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Introduction to statistical method in economics Problem Problem 1

. Σ Suppose that the X sample statistics for a random sample of fi0 observations from a N µ, a population are the following: E = † = 4 SX a.

Construct a 9†% confidence interval for µ. b. Suppose that you want the length of the confidence interval to be no greater than 3.†F. What is the associated confidence level?

Problem 2 . Σ X Assume that E , E fi , ...,X E isna random sample from a N µ, a population.

Write down the distribution of a. ,

b.

Write down the distribution of n

n

.

E —µ S E —µ

ΣX

SX

c. What do your answers imply about the relationship between the t-student and F distributions?

Problem 3 . Σ X Suppose t hat a random sample is drawn from a N µ, a population, X where a is known.

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a.How large of a sample must be drawn so that the confidence interval for µ has a length less than O.Ofia? b.Suppose that the known value of aX is 2 and that you draw a sample with E = †. However, due to time and budgetary constraints, you were only able to draw a sample half as large as the sample you came up with in part a. If we hold the confidence interval length constant, what is our new confidence level? c.What is the 9†% confidence interval for the sample you used in part b? How does it compare to the desired length from part a?

Problem 4 A Cambridge landlord is trying to decide what rent to charge for her apartments. She decides to poll a group of students about how much rent they are able to pay. She asks nine students what they are able to pay per month, E t , and finds the following: fi Σ 9 E = x t= 6OO 9 t=fi fi Σ 9 . Σ X = x t — E = fiOOOO SX 8 t=fi She infers that 9†% of the students in the area can pay between 8400 and 8800 per month. a.

What assumptions are underlying this inference?

What are the potential problems with her inference? How will the size of her confidence interval change as she polls more students? b.Now suppose that the assumptions made in part a are true. Con- struct a 9†% confidence interval for the mean of the underlying distribution of E t . How will the size of this confidence interval depend on the number of students that are polled? How many students would the landlord need to poll to get a confidence interval with a width of 8fi00 or less?

Problem 5 Suppose that you are going to draw a random sample from a Bernouli distribution in an effort to estimate the parameter p. How large of a sample will you need to draw in order to ensure that the width of the 9†% confidence interval for your estimate no greater than 0.0fi? (Hint: assume that n will be large and consider bounds on the quantity p (fi — p).)

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Solution Problem 1 Suppose that the sample statistics for a random sample of fiO observations . Σ X from a N µ, a population are the following: E a.

SX

We know that

= † = 4

E —µ ~ t 9 S Since n is not very large, we cannot assume normality, but must work with the t distribution.

We need to find the number tO.OX†,9 such that . Σ , E —µ Pr —t O.OX†,9 ≤ n ≤ t O.OX†,9 = 0.9† S ,

n

From our tables we find that tO.OX†,9 = 2.262. Then, . Σ , E —µ Pr —2.262 ≤ n ≤ 2.262 = 0.9† S Σ S S = 0.9† . — 2.262 , Pr E ≤ µ ≤ E ‡ 2.262 , n n Σ Σ ¸S ¸S Our random interval is thus E — 2.262 , E ‡ 2.262 , which for this n n Σ Σ X X ¸ sample is equal to † — 2.262 , † ‡ ¸ 2.262 or [3.†F, 6.43]. fiO fiO Σ b.

In general, our confidence level takes the form E — t which is merely a generalization from part a. The length of the interval is S S . E ‡ t Σ —. E —t Σ αƒX,9 , αƒX,9 , n S n = , 2t αƒX,9

S αƒX,9 ¸ n

,E ‡ t

S αƒX,9 ¸ n

Σ ,

n

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Given the length, S, and n, we can solve for tαƒX,9: 3.†F = ,

2tαƒX,9 ,

2 fi0

=

3.†F fi0 4 t 2.82 = Then from the table, we can see that we must chooseαƒX,9 α = 0.02, so we have a 98% confidence level. t αƒX,9

Problem 2 a.

,

n

E —µ S

~ t

n—fi

We can see that this is the case because E —µ , n E . — µΣ/ a , n = √ = .

S (n — fi) SX /aX (n — fi) zX which has a t distribution with n — fi degrees of freedom. .

b.

We can derive this by . Σ E —µ n

n

Σ E —µ SX

Z n—fi / (n — fi)

X

~ F(fi,n—fi)

. Σ X n E — µ / a (fi)X z Xfi/fi = X X / (n — fi) (n — fi) S /a (n — fi) SX zXn—fi which has an F distribution with fi and n — fi degrees of freedom.

X

=

c. The square of a statistic that has a t distribution with h degrees of freedom will have an F distribution with fi and h degrees of freedom.

Problem 3 Since aX is known, we can use the normal distribution rather than the t distribution. . Σ , E —µ a.

To construct find a number Pr the —xconfidence ≤ n interval, we need ≤ xtoO.OX† = 0.9† O.OX† a xO.OX† such that

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From our normal distribution table, we find that xO.OX† = fi.96. Then, we have , E —µ Σ Pr —fi.96. ≤ n ≤ fi.96 = 0.9† a Σ a a Pr .E — fi.96 , ≤ µ ≤ E ‡ fi.96 , = 0.9† n n Σ Σa ¸ Thus our confidence interval is E — fi.96 , E ‡ fi.96¸ an , which has na ¸ a want to find n such that this lenght is equal to length of 2 × fi.96 . We n 0.0fia, so we set the values equal, cancel the aon both sides, and solve for n: a = 0.0fia 2 × fi.96 , n 3.92 0.0fi = 392 , fi†3664 n = Thus we will need quite a large sample! n =formula for the length of the confidence interval is The general b. x αƒX ¸Xan . Given the length, n, and a, we can solve for xαƒX, which will determine the confidence level.

2a = 0.0fia xαƒX , n 0.0fi √ x αƒX fi†3664/ 2 = 2 We then turn to our table to see that this = value fi.39corresponds to α/2 = 0.0823, or a confidence level of 83.†%. a ¸ c.

Again our confidence interval will be Σ E — fi.96 , E ‡ fi.96 ¸ ,an n Σ Σ X we can substitute in to get E — 0.0fi, E ‡ 0.0fi , so the and using a = 2, Σ length is O.O2.

Problem 4 a. Notice that the landlord is constructing a confidence interval for E s , not E . To arrive at 600 T 200, she first assumes that the Es are iid normally distributed, with mean 600 and variance fi00X, and that this variance is known. She then approximates the true confidence interval of 600 T fi.96a by 600 T 2 (fi00). Because the confidence interval is for Es , not E , we cannot appeal to a CLT (and her sample size would be too small in any case), so if the Es are not normally distributed, the confidence interval is wrong. In addition, because she is actually using an estimate of the variance, she ought to be using a t distribution with 8 degrees of freedom instead of a x, which would produce a confidence interval of 600 T 2.306 (fi00). Because

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the landlord is constructing a confidence interval interval of E s , not E , the size of the confidence interval will not decrease as n increases. . b. Here we will construct a 9†% confidence interval for the population mean based on the assumptions that the Es are normally distributed and the variance is known to be fi00.X. Pr —x

≤ O.OX†

n

,

E —µ a

≤ x

Σ O.OX†

=

0.9†

a a Σ = 0.9† ≤ µ ≤ E ‡ xO.OX† , n n . Σ fi00 fi00 = 0.9† Pr 600 — (fi.96) , ≤ µ ≤ 600 ‡ (fi.96) , 9 9 Pr (†34.6F ≤ µ ≤ 66†.33) = 0.9† 39X ¸a ¸ The width of the confidence interval is 2 (fi.96) = . n n want the length of the confidence interval to be less than fiOO, we manipulate ¸ X , which implies n ≤ fi6.

39X n c fi00 to get n > (3.92) . — xO.OX† , Pr E

Thus if we

Problem 5 A confidence interval of O.Ofi is very small, as we have seen before, so we will probably need a very large sample. When n is large, the sample mean, which in this case will be the number of successes divided by the total number of observations (i.e. the percentage of the sample that is a success), will be approximately normally distributed, by the central limit theorem: E —p(fi—p) p . n ~ N (0, fi) (Recall that the variance of a Bernouli random variable is p (fi — p)). We have seen that the length of a 9†% confidence interval will in general have . p(fi—p) the form 2 (fi.96) =¸ a2 (fi.96) . The difficulty in this case is that n n this expression depends on p. However, you can easily show that p (fi — p) . fi = is maximized at p = 0.†, so the length will be at most 2 (fi.96) 4n Thus to guarantee that our confidence interval will be no more than O.Ofi in ¸ X fi.96 width, we set 0.0fi ≤ , andnfind that we must have n ≤ (fi96) = 384fi6.

This number is sufficiently large to justify the use of the CLT.

fi.96̧ n

.

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