535. This indicates that we cannot reject the null hypothesis that β 1 = 0.

Figure 12: A plot of the data and the model for Question 11.3.2.

Question 11.3.2

Part (a): I compute a 95% confidence interval on β 1 of [1] -0.835352

1.659352

Part (b): The fact that zero is a member of the above confidence interval means that we cannot reject the null hypothesis and can’t conclude that β 1 ≠ 0.

Part (c): A plot of the data and linear model are given in Figure 12 . The fact that the slope is not significant means that we can’t make policy decisions on the basis that there is an effect between spending and graduation rate.

Question 11.3.3

We want to determine if the estimate of β 1 found is significant. From the output of the lm command we find a P-value for this parameter of 0.000107. As this is less than 0.01 we can reject the null hypothesis (of no temperature effect) at the 1% level.

Question 11.3.4

The precision of the slope estimate is given by its standard error which is the square root of the estimators variance which is

To make our estimate of β 1 as small as possible we want to make the denominator above as larger as possible. This means that we want our points x i to be as far from the mean x as possible. To do that we could put n points at x = 0 and the other n points at x = 5.

Question 11.3.5

From the text we know that

which means that we can compute

using the normal distribution if we can evaluate the standard error of 1 . Using what we are told about x i we compute this to be 0.8660254. Using this and the above we compute the probability to be 0.9167355.

Question 11.3.6

We start by writing the sum we are given as the following

∑ i =1 n (y i −

0

−

1

x i ) 2 = ∑ i =1 n y i (y i − = ∑ i =1 n y i 2 −

0 0

−

1

x i ) − ∑ i =1 n (

∑ i =1 n y i −

1

0

+

1

x i )(y i −

0

−

1

x i)

∑ i =1 n x i y i

− 0 ∑ i =1 n (y i − 0 − 1 x i ) − 1 ∑ i =1 n x i (y i − 0 − 1 x i ).

This has taken the sum we are given and transformed it into “the sum we want” plus “extra stuff”.

Our goal is now to show that this extra stuff is actually zero. If we consider the fourth sum

we note that this is zero because it is the expression

which is the derivative we take of the likelihood (and require to be zero) in the proof of the least squares estimates for β 0 and β 1 . In the same way the fifth sum

is zero because it is the expression

Question 11.3.7

The formula for the confidence interval for β 0 is given in the book. In the R code for this problem we use that formula to find (66. 55013, 68. 46546).

Question 11.3.8

We would want to test H 0 : β 1 = 0 vs H a : β 1 ≠ 0 which would determine if cancer mortality was related to the index of exposure. In the summary command app1.ed to this linear model we see that the P-value for β 1 is 0.000332 indicating that we can reject the null hypothesis and conclude that there is an effect of exposure on cancer mortality.

Question 11.3.9

For the suggested model we find a P-value for H 0 : β 1 = 0 of 0.00177 indicating that we can reject the null hypothesis.

Question 11.3.10

Recall that Y is given by

Taking the expectations of both sides we get

Now E (Y i ) = β 0 + β 1 X i under the assumptions of the linear model so the above becomes

Question 11.3.11

Recall that the estimate of β 0 is unbiased E [

0

] = β 0 and is given by

As Y and

0

is given by

1

are independent the variance of Var

= Var =

+

+ x 2 Var =σ2

.

Note that this is the proof of Part (d) in Theorem 11.3.2 where this expression is also shown in a slightly simpler form. Using the above expression and the fact that

is a standard normal random variable we can derive a confidence interval for β 0 along the lines of what has been done many times in the text.

Question 11.3.12

Notice that the variability of points for a fixed value of x seems to increase as x does. This violates the assumption that σ 2 is not a function of x . The “center” of the distribution of points at each x value seems to be centered on the simple linear model line which is to be expected if its assumptions are true.

Question 11.3.13

Using the statistics derived in the text for testing the hypothesis that σ 2 = σ 0 2 we find [1] "H0: acceptance region (10.982, 36.781); observed value: 31.778"

as our observed value is inside the acceptance region we cannot reject H 0 .

Question 11.3.14

I compute this to be (1375. 140, 4097. 214).

Question 11.3.15

The model I computed for this data was slightly different than what the book gave. If anyone sees anything wrong with my arguments please let me know. Using my numbers I compute the confidence interval to be (1. 543779, 11. 886283).

Question 11.3.16

Both of these can be easily computed with the R command predict . Using this I compute fit lwr upr 1 13.72868 13.61877 13.83859 fit lwr upr 1 13.72868 13.2871 14.17027 Question 11.3.17

In the R code for this problem I find fit lwr upr 1 2.073504 2.059622 2.087387

Question 11.3.18

In the R code for this problem I find fit lwr upr 1 166.0153 134.0828 197.9479

A health official is likely to be interested in the confidence interval as that would allow him the ability to make policy decisions.

Question 11.3.19

In the R code for this problem I find fit lwr upr 1 194.0105 173.8909 214.1301

The interval above does not includes the specific salary. The prediction interval fit lwr upr 1 194.0105 147.4049 240.6161

does however.

Question 11.3.20

In the R code for this problem I find confidence and prediction intervals given by fit lwr upr 1 197.7987 179.4116 216.1859 fit lwr upr 1 197.7987 159.9099 235.6876 Question 11.3.21

In the R code for this problem I compute [1] "t= -1.424; p-value 0.185"

Thus we cannot conclude there is enough evidence to claim discrimination.

Question 11.3.22

For the R code for this problem I compute [1] "t= -2.588; p-value= 0.036"

thus we can conclude that the rates at which the two administrations lost support are different (at least at the 5% level).

Question 11.3.23

Using the expression for

Now as Y and

1

0

in terms of

1

we have

are independent, by using formulas from the book the variance of Ŷ is given by Var

= Var =

+ (x − x ) 2 Var

+ (x − x ) 2

=σ2

.

The numerator N in this expression can be written as N = ∑ i =1 n (x i 2 − 2x i x + x 2 ) + nx 2 − 2nx x + x 2 n = ∑ i =1 n (x i 2 − 2x i x + x 2 + x 2 − 2x x + x 2 ) = ∑ i =1 n (x i 2 − 2x i x + x 2 ) + ∑ i =1 n (2xx i − 2x i x − 2x x + 2 x 2 ) = ∑ i =1 n (x i − x ) 2 + 2xn x − 2 x (n x ) − 2nx x + 2n x 2 = ∑ i =1 n (x i − x ) 2 .

Using this we have shown that

which is the desired expression.

Question 11.3.24

To start we write the sum on the left-hand-side as ∑ i =1 n (Y i − Y ) 2 = ∑ i =1 n (Y i − Ŷ i + Ŷ i − Y ) 2 = ∑ i =1 n (Y i − Ŷ i ) 2 + 2 ∑ i =1 n (Y i − Ŷ i )(Ŷ i − Y ) + ∑ i =1 n (Ŷ i − Y ) 2 Next consider the second sum. We have

Now in Question 11.3.6 we showed that ∑ i =1 n (Y i − Ŷ i ) = 0 so the second sum is zero and we are left with

Now recall that Ŷ i =

0

+

1

x i so that the above sum can be written as

The first sum is zero from the previous argument and in Question 11.3.6 we showed that ∑ i =1 n x i (Y i − Ŷ i ) = 0. All of this together gives

the desired expression.

Question 11.4.1

To compute

we first will compute E (X ) and E (Y ). The marginal densities f X (x ) is

From this we have E (X ) = 1 E (X 2 ) = 1 2 Var

+2

=

+22

=

= E (X 2 ) − E (X ) 2 =

.

For the Y random variable we have

From this we have E (Y ) = 1 E (Y 2 ) = 1 2 Var

+3

=

+32

= E (Y 2 ) − E (Y ) 2 =

= .

Using these expressions we have Cov

=

+

+

+

=−

= − 0. 01653,

and

Question 11.4.2

We will first compute the marginal distribution of X where we find f X (x ) = ∫ 0 1 f X,Y (x,y )dy = ∫ 0 1 (x + y )dy =

0

1

= x (1 − 0) + (1 − 0) = + x .

The expectation of X and X 2 is then E (X ) = ∫ 0 1 x

dx =

E (X 2 ) = ∫ 0 1 x 2

dx =

0

1

0

= (1 − 0) + (1 − 0) = 1

= + =

.

2 The variance is then Var = − = . As X and Y are symmetric in the density f X,Y (x,y ) the densities and expectations of the random variable Y are the same as that for the random variable X .

To compute Cov

we have

Using this we have

Question 11.4.3

The marginal densities from this joint density are given by f X (x ) = ∫ y =0 x 8xydy =

y =0

f Y (y ) = ∫ x = y 1 8xydy = 8y

x

= 4x 3

x=y

1

= 4y (1 − y 2 ).

The expectations over X are then E (X ) = ∫ 0 1 x (4x 3 )dx = E (X 2 ) = ∫ 0 1 x 2 (4x 3 )dx = Var The expectations over Y are then

= −

2

=

.

0

1

0

= 1

=

E (Y ) = ∫

0

1

y (4y (1 − y 2 ))dy =

E (Y 2 ) = = −

Var

2

=

.

This means that = ∫ x =0 1 ∫ y =0 x

Cov

(8xy )dydx

= 8 ∫ x =0 1 x

∫ y =0 x y

= ∫ x =0 1 x 3

2

This fraction is equivalent to

dydx

dx =

=

.

. From this we have

Question 11.4.4

From the joint density we are given we can compute f X (x ) = f Y (y ) =

.

From these we have

We now compute Cov

Finally we have

E (X ) = 1

+2

=

E (Y ) = 1

+2

+3

+4

E (X 2 ) = 1 2

+22

=

E (Y 2 ) = 1 2

+22

+32

Var

=

Var

=

=

+42

=

.

= E [(X − E (X ))(Y − E (Y ))] =

+

+

+

=

.

Question 11.4.5

To start note that

as Cov Var so

= 0 when a is a constant for any random variable V . Next recall that Var

=b2

as b and d are positive.

Question 11.4.6

To start, using formulas for the sums of powers of integers we have E (X ) = ∑ i =1 n = ∑ i =1 n i = E (Y ) = ∑ i =1 n

=

= ∑ i =1 n i 2 =

=

.

From the definition of Y in terms of X we have E (X 2 ) = E (Y ). Now E (Y 2 ) = ∑ i =1 n

=

∑ i =1 n i 4 =

=

.

Using these we have

and Var

=

−

=

.

Next we have Cov

= E [(X − E (X ))(Y − E (Y ))] = E [(X − E (X ))(X 2 − E (X 2 ))] = ∑ i =1 n

.

Using Mathematica we can evaluate this summation (we can also do it by expanding the terms and using formulas). This gives

Finally ρ (X,Y ) = =

.

If we take the limit as n →∞ we get

Question 11.4.7

Part (a): We have Cov

= Cov = Cov = Var

+ Cov − Cov + Cov − Var .

− Cov

Part (b): For this we have

Note that

Thus using this result and Part (a) we have shown that

Question 11.4.8

Using the fact from the proof of Theorem 11.3.2 in the previous section we have that (69) We will use this expression in the the square root in the two factors in the denominator of the expression for R . For example the first factor becomes

Using an expression like this for the Y variables as well and moving the to an equivalent position in the numerator we get the desired expression.

Question 11.4.9

Now the sample correlation coefficient is given by replacing X i and Y i with the sample values x i and y i in Eq. 11.4.2 in the book. From the formula for 1 in Theorem 11.3.1 we see that

Question 11.4.10

From the summary command we compute R 2 = 0. 5322 or about 53%.

Question 11.4.11

I get a correlation of r = − 0. 02961141 which is negative implying that if there is a effect, then stadiums at higher altitudes should have fewer home runs. It can be shown (using cor.test ) that this correlation is not significant.

Question 11.4.12

I get a correlation of r = 0. 72635895. Using cor.test we can show that this correlation is significant.

Question 11.4.13

On this data I compute r 2 = 0. 5811.

Question 11.4.14

For this data I compute r = 0. 3367317 which would explain r 2 = 0. 1134 of the variability in the yearly return.

Question 11.5.1

From the properties of the bivariate normal distribution f Y (y ) is normal with a mean μ Y and variance σ Y 2 . The we have

Also f Y (y | x ) is normal with a mean and a variance given by the formulas in the book. Namely μ Y|x = μ Y+

(x − μ x ) = 5. 2

σ Y | x 2 = (1 − ρ 2 )σ Y 2 = 7. 4529, using the numbers given in the question. Using these we have

Question 11.5.2

Part (a): As both X and Y are normal random variables so is Y − ρX . Next using formulas for working with covariances we have

As σ Y = σ X the above is zero. As the covariance between X and Y − ρX is and these variables are normally distributed they are independent.

Part (b): We have

by Question 11.4.7. As this is zero by the assumptions in this question we have that X + Y and X − Y are uncorrelated. Because they are also normal random variables they are independent.

Question 11.5.3

Part (a): Given that μ X = μ Y = 0 and σ X 2 = σ Y 2 = 1 we have that the joint density of (X,Y ) is given by

Now to answer this question we will introduce the random variables U and V defined as U =X+Y V =X−Y, and ask what the probability density function of (U,V ) is. The inverse transform of the above is X = (U + V ) Y = (U − V ).

The probability density function of (U,V ) is given by

From the above transformations we have

Taking the absolute value of this we have that p ( U,V ) (u,v ) = p ( X,Y ) (x (u,v ),y (u,v )). The argument of the exponential in the expression for p ( X,Y ) (x (u,v ),y (u,v )) is given by

We can expand the above and simplify to get

This means that our density p ( U,V ) (u,v ) can be written as p ( U,V ) (u,v ) =

exp

exp

=

exp

exp

=

.

Notice that this is the product of the probability density function for two normal random variables with means zero and variances 2(1 + ρ ) and 2(1 − ρ ) respectively. Thus U and V are independent normal random variables. As U = X + Y we have shown that X + Y has a normal distribution.

Part (b): We have

Next Var

= c 2 Var

+ d 2 Var

+ 2cd Cov

= c 2 σ X 2 + d 2 σ Y 2 + 2cdρ (X,Y )σ X σ Y .

Question 11.5.4

From the formulas in the book (and the numbers from this question) we have μ Y|x = μ Y+

(x − μ X ) = 10. 11682

σ Y | x 2 = (1 − ρ 2 )σ Y 2 = 1. 664.

This means that

Next recalling that when we take the average of n observations our variance decreases as Y ∼ N using formulas like the above we find

Question 11.5.5

Note that if we compare this expression to that of a bivariate normal distribution given in Definition 11.5.1 we see that as there are no terms linear in x or y in the argument of the exponent that we must have μ X = μ Y = 0. Because of this the density f ( X,Y ) (x,y ) reduces to (70) If we compare the argument of the exponential in this expression to that given in the question we see that −

=− =

−

=−

(

=

(

=− .

(

These are three equations in three unknowns. From Equation 71 we have (74) From Equation 73 we have

Setting these two expressions equal we get that σ X 2 = 4σ Y 2 or σ X = 2σ Y . Putting Equation 74 into Equation 72 we get

As we know that the ratio of = we can solve for ρ to find ρ = . Using this value of ρ from Equation 71 we get σ X 2 = 4 and from Equation 73 we get σ Y 2 = 1.

Finally we get

when we simplify.

Question 11.5.6

Since the means are known to be zero the joint density takes the form of Equation 70 . If we compare the arguments of the exponential with the expression given in the text we must have −

−

=−a

(

= 2u

(

=−b.

(

These are three equations in three unknowns. From Equation 75 we have (78) From Equation 77 we have

Setting these two expressions equal we get that 2aσ X 2 = 2bσ Y 2 or σ X = into Equation 76 we get

σ Y . Putting Equation 78

or

As we know an expression for the ratio

we can solve for ρ to find

As we must have | ρ |≤ 1 the above expression places constraints on a , b , and u . Using this value of ρ in Equation 75 we get

In the same way using our expression for ρ in Equation 77 we get

Question 11.5.7

If we make α = 0. 01 then we only reject H 0 if the observed t value is outside of − t 0 . 005 , 18 = − 2.

87844 and t 0 . 005 , 18 = 2. 87844. As the observed t value is not outside of this range we cannot reject H0.

Question 11.5.8

We can use the cor.test function in R to determine if the two variables are independent. Using that function we get a P-value of 0.2776. From this we conclude that we cannot reject the null hypothesis that there is no correlation between these two variables.

Question 11.5.9

The cor.test function gives a P-value of 0.9272 indicating that these two variables are not significantly correlated and can assumed to be independent.

Question 11.5.10

For this data the cor.test function gives a P-value of 0.0003742 indicating that these two variables are correlated and cannot be considered independent.

Question 11.5.11

For this data the cor.test function gives a P-value of 0.7564 indicating that these two variables are not correlated and can be considered independent. Note that the sample correlation I get for this data is -0.249395 which is the negative of the number given in the text and if true would indicate that increasing tutoring money would actually decrease the graduation rate.

Chapter 12 (The Analysis of Variance) The F Test The computational problems for this section are worked in the R code section _2.R and may call functions defined in utils.R .

Question 12.2.1

We implement an ANOVA test in the R code for this section. When we run that code we get [1] "alpha: 0.072267"

which indicates that we cannot reject the null hypothesis (equal means for each model) at the 5% level but we can at the 10% level.

Question 12.2.2

When we run the R code for this problem we get α = 0. 00441884 indicating that there is a significant difference between the means of the direction of the earth’s magnetic field at the 5% level.

Question 12.2.3

When we run the R code for this problem we get F = 3. 447365 and α = 0. 04641158 indicating that there is not a detectable difference at the 1% level between the means of the price-to-earnings ratio.

Question 12.2.4

When we run the R code for this problem we get F = 6. 393578 and α = 0. 008064081 indicating that there is a strong detectable difference (at the 1% level) between the means of the yields for different corn varieties.

Question 12.2.5

When we run the R code for this problem we get F = 3. 700306 and α = 0. 06167147 indicating that there is some detectable difference between the means of the ages of pottery shards found at different locations. Thus we can conclude that there is a difference in ages between types (at the cost of making an error about 6% of the time). Note that this result is not strong enough to reject the null hypothesis at the 1% level.

Question 12.2.6

When we run the R code for this problem we get F = 7. 370205 and α = 0. 008167311 indicating that there is a detectable difference between the mean IQ change between the different groups at the 5% level (and also for more stringent criteria).

Question 12.2.7

In the R code for this section we can use the given numbers to compute the missing numbers in the given ANOVA table. The formulas to fill the numbers in this table are presented in the R code. For example, we find SSTR = 271. 36 and MSTR = 67. 84 etc.

Question 12.2.8

Under the assumptions of the analysis of variance we expect to have the same variance in each group. Using R we compute the sample standard deviation in each group to be > tapply( DF$value, DF$treatment, sd ) A B C D 0.5773503 10.8934231 1.7078251 1.4142136

Notice that treatment B has a very large standard deviation and treatment A has a very small standard deviation. Both are very different than the values observed for treatment C and D indicating that the assumptions of ANOVA in this case might be incorrect.

Question 12.2.9

We first recall that the books equation 12.2.4 is given by (79) with C given by (80) Next from the the definition of SSTOT given in the book we have

We can simplify this if we recall that Y ⋅⋅ is given by

thus squaring this we have

Using these two expression for Y ⋅⋅ in terms of T ⋅⋅ in what we have for SSTOT we find SSTOT = ∑ j =1 k ∑ i =1 n j Y ij 2 − 2

T ⋅⋅ + Y ⋅⋅ 2 n

= ∑ j =1 k ∑ i =1 n j Y ij 2 − T ⋅⋅ 2 + T ⋅⋅ 2 = ∑ j =1 k ∑ i =1 n j Y ij 2 − = ∑ j =1 k ∑ i =1 n j Y ij 2 − C , which is the first expression we were to show.

Next recall that the books equation 12.2.5 is (81) To show this we start by recalling the definition of SSTR or

Using

Y⋅j =

∑ i =1 n j Y ij =

T ⋅ j and

Y ⋅⋅ = T ⋅⋅ , we can write SSTR in terms of T ’s as SSTR = ∑ j =1 k n j

2

= ∑ j =1 k n j = ∑ j =1 k

−

∑ j =1 k T ⋅ j +

= ∑ j =1 k

−

+

= ∑ j =1 k

−C,

= ∑ j =1 k

T ⋅⋅ 2 ∑ j =1 k n j −

which is the second expression we were to show.

Question 12.2.10

The trick to problems like this is to find an orthogonal matrix A that transforms a vector x of independent standard normal random variables into another vector z such that one component of z takes the form of an expression we want to prove something about. We can then use Fisher’s Lemma to determine properties of the components of z . For this problem we can get an understanding of how this is done by writing SSTR as SSTR = ∑ j =1 k n j ( Y ⋅ j − Y ⋅⋅ ) 2 by the definition of SSTR = ∑ j =1 k n j ( Y ⋅ j − μ )) 2 − n ( Y ⋅⋅ − μ ) 2 as shown in the book.

We can take the above expression and divide by σ 2 to get

Now consider a vector x with components x j given by

for j = 1, 2, … ,k − 1,k . Note that these are standard normal random variables. Next create an orthogonal matrix A that has a last row given by the vector

and consider the transformation z = Ax . Note that we don’t care what the other elements of the matrix A are only that we can control that last row. With this A the last component of the vector z (i.e. z k ) takes the form

zk =

x1+

x2+

=n1

+

+n2

xk +

= =

+nk

− =

.

Then by Fishers lemma we know that z j are independent standard normal random variables and that

Thus writing the term z k 2 by itself (since we know an expression for it) we have

or if we put in the definition of x j we would have that

Thus as is the sum of k − 1 independent standard normal random variables it has a χ 2 distribution with k − 1 degrees of freedom (this is the definition of a χ 2 random variable) and we have shown the desired result.

Questions 12.2.11-12.2.13

These questions are worked in the R code section _2.R . In that R code we use the aov command to get the F-value and then the t.test command to get the t-value for the different datasets. When we run that code we find [1] "Question 11: ANOVA F: [1] "Question 12: ANOVA F: [1] "Question 13: ANOVA F:

2.035906; t^2= 15.039951; t^2= 1.580811; t^2=

2.035906" 15.039951" 1.580811"

Notice that in each case the square of the t-value is the F-value. One can look at the output from the aov function call in each case to determine if we would accept or reject the null hypothesis of equal means. For the three problems above we would accept, reject, and accept H 0 respectively.

Multiple Comparisons: Tukey’s Method The computational problems for this section are worked in the R code section _3.R and may call functions defined in utils.R .

Question 12.3.1

We implement some R code to study pairwise differences in the mean in the R code for this section.

When we run that code on the data for this problem one of the outputs of the code mean _diff _CI _utils is $CI_ints group_I group_J mean_differences ci_lower_limit ci_upper_limit 1 non light -0.8333333 -15.26362 13.596955 2 non moderate -9.3333333 -23.76362 5.096955 3 light moderate -8.5000000 -22.93029 5.930289 4 non heavy -19.3333333 -33.76362 -4.903045 5 light heavy -18.5000000 -32.93029 -4.069711 6 moderate heavy -10.0000000 -24.43029 4.430289

The pairwise differences that have confidence intervals that don’t include zero are the ones where we can reject the null hypothesis. These would be the comparisons between non and heavy and light and heavy .

Question 12.3.2

When we run the code for this section on the data for this problem one of the outputs of the code mean _diff _CI _utils is $CI_ints group_I group_J mean_differences ci_lower_limit ci_upper_limit 1 Financial Utility -0.40 -6.158472 5.3584719 2 Financial Industrial -5.47 -11.228472 0.2884719 3 Utility Industrial -5.07 -10.828472 0.6884719

Notice that all of the pairwise differences have confidence intervals that include zero and we cannot reject the null hypothesis for any pair.

Question 12.3.3

When we run the code for this section on the data for this problem the res structure gives an F-value of 5.808015 and a P-value of 0.01355532. Thus we can reject the hypothesis of all equal means at the 5% level but not the 1% level. Next note that from of the outputs of the code mean _diff _CI _utils we have $CI_ints group_I group_J mean_differences ci_lower_limit ci_upper_limit 1 Abbott Cutter -69.33333 -217.6667 79.00002 2 Abbott McGaw -192.16667 -340.5000 -43.83332 3 Cutter McGaw -122.83333 -271.1667 25.50002

Notice that the only confidence interval that does not contain zero is the difference between Abbott and McGaw . This is the only pair for which we can reject the null hypothesis of no difference between the means.

Question 12.3.4

When we run the code for this section on the data for this problem we find a P-value of 0.008064081 indicating that we can reject the hypothesis of all equal means. The confidence intervals for the pairwise differences (one of the outputs of the code mean _diff _CI _utils ) are given by $CI_ints group_I group_J mean_differences ci_lower_limit ci_upper_limit 1 Variety_4 Variety_1 -0.6333333 -9.368868 8.1022015 2 Variety_4 Variety_5 -3.6333333 -12.368868 5.1022015 3 Variety_1 Variety_5 -3.0000000 -11.735535 5.7355349 4 Variety_4 Variety_2 -6.7666667 -15.502202 1.9688682 5 Variety_1 Variety_2 -6.1333333 -14.868868 2.6022015 6 Variety_5 Variety_2 -3.1333333 -11.868868 5.6022015 7 Variety_4 Variety_3 -11.5000000 -20.235535 -2.7644651 8 Variety_1 Variety_3 -10.8666667 -19.602202 -2.1311318 9 Variety_5 Variety_3 -7.8666667 -16.602202 0.8688682 10 Variety_2 Variety_3 -4.7333333 -13.468868 4.0022015

Note that there are only significant differences between Variety _4 and Variety _3 and Variety _1 and Variety _3 .

Question 12.3.5

When we run the code for this section on the data for this problem we find a P-value of 0.0007604957 indicating that we can reject the hypothesis of all equal means. The confidence intervals for the pairwise differences (one of the outputs of the code mean _diff _CI _utils ) are given by $CI_ints group_I group_J mean_differences ci_lower_limit ci_upper_limit 1 In_US Abroad -13.33333 -29.50456 2.837893 2 In_US Foreign_National -40.00000 -56.17123 -23.828774 3 Abroad Foreign_National -26.66667 -42.83789 -10.495441

Note that there are significant differences between In _US and Foreign _National and Abroad and Foreign _National .

Question 12.3.6

No. Speaking loosely, we can imagine a situation where μ 1 is ”small” (say zero) while μ 2 and μ 3 are ”large” and both μ 2 and μ 3 take the same value (say 10 6 ). Then as μ 2 and μ 3 are equal we should not reject the hypothesis that μ 2 = μ 3 while we can reject the other two hypothesis.

Question 12.3.7

The Tukey intervals should increase. Recall that k is the number of classes and is the number of data points used to estimate the class means. Now as k increases we will have more ways to incorrectly include that some pairs of the sample means are different when in fact they are not. Thus the width of the confidence interval needs to increase. We can verify this with the following R code

n_over_k = 30 # fixed number of samples to use in estimating the mean q_alpha = 0.05 ks = seq( 2, 10, 1 ) # the number of classes Qs = qtukey( 1-q_alpha, ks, ks * (n_over_k-1) ) # proportional to the CI width plot( ks, Qs, type=’l’, xlab=’k (number of classes)’, ylab=’Q’ ) grid()

Running the above code we indeed see that the widths of the confidence intervals increase.

Question 12.4.1

We can use the R code wANOVA perform an ANOVA test on this data. When we do that we find a Pvalue of 0.0001000303 indicating that we can reject the null hypothesis and conclude that there are differences among the means.

The first contrast we are asked to consider can be written as the coefficients c 11 = 1, c 12 = 0, and c 13 = − 1. A second orthogonal contrast could be c 21 = , c 22 = − 1, and c 23 = . Then this contrast expresses

The null hypothesis, that this is zero, is equivalent to the statement that

When we run the R code for this problem we get [1] "SSTR= 510.711111; SSc1+SSc2= 510.711111" [1] "Contrast 1: SS= 264.033333 f= 11.953219 P-value= 0.000078" [1] "Contrast 2: SS= 246.677778 f= 11.167505 P-value= 0.000129"

The first line shows that the treatment sum of squares is equal to the sum of the two given orthogonal contrasts (as it should be) and the second two lines perform hypothesis tests on the given contrasts.

We would reject the null hypothesis (the contrast equals zero) in both cases.

Question 12.4.2

If we run the R code for this section we get the following [1] "SS=

72.361000 f=

6.847180 P-value= 0.006382"

thus for this contrast we would the null hypothesis.

Question 12.4.3

If we run the R code for this section we get the following [1] "SS= 812.250000 f=

10.186017 P-value= 0.000278"

thus for this contrast we would reject the null hypothesis.

Question 12.4.4

If we run the R code for this section we get the following [1] "SS=

29.500952 f=

3.608084 P-value= 0.048122"

thus for this contrast we would reject (barley) the null hypothesis.

Question 12.4.5

If we run the R code for this section we get the following [1] "Contrast 1: c= -1.250 SS= 4.687500 P-value= 0.142563" [1] "Contrast 2: c= -0.642 SS= 1.122917 P-value= 0.694462" [1] "Contrast 3: c= -2.292 SS= 8.967391 P-value= 0.025079" [1] "SSTR= 14.777808; sum of all contrast SSS= 14.777808"

Here we have reproduced the results from the book on the first two contrasts and introduced the new contrast. We see that for this new contrast we should not reject the null hypothesis.

Question 12.4.6

To find the coefficients c j of a third orthogonal contrast, recall that the coefficients must sum to zero or

They must also satisfy

for all s ≠ t . In this problem we have n j = 3 for all j so the condition that the first contrast is orthogonal of the third means that

The orthogonality of the second contrast with the third means that

If we write these three equations in matrix form we get

Thus the values of the coefficients c j for the third contrast must come from a vector c that in the null space of the above matrix. Using the MASS function Null (we could have done this by hand) we find that the coefficients c1 =0 c2 =0 c3 =− c4 = , will work. For this contrast we find SS C 3 = 66. 66667 and [1] "Contrast 1: SS= 308.166667 P-value= 0.403799" [1] "Contrast 2: SS= 184.083333 P-value= 0.600944" [1] "Contrast 3: SS= 66.666667 P-value= 0.867514"

Thus none of the contrast are significant.

Question 12.5.1

As we are assuming that the data Y ij , comes from a Poisson distribution the transformation we need to apply to stabilize the variance is the square root. In the R code for this section we do that, and then run an normal ANOVA test. From that ANOVA test we find F = 6. 228262 and a P-value of 0.03168215 indicating that we can reject the null hypothesis of equal means.

Question 12.5.2

One thing to check is to make sure that the variance of each group is approximately the same. For the data given here the fifth class with an estimated mean of 1.0 and an estimated standard deviation of 1.0 looks to have a different variance than the other groups (it is equal to 100% of the value of the estimated mean). The statistician using this data would have to decide if this discrepancy is large enough to be a problem.

Question 12.5.3

As this data comes from a binomial random variable we need to apply an arcsin transform to stabilize the variance. When we do that and compute an ANOVA test on these transformed variables we find F = 22. 3378 and a P-value of 0.0003962401. Thus we can easily reject the null hypothesis at the 5% level.

Question 12.A.3.1

Recall that the noncentrality parameter is given by

Here we have k = 5, n j = 3 for all j and μ = 0. This last fact is because

for both H 1 ∗ and H 1 ∗∗ . Thus the above expression for γ simplifies to

If we compute this for the two hypothesis H 1 we find

for H 1 ∗ and

for H 1 ∗∗ . As this second number is larger than the first that F test will have greater power.

Question 12.A.3.2

No. In this case we have

Question 12.A.3.3

The expression in 12.A.3.1 should be

i.e. the numerator in the fraction in the exponent is γt and not γ t . To compute the expectation we will use

Taking this derivative we get

Now we could simplify this but we want to evaluate it at t = 0 so we will just perform that evaluation and then simplify. We find

Question 12.A.3.4

From the previous problem we have

Question 12.A.3.5

Since the moment generating function of the sum of independent random variables is the product of the individual moment generating functions we have

where R = ∑ i =1 n r i and G = ∑ i =1 n γ i . This form tells us that V is a noncentral χ 2 random variable with R degrees of freedom and noncentrality parameter G .

Chapter 13 (Randomized Block Designs) Notes on the F Test for a Randomized Design

In this section there is a discussion on when it helpful and useful to consider a blocked experiment as opposed to using simply a one-way ANOVA table (i.e. ignoring the blocking). On first reading I found that section difficult to understand. On a second reading I think the main point is that if one has the information for blocking one can look at the value of SSB relative to SSTOT and SSE to determine if it is helpful to consider the experiment blocked. Since blocking will increase SSB (from zero) it must reduced the SSE. If blocking reduces SSE quite a bit, meaning that from the total “amount” in SSTOT we can put a large amount into SSB we much will make SSE much smaller and will have reduced the uncertainty by a large amount. If on the other hand SSB is not that large relative to SSTOT there maybe less help in performing the blocking in that particular situation and perhaps future experiments could skip this step.

For many of the problems in this chapter we could try to implement the statistical methods directly but will instead use the canned R procedures to solve them. These routines are standard and error free. For most applications there is no reason to “reinvent the wheel”.

The questions for the book are worked in the R code section _X.R where X is the section number.

Question 13.2.1

In this question the blocking is done via student and then the treatment is whether or not the student is hypnotized. We can use the R function aov to perform this analysis. The summary command on its output gives state student Residuals

Df Sum Sq Mean Sq F value Pr(>F) 1 61.6 61.63 7.199 0.0178 * 14 400.8 28.63 3.344 0.0155 * 14 119.9 8.56

Note that both effects are significant at the 5% level and we would reject the null hypothesis.

Question 13.2.2

Here the blocking factor is the city. We can solve this question using the aov function. When we do that we get the following output network city Residuals

Df Sum Sq Mean Sq F value Pr(>F) 2 17.112 8.556 8.718 0.0168 * 3 4.689 1.563 1.593 0.2868 6 5.888 0.981

Note that the blocking factor city is not significant and perhaps we don’t need to block by this variable. The network variable is significant at the 10% level. Thus we would reject the hypothesis that viewing levels are the same across networks.

Question 13.2.3

The output of the aov command on this data gives additive batch Residuals

Df Sum Sq Mean Sq F value Pr(>F) 1 0.03401 0.03401 4.195 0.0865 .

6 0.01999 0.00333 0.411 0.8483 6 0.04864 0.00811

Which indicates that blocking (the batch number) is not significant and neither is the treatment (at least at the 5% level) and we would not reject the null hypothesis.

Question 13.2.4

The output of the aov command on this data gives

region year Residuals

Df Sum Sq Mean Sq F value Pr(>F) 2 4.042 2.0208 8.276 0.0188 * 3 3.822 1.2742 5.218 0.0414 * 6 1.465 0.2442

Which indicates that both the blocking variable (the year) and the treatment variable (the region) are significant both at the 5% and 1% level respectively. We would reject the null hypothesis in this case.

Question 13.2.5

For this question we need to use the TukeyHSD function. Doing this gives the following table Fit: aov(formula = admission_rate ~ time + month, data = DF_melt) $time

diff lwr upr p adj during-before 2.4166667 -0.09605719 4.9293905 0.0608911 after-before 0.5416667 -1.97105719 3.0543905 0.8518081 after-during -1.8750000 -4.38772386 0.6377239 0.1697362

Here time is the factor representing the time relative to the full moon. As the confidence intervals for each difference include zero we can conclude that there are no significant differences in the means.

Question 13.2.6

The output of the aov command on this data gives quarter year Residuals

Df Sum Sq Mean Sq F value Pr(>F) 3 14.05 4.682 0.249 0.860 4 43.56 10.889 0.580 0.683 12 225.30 18.775

This indicates that both the quarter and the year are not significant in determining the monthly return.

Question 13.2.7

Using the TukeyHSD command we have that Fit: aov(formula = share ~ network + city, data = DF_melt) $network

diff lwr upr p adj cbs-abc -2.925 -5.0743135 -0.7756865 0.0137935 nbc-abc -1.475 -3.6243135 0.6743135 0.1687873 nbc-cbs 1.450 -0.6993135 3.5993135 0.1766083

As the difference between CBS and ABC has a confidence interval that does not include zero we can conclude that there is a difference between these two networks.

Question 13.2.8

Part (a): In the associated R code we find system subject Residuals

Df Sum Sq Mean Sq F value Pr(>F) 3 1709.6 569.9 100.590 9.07e-15 *** 9 193.5 21.5 3.796 0.00337 ** 27 153.0 5.7

Thus we see that both the type of system and the subject affects the injection time.

Part (b); The Tukey confidence intervals are given by Fit: aov(formula = time ~ system + subject, data = DF_melt) $system

diff lwr upr p adj vari_ject-standard -18.30 -21.212933 -15.387067 0.0000000 unimatic-standard -11.42 -14.332933 -8.507067 0.0000000 tubex-standard -9.58 -12.492933 -6.667067 0.0000000 unimatic-vari_ject 6.88 3.967067 9.792933 0.0000036 tubex-vari_ject 8.72 5.807067 11.632933 0.0000000 tubex-unimatic 1.84 -1.072933 4.752933 0.3290943

Notice that several systems have significantly different mean injection times. The only difference in means that is not significant (at most reasonable α levels) is the mean difference between the tubex and unimatic systems. The largest difference seems to be between the vari _ject and the standard systems.

Question 13.2.9

Part (a): The aov command on this data gives Df Sum Sq Mean Sq F value Pr(>F) sleep_stage 2 16.99 8.49 4.129 0.0493 * shrew 5 195.44 39.09 19.000 8.12e-05 *** Residuals 10 20.57 2.06

Note that the blocking factor (here shrew ) is very significant. The treatment factor (here the stage of sleep) is also significant at the 5% level.

Part (b): In this case we want to test for rejection of the null hypothesis

which we can do using contrasts. To use the order of the treatments given in the text we will write this as the contrast C given by

We will estimate this using

I find Ĉ = 1. 116667. The formula for SS C is given in the book and I compute SS C = 4. 987778.

Using these two we can compute an F-value and a P-value for the test of the above null hypothesis. Before presenting those note that an orthogonal contrast is given by a set of coefficients c j such that ∑ j =1 3 c j = 0 and

One set of numbers of this form is c 2 = (0, 1⁄ 2, − 1⁄ 2). For the significance of these two contrasts we find [1] "Contrast 1: SS= 4.988 f= [1] "Contrast 2: SS= 12.000 f=

0.346 P-value= 0.833 P-value=

0.713" 0.454"

For these two contrasts when we add we get

which equals the value res$SSTR as it should.

Question 13.2.10

The first contrast is that

The second contrast is that

The third contrast compares the average of the Bread and Butter responses with that of the average of the Plain and Roast Beef responses. When we run this R code we get the following [1] "Contrast 1: SS= 18.225 f= [1] "Contrast 2: SS= 38.025 f= [1] "Contrast 3: SS= 0.128 f=

0.555 P-value= 1.159 P-value= 0.004 P-value=

0.652" 0.356" 1.000"

From this we see that none of the contrasts are significantly different.

Question 13.2.11

Recall from Theorem 13.3.1 we have (82)

with SSTOT = ∑ i =1 b ∑ j =1 k (Y ij − Y ⋅⋅ ) 2 SSTR = ∑ i =1 b ∑ j =1 k ( Y ⋅ j − Y ⋅⋅ ) 2 SSB = ∑ i =1 b ∑ j =1 k ( Y i ⋅ − Y ⋅⋅ ) 2 SSE = ∑ i =1 b ∑ j =1 k (Y ij − Y ⋅ j − Y i ⋅ − Y ⋅⋅ ) 2 .

Lets start with the expression for SSTOT which we can write as

Now

as Y ⋅⋅ is the grand mean. Using this we have SSTOT = ∑ i =1 b ∑ j =1 k (Y ij − Y ⋅⋅ )Y ij = ∑ i =1 b ∑ j =1 k Y ij 2 − Y ⋅⋅ ∑ i =1 b ∑ j =1 k Y ij = ∑ i =1 b ∑ j =1 k Y ij 2 − If we let C =

.

we get the expression in the book.

Next consider the expression for SSTR. We have

Now we recall that T ⋅ j = b Y ⋅ j and T ⋅⋅ = kb Y ⋅⋅ so we can write the above as SSTR = b ∑ j =1 k

2

= ∑ j =1 k = ∑ j =1 k

T ⋅j−

Now

so the second sum vanishes. With this we have

∑ j =1 k

.

SSTR = ∑ k j =1 = ∑ j =1 k T ⋅ j 2 −

T ⋅ j = ∑ j =1 k T ⋅ j 2 −

∑ j =1 k T ⋅ j

,

which is the result in the book.

Next consider the expression for SSB. We have

Now we recall that T i ⋅ = k Y i ⋅ and T ⋅⋅ = kb Y ⋅⋅ so we can write the above as SSB = k ∑ i =1 b

2

= ∑ i =1 b = ∑ i =1 b

T i⋅−

∑ i =1 b

.

Now

so the second sum vanishes. With this we have SSB = ∑ i =1 b = ∑ i =1 b T i ⋅ 2 −

T i ⋅ = ∑ i =1 b T i ⋅ 2 −

∑ i =1 b T i ⋅

,

which is the result in the book.

Question 13.2.12

We start with

To find the maximum likelihood estimate of L with respect to β i and μ j we take the derivatives of L with respect to these parameters, set the results equal to zero, and then solve for them. One of these conditions is

or solving for β i we get (83) Which is valid for i = 1, 2, … ,b − 1,b . The other condition is

or solving for μ j we get (84) Which is valid for j = 1, 2, … ,k − 1,k . Taken together Equations 83 and 84 are a system of kb equations in kb unknowns.

I think the equations as they are given are not of full rank and thus an additional constraint needs to be imposed before they will have a unique solution. To see this consider the case where k = 2 and b = 3. Then the equations above are given by β 1 = Y 1 ⋅ − (μ 1 + μ 2 ) β 2 = Y 2 ⋅ − (μ 1 + μ 2 ) β 3 = Y 3 ⋅ − (μ 1 + μ 2 ) μ 1 = Y ⋅ 1 − (β 1 + β 2 + β 3 ) μ 2 = Y ⋅ 2 − (β 1 + β 2 + β 3 ).

As a linear system these equations take the form

One can show that the rank of the coefficient matrix in the above is four and not five as would be required to have a unique inverse. I’m not sure exactly what constraint we should add to get a unique solution. If anyone sees how to proceed with this question please contact me.

Question 13.2.13

Part (a): This is false. Notice that the left-hand-side and the right-hand-side of this proposed equality can be written as ∑ i =1 b Y i ⋅ = ∑ i =1 b

= ∑ i =1 b ∑ j =1 k Y ij

∑ j =1 k Y ⋅ j = ∑ j =1 k

= ∑ i =1 b ∑ j =1 k Y ij .

These are not equal unless k = b .

Part (b): If this is true then we can never have a situation where the MSE is larger than both MSTR and MSB. As the only constraint on our mean square error (or sum of square errors) is that they satisfy Equation 82 . Thus we can imagine a situation where there is no information due to the treatment or due to the blocking and thus SSTR and SSB are very small. Then the rest of the “variance” in SSTOT must be in SSE and for appropriate choices for k and b the given statement would be false.

Question 13.3.1

Using the code from the R function t.test we can test to see if on a yearly basis the Global Rock returns are larger than the Lipper returns. The confidence interval for the mean difference does not have both endpoints positive and thus we cannot conclude that Global Rock returns are larger than the Lipper returns.

Question 13.3.2

Using the R function t.test we conclude that the mean difference between the number of trials needed to learn depth perception is smaller for mothered lambs (as the confidence interval for the mean difference has both endpoints negative).

Question 13.3.3

Using the R function t.test we conclude that our confidence interval on the mean of the difference (based on α = 0. 05) includes zero and we cannot reject H 0 . The confidence interval for a value of α = 0. 01 will be larger than the one for α = 0. 05 and thus also include zero yielding the same conclusion.

Question 13.3.4

Using the R function t.test to test the differences between the number of correct responses when waking vs. hypnotic. We perform a two sided test to see if there is any difference. The confidence interval for this difference is negative which means that we can reject H 0 and conclude that when hypnotized a person is able to correctly recognize more cards. This is the same conclusion obtained in the previous question that used this data.

Question 13.3.5

We will use the R function t.test to perform this test. When we run the R code for this problem we see that the confidence interval for the difference contains zero and therefore we cannot reject the null hypothesis at this level of significance. This result agrees with the one that we found in the previous question that used this data (i.e. Question 13.2.3). Note that if we square the t value for this test we get (− 2. 0481) 2 = 4. 194714 which is the F value found in the previous section with this data. The same conclusion would hold for the t and F critical values.

Question 13.3.6

The statement needed to derive confidence bounds on the mean μ D is that

This can be used to derive the 100(1 − α ) % confidence interval for the true paired difference mean or

This interval is in the output from the function t.test . Using that routine on this data we get 95 percent confidence interval: -0.1121851 1.0521851 Question 13.3.7

Using the output from the t.test from the previous question 13.3.3 we find 95 percent confidence interval: -0.2138918 0.4305585 Question 13.3.8

Warning: I ran out of time working on this problem. What you will find below is my attempt at it to the point where I got stuck. If anyone sees anyway of continuing my work to get the desired expression please contact me. I’m including these notes in the hopes that they might help someone find a solution.

The goal for this problem is to show that

Where the fact that there are only two treatment types k = 2 has been used. To start we will derive some expressions for Y ⋅⋅ in terms of Y ⋅ j and Y i ⋅ . From the definition of Y ⋅ j (the average down the columns) we have that

(85) This is because the expression b Y ⋅ 1 is the sum of the elements in the first column. From this we see (to be used later) that Y ⋅ 1 − Y ⋅⋅ = Y ⋅ 1 − Y ⋅ 2

(

Y ⋅ 2 − Y ⋅⋅ = Y ⋅ 2 − Y ⋅ 1 .

(

Next in the same way, from the definition of Y i ⋅ we have that (88) Now lets consider the expression for SSTR. We find

Using Equations 86 and 87 the above we can write SSTR = b = = = ( Y⋅1−Y⋅2)2= D2.

This is because D is the sample average of the pairwise difference i.e.

This is one-half of the expression we will need.

As an aside we consider an expression for the sample variance. Starting with its definition

Then we can write this as S2 = =

∑ i =1 n Y i (Y i − Y ) − ∑ i =1 n Y i 2 −

∑ i =1 n Y (Y i − Y )

∑ i =1 n Y i (89)

=

∑ i =1 n Y i 2 −

Y2.

This means that the sample variance of the pairwise differences is given by SD2 =

∑ i =1 b (Y i 1 − Y i 2 ) 2 −

D2

=

∑ i =1 b (Y i 1 − Y i 2 ) 2 −

( Y⋅1−Y⋅2)2.

(90)

Next we expand the expression for SSE. With each term on its own row for better readability (perhaps) we have SSE = SSTOT − SSTR − SSB = ∑ i =1 b ∑ j =1 2 (Y ij − Y ⋅⋅ ) 2 − ∑ i =1 b ∑ j =1 2 ( Y ⋅ j − Y ⋅⋅ ) 2 − ∑ i =1 b ∑ j =1 2 ( Y i ⋅ − Y ⋅⋅ ) 2 = ∑ i =1 b

−b

− 2 ∑ i =1 b ( Y i ⋅ − Y ⋅⋅ ) 2

= ∑ i =1 b (Y i 1 2 − 2 Y ⋅⋅ Y i 1 + Y ⋅⋅ 2 + Y i 2 2 − 2 Y ⋅⋅ Y i 2 + Y ⋅⋅ 2 ) −b − 2 ∑ i =1 b ( Y i ⋅ 2 − 2 Y ⋅⋅ Y i ⋅ + Y ⋅⋅ 2 ) = ∑ i =1 b (Y i 1 2 + Y i 2 2 ) − 2 Y ⋅⋅ ∑ i =1 b (Y i 1 + Y i 2 ) + 2b Y ⋅⋅ 2 − 2b Y ⋅⋅ 2 − b ( Y ⋅ 1 2 + Y ⋅ 2 2 ) + 2b Y ⋅⋅ ( Y ⋅ 1 + Y ⋅ 2 ) − 2 ∑ i =1 b Y i ⋅ 2 + 4 Y ⋅⋅ ∑ i =1 b Y i ⋅ − 2b Y ⋅⋅ 2 In the second term we have

and in the sixth term we have

and in the eighth term we have

by Equation 88 . Using all of these our expression for SSE now becomes

If we introduce terms to make this better match the right-hand-side of Equation 90 (multiplied by b − 1) we have SSE = ∑ i =1 b (Y i 1 2 − 2Y i 1 Y i 2 + Y i 2 2 ) − b ( Y ⋅ 1 2 − 2 Y ⋅ 1 Y ⋅ 2 + Y ⋅ 2 2 )

+ 2 ∑ i =1 b Y i 1 Y i 2 − 2b Y ⋅ 1 Y ⋅ 2 − ∑ i =1 b (Y i 1 + Y i 2 ) 2 + 2b Y ⋅⋅ 2 Simplifying this some SSE = ∑ i =1 b (Y i 1 − Y i 2 ) 2 − b ( Y ⋅ 1 − Y ⋅ 2 ) 2 + 2 ∑ i =1 b Y i 1 Y i 2 − ∑ i =1 b (Y i 1 2 + 2Y i 1 Y i 2 + Y i 2 2 ) − 2b Y ⋅ 1 Y ⋅ 2 + 2b

2

.

Simplifying the second line SSE = ∑ i =1 b (Y i 1 − Y i 2 ) 2 − b ( Y ⋅ 1 − Y ⋅ 2 ) 2 − ∑ i =1 b (Y i 1 2 − 2Y i 1 Y i 2 + Y i 2 2 ) + ( Y ⋅ 1 2 − 2 Y ⋅ 1 Y ⋅ 2 + Y ⋅ 2 2 ), and finally

If we could argue that this last two terms vanish it seems like we almost have the desired expression for F = T 2 (except for perhaps a factor of two). At this point I had to stop working on this problem. If anyone sees how to continue it further please contact me.

Chapter 14 (Nonparametric Statistics) The questions for the book are worked in the R code section _X.R where X is the section number.

Question 14.2.1

For this question using the exact binomial PDF I find [1] "k= 8; p_value=

0.109375"

Question 14.2.2

Running the R code for this problem I get [1] "exact_prob= 0.003693; norm_approx= 0.002254"

Using either of these probabilities we would reject the null hypothesis.

Question 14.2.3

For this problem we want to perform the test

Here log(2) is the true median for an exponential distribution as suggested. Running the R code for this problem I get [1] "p_value= 0.396144"

which indicates that we cannot reject the null hypothesis and there is not strong enough evidence to conclude that the true median of this data is not log(2).

Question 14.2.4

As the normal distribution is symmetric under the null hypothesis that μ = μ 0 we have that

Thus if we let X be the random variable recording the number of times that Y i ≥ μ 0 then in the limit of large n we have that is approximately standard normal. Following the ideas in Theorem 14.2.1 from the book for the alternative hypothesis that μ > μ 0 we have that we reject H 0 is Z ≥ z α for α = 0. 05. For this value of α we find z α = 1. 644854. The power of this test is the probabilities that we reject the null hypothesis when μ = 11. This is the probability that

For this problem we have that n = 22 so we can evaluate the right-hand-side of this to get 14.85752. Under H a the variable X is a binomial random variable with parameter p given by

This later expression is given by

using R . We then want to ask what is the probability that X > 14. 85 or X ≥ 15. Again using R I compute this to be 0.5071814 which is the expression for the power.

Question 14.2.5

The values of Y + and their probability are given by a binomial random variable with p = 1⁄ 2 and n = 7. Its values are given by > dbinom( 0:n, n, 1/2 ) [1] 0.0078125 0.0546875 0.1640625 0.2734375 0.2734375 0.1640625 0.0546875 [8] 0.0078125

To determine the various α levels we would pick a threshold t between 0 and n and then determine what P (Y + ≥ t ) would be for each of these thresholds. Using R I compute these values

to be [1] 0.9921875 0.9375000 0.7734375 0.5000000 0.2265625 0.0625000 0.0078125 [8] 0.0000000 Question 14.2.6

As the data for this problem is a ratio it may not have a Gaussian distribution and as such I will use the large sample sign test to see if the median value for this data is close to the proposed value of 0.618. In the R code I compute [1] "p_value= 0.654721"

Indicating that we cannot reject the null hypothesis that the true median of this data is 0.618.

Question 14.2.7

Running the R code written for this problem I find [1] "large sample sign test p_value= 0.409273" t = -1.7091, df = 18, p-value = 0.05231

It is somewhat troubling that the P-values are so different between the tests. This would indicate conflicting conclusions. If anyone sees anything wrong with what I have done please contact me.

Question 14.2.8

Our hypothesis test to use is given by

Here H stands for hypnotic and W stands for waking. Using a large sample sign test we get [1] "p_value= 0.035351"

This indicates that we should reject the null hypothesis. This is the same conclusion obtained in Question 13.3.5 when using this data.

Question 14.2.9

In this case we have n = 28 and k = 19. This gives a P-value of 0.058782 and thus we cannot reject H 0 .

Question 14.2.10

The probability that any sample Y i is larger than a threshold t when Y i is given by a uniform random variable over (0,𝜃 ) is

for 0 ≤ t ≤ 𝜃 . If we want the 25th percentile then we would have t and 𝜃 such that p = 0. 25.

Under H 0 we would have t = 6 and = 0. 25 so 𝜃 = 24. To set up a large sample test let Y + be defined as

or the count of the number of times that our sample is larger than the value of 6. Then Y + is a binomial random variable with p = 0. 25 and n = 36. For a large enough sample the mean of Y + would be np = 9 and the variance of Y + would be np (1 − p ) = 6. 75. Then the variable Z

is approximately standard normal. We can use Z to determine a sign test. Specifically we would reject H 0 if

If 7 is the true 25th percentile the we have 𝜃 = 28 and Y + is a binomial random variable now with

To determine the probability we commit a Type II error we need to know what is the probability that | Z | < z α in this case. In the R code for this problem I get a probability of a type II error of zero.

Question 14.2.11

Running the R code for this problem we find [1] "k= 3; p_value= 0.343750"

This agrees with the result in Chapter 13 Section 3 Question 6 that used the t-test in that both test conclude not to reject the null hypothesis.

Question 14.3.1

Running the R code for this problem we get [1] "w= 9; p_value= 0.242188"

The critical region (for rejection of H 0 ) is

From the book to evaluate α we need p W (w ). To get that we need to compute the coefficients of e it in ∏ i =1 8 (1 + e it ). Using Mathematica we find

With a P-value this large we cannot reject H 0 .

Question 14.3.2

To compute p W (w ) we need to compute

For the potential different α values we could use to test H 0 : evaluate

for w 2 ∗ taking its values from

,

− 1,

=

0

vs. H 1 :

>

0

we need to

− 2, … . Using R we get the values

> print( cumsum( rev(pw) ) ) [1] 0.03125 0.06250 0.09375 0.15625 0.21875 0.31250 0.40625 0.50000 0.59375 [10] 0.68750 0.78125 0.84375 0.90625 0.93750 0.96875 1.00000 Question 14.3.3

Using the Wilcoxon signed rank test we compute [1] "w= 76; p_value= 0.363488"

and we cannot reject the null hypothesis.

Question 14.3.4

To see if there was any effect at all we want to consider a two sided test Running the R code for this problem we get [1] "w= 34; p_value= 0.694887"

Given how large the P value is we cannot reject the null hypothesis and must conclude that alcohol has no effect on writing speed.

Question 14.3.5

For this problem I find [1] "w= 58; p_value= 0.068249"

As this P-value is less than the required 10% we can reject the null hypothesis. Comparing this result to that from Question 14.2.7 we see that the t-test in that problem gave the same conclusion as above. Note the answer in the back of the book gets w = 61 which I think is a typo.

Question 14.3.6

For this problem I find [1] "w= 13; p_value= 0.139414"

which indicates we would not reject the null hypothesis at the α = 0. 05 level. This conclusion agrees with that of both Question 14.2.11 (using the sign test) and Question 13.3.6 (using the ttest).

Question 14.3.7

The signed rank test would have larger power as it assumes more information about our distribution than does the sign test. In other words, it imposes more constraints on the type of distributions it can work with. In general, a statistical test that requires more information (like the fact that a t-test is assuming normally distributed random variables) will have a greater power (when that information is true) than a test that does not require that information. The sign test is only using yes/no or binary values while to use the signed rank test we must be able to rank (requiring more information) our data. As our distribution is assumed symmetric we can assume that we can rank the data and we have that required property and the signed rank test would have more power.

Question 14.3.8

For this problem I find [1] "w= 0; p_value= 0.001474"

which indicates we would reject the null hypothesis at the α = 0. 05 level. This conclusion agrees with what we get if we use a t-test on this data.

Question 14.3.9

For this problem and using a one-sided test assuming alcohol shortens lifespan I find [1] "w_prime= 72.50; p_value= 0.029832"

which indicates we would reject the null hypothesis. This conclusion also agrees with what we get when we use a t-test on this data. In that case I’m finding (partial)

t = -1.8341, df = 17.134, p-value = 0.04203

which again indicates that the mean age at death is significant at the 5% level.

Question 14.3.10

For this problem I find [1] "w_prime= 140.50; p_value= 0.007285"

which indicates that we can reject the null hypothesis. This result also agrees with a t-test of the data.

Question 14.4.1

For the problems in this section we will use some of the R code written for the excellent book [1 ] . The formulas for the Kruskal-Wallis test presented in that book are the same (with the exception of changes to notation) and using this code will speed up the development of these solutions.

When we run the R associated with this problem we get [1] "B= 5.637; B_crit= 5.991; p_value= 0.059693"

As our B value is not larger than the critical value we cannot reject the null hypothesis.

Question 14.4.2

For this problem I get [1] "B= 0.026; B_crit= 3.841; p_value= 0.870991"

and we would not be able to reject the null hypothesis.

Question 14.4.3

For this problem I get [1] "B= 1.681; B_crit= 3.841; p_value= 0.194851"

and we can reject the null hypothesis.

Question 14.4.4

For this problem I get [1] "B= 9.126; B_crit= 3.841; p_value= 0.002520"

and we can reject the null hypothesis. This result agrees with the analysis done in Question 12.2.12.

Question 14.4.5

For this problem I get [1] "B= 10.715; B_crit= 7.815; p_value= 0.013371"

and we can reject the null hypothesis.

Question 14.4.6

Part (a): For this problem I get [1] "B= 8.106; B_crit= 5.991; p_value= 0.017372"

and we can reject the null hypothesis.

Part (b): Using an ANOVA test for the means > summary(am) Df Sum Sq Mean Sq F value Pr(>F) Plant 2 403931 201966 1.401 0.264 Residuals 27 3891027 144112

Which indicates that the difference in the means due to the Plant variable is not significant.

Part (c): Performing the indicated change and rerunning the Kruskal-Wallis test I find [1] "2246 => 1500: B= 8.106; B_crit= 5.991; p_value= 0.017372"

which again indicates that the medians are different (i.e. we can reject the null hypothesis).

Part (d): Performing the indicated change and rerunning the Kruskal-Wallis test I find Plant Residuals

Df Sum Sq Mean Sq F value Pr(>F) 2 678907 339453 4.241 0.025 * 27 2161203 80045

which now indicates that the means are different (i.e. we can reject the null hypothesis).

Question 14.4.7

For this problem I get [1] "B= 12.481; B_crit= 5.991; p_value= 0.001949"

and we can reject the null hypothesis.

Question 14.4.8

To start, consider the expression Z j 2 . We have

Thus the product we need to sum is given by Zj2 = = If we sum this expression for j = 1, 2,

−

+

.

k − 1,k using the facts that ∑ j =1 k n j = n and

This last expression is true because the ranks are the numbers 1, 2, … ,n − 1,n and in the above expression we are sum all of these numbers. Using this we get ∑ j =1 k

Zj2 = =

∑ j =1 k R ⋅ j 2 −

+ 3(n + 1)

∑ j =1 k R ⋅ j 2 − 3(n + 1),

which is the expression for B in the book.

Question 14.5.1

Running the R code for this question gives [1] "g= 8.800; g_crit= 9.488; p_value= 0.066298"

As our P-value is larger than 0.05 we cannot reject the null hypothesis.

Question 14.5.2

Running the R code for this question gives [1] "g= 1.500; g_crit= 5.991; p_value= 0.472367"

With a P-value this large we cannot reject the null hypothesis. This agrees with the analysis in Case Study 13.2.3.

Question 14.5.3

Running the R code for this question gives [1] "g= 17.042; g_crit= 5.991; p_value: 0.000199"

With this P-value we would reject H 0 in favor of H a .

Question 14.5.4

Running the R code for this question gives [1] "g= 6.200; g_crit= 5.991; p_value= 0.045049"

With this P-value we would reject H 0 in favor of H a at the 5% level (but just barely). Using an ANOVA test (which assumes normal random variables) we get Group density Residuals

Df Sum Sq Mean Sq F value Pr(>F) 9 0.04565 0.005073 1.226 0.3393 2 0.02945 0.014723 3.558 0.0499 * 18 0.07449 0.004138

which shows that the density variable is significant at the 5% level also (but again just barely).

Question 14.5.5

Running the R code for this question gives [1] "g= 8.400; g_crit= 9.210; p_value= 0.014996"

With this P-value we would not reject H 0 in favor of H a at the 1% level. Using an ANOVA test (which assumes normal random variables) we get Block method Residuals

Df Sum Sq Mean Sq F value Pr(>F) 4 438.0 109.50 12.81 0.00148 ** 2 260.9 130.47 15.26 0.00186 ** 8 68.4 8.55

which shows that the method variable is significant at the 1% level.

Question 14.5.6

To solve this problem we start by noting that

To evaluate this we note that

because when we summing over the ranks of the i th treatment block we are summing over the integers from 1, … ,k . Thus

Now r ⋅ j is the sum of the j th treatment levels or ∑ i =1 b r ij and by the definition of the “bar” notation r ⋅ j = r ⋅ j . Next we consider the given expression for g where we have g =

∑ j =1 k ( r ⋅ j − r ⋅⋅ ) 2

=

∑ j =1 k

=

∑ j =1 k

=

∑ j =1 k r ⋅ j 2 −

2

∑ j =1 k r ⋅ j +

.

Now in the same way as above we can evaluate ∑ j =1 k r ⋅ j as

Using this in the expression for g we get g = =

∑ j =1 k r ⋅ j 2 −

+ 3b (k + 1)

∑ j =1 k r ⋅ j 2 − 3b (k + 1)

the expression we wanted to show. This expression looks like the treatment sum of square in randomized block design.

Question 14.6.1

Running the R code for this question gives [1] "January Changes: w= 23; z= -0.533; p_value= 0.593967" [1] "Annual Changes: w= 21; z= -1.333; p_value= 0.182617"

In each of these cases we cannot reject the null hypothesis that the percent changes are random.

Question 14.6.2

Running the R code for this question gives [1] "w= 17; z= 0.671; p_value= 0.502001"

We conclude that we cannot reject the null hypothesis.

Question 14.6.3

Running the R code for this question gives [1] "w= 19; z= 1.678; p_value= 0.093276"

We conclude that we cannot reject the null hypothesis.

Question 14.6.4

Running the R code for this question [1] "w= 22; z= 1.042; p_value= 0.297254"

We conclude that we cannot reject the null hypothesis.

Question 14.6.5

Running the R code for this question [1] "w= 25; z= -0.512; p_value= 0.608841"

We conclude that we cannot reject the null hypothesis.

Question 14.6.6

Running the R code for this question [1] "w= 11; z= -1.112; p_value= 0.266028"

We conclude that we cannot reject the null hypothesis.

References [1]

J. L. Devore. Probability and Statistics for Engineering and the Sciences . Pearson Education, 6 edition, 2003.

[2]

S. Ross. A First Course in Probability . Macmillan, 3rd edition, 1988.