SOLUTION MANUAL FOR Statistical Methods 4th Ed Instructor's Donna L. Mohr, William J. Wilson, Rudol

SOLUTION MANUAL COMPLETE CHAPTERS UPDATE

CHAPTER 1

EXERCISE 1 a) Mean = 17.00, SD = 5.53, median = 17.00, Range = 22, IQR = 20 – 13 = 7. b) See boxplot.

c) A typical value for X is 17 which is both the mean and the median. About half the values are between 13 and 20. The boxplot shows that the upper tail is slightly longer, but difference is small and mean and median are equal. This distribution is roughly symmetric.

Chapter 1 – Page 1

EXERCISE 2 Variable

Mean

Median

Variance

Standard Deviation

Shape

WATER

7.125

1.500

452.864

21.28

extremely positively skewed

VEG

1.120

0.00

4.327

2.080

extremely positively skewed

FOWL

75.635

11.5

42197.33

205.420

extremely positively skewed

WATER Stem Leaf

#

Boxplot

14 9 13 12 11 10 9 8 7 6 5 4 3 13 2 1 05667 0 00000000000001111111111112222222222345556779 ----+----+----+----+----+----+----+----+---Multiply Stem.Leaf by 10**+1

1

*

2

*

5 44

0 +--+--+

VEG Stem Leaf 9 0 8 8 0 7 7 0 6 6 5 5 2 4 4 0 3 3 0 2 8 2 00002 1 58 1 0002 0 58 0 00000000000000000000000000000002 ----+----+----+----+----+----+--

# 1

Boxplot *

1

*

1

*

1

0

1

| | | | | +-----+ | + | | | *-----*

1 1 5 2 4 2 32

Chapter 1 – Page 2

FOWL Stem Leaf 14 1 13 12 11 10 9 8 7 6 5 4 3 6 2 124 1 227888 0 00000000000000000000001111112222223335678 ----+----+----+----+----+----+----+----+Multiply Stem.Leaf by 10**+2

# 1

Boxplot *

1 3 6 41

* 0 0 +--+--+

b) Frequency distribution for FOWL Fowl 0 ≤ x < 100 100 ≤ x < 200 200 ≤ x < 300 300 ≤ x < 400 1400 ≤ x < 1500

Frequency 41 6 3 1 1

Midpoint 50.0 150.0 250.0 350.0 1450.0

3

5

approximate mean = 5500/52 = 105.8 approximate variance = 50746

c) Scatterplots fow l 1500 1400 1300 1200 1100 1000 900 800 700 600 500 400 300 200 100 0 0

1

2

4

6

7

8

9

veg

Aside from the single point with extraordinary number of waterfowl, there is little relationship.

Chapter 1 – Page 3

EXERCISE 3 a) INDEX: mean = -0.149 median = -0.155

variance = 1.771

IQR = 1.59 (show stem and leaf, histogram, or boxplot as part of answer)

FUTURE: mean = -0.208 median = -0.3 variance = 0.601 IQR = 0.81 (show stem and leaf, histogram, or boxplot as part of answer) b) The Scatter plot is shown below. There is a clear trend for positive changes in the FUTURE contract value to be associated with positive changes in the NYSE INDEX. Yes, FUTURE can be used to help predict changes in the NYSE index.

index 3 2 1 0 -1 -2 -3 -4 -3

-2

-1

0

1

2

future

Chapter 1 – Page 4

EXERCISE 4 Y1

Stem Leaf 8 14 7 3 6 1237889 5 338 4 025 3 112 2 478 1 668 ----+----+----+----+

# 2 1 7 3 3 3 3 3

Boxplot | | +-----+ *-----* | + | +-----+ | |

Y2

Stem Leaf 6 5679 6 4 5 789 5 22 4 678 4 0024 3 559 3 12333 ----+----+----+----+

# 4 1 3 2 3 4 3 5

Boxplot | | +-----+ | | *--+--* | | +-----+ |

Y3

Stem Leaf 9 4 8 0 7 6 777 5 444 4 000 3 2 7777777 1 333333 0 0 ----+----+----+----+

# 1 1

Boxplot | | | | +-----+ | | | + | *-----* +-----+ |

3 3 3 7 6 1

Chapter 1 – Page 5

Y4

Stem Leaf 7 4 6 5 0 4 578 3 99 2 123357 1 01468 0 1113699 ----+----+----+----+

# 1

Boxplot | | | | +-----+ *--+--* | | +-----+

1 3 2 6 5 7

Y1 is slightly negatively (or left) skewed; Y2 is nearly symmetrically distributed, though without a well defined peak; Y3 is positively (or right) skewed; Y4 is extremely positively skewed. comparison to empirical rule, actual percentage of observations in interval mean

Std dev

Shape

±1 SD (expect 68%)

±2 SD (expect 95%)

±3 SD (expect all)

Y1

4.840

2.107

slight left skew

68%

100%

100%

Y2

4.748

1.262

near symm.

60%

100%

100%

Y3

3.696

2.434

right skew

76%

96%

100%

Y4

2.328

1.888

extreme right skew

64%

96%

100%

The empirical rule works reasonably well even for the skewed distributions.

Chapter 1 – Page 6

EXERCISE 5 a) DAYS: Mean= 15.854 median= 17, variance= 24.324

TEMP: Mean= 39.348 median= 40 variance= 11.152

b) Scatterplot:

days 30

20

10

0 30

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

46

47

temp

Chapter 1 – Page 7

From the scatterplot, there appears to be no definitive relationship between the average temperature and the number of rainy January days.

EXERCISE 6 a) The distribution of EXPEND is extremely right skewed with a mean ($1816M) that is much greater than the median ($1169M), so the best measure of location is the median and the best measure of dispersion is the IQR ($1938M). There are three outlier states with California being an extreme outlier at $15348M. b) Per Capita expenditures are also right skewed, but slightly less so than expenditures. The median is $0.269k ($2,690) and the IQR is $0.139k. c) For Per Capita expenditures, there is one state that is an extreme outlier (Alaska) and another that is a moderate outlier (Delaware). Both these are states with small populations, which may mean that there are a lot of fixed costs in the systems.

EXERCISE 7 Scatter of DFOOT vs. HCRN:

dfoot 6

5

4

3

2 1

2

3

4

5

hcrn

6

7

8

9

Scatter of

Chapter 1 – Page 8

dfoot vs. ht:

dfoot 6

5

4

3

2 20

21

22

23

24

25

26

27

28

29

30

27

28

29

30

31

32

ht

Scatter of HCRN vs. HT:

hcrn 9 8 7 6 5 4 3 2 1 20

21

22

23

24

25

26

ht

31

32

The

strongest relationship exists between DFOOT, the diameter of the tree at one foot above ground level and HT, the total height of the tree. (notice from the scatter plot that it is closest to a linear relationship) One would expect that as the base of the tree increases in diameter the tree would increase in height as well.

Chapter 1 – Page 9

EXERCISE 8 a and b) The histogram for the ages of the cases slightly left-tailed (it would be less so except for the artificial truncation of the uppermost class). But the histogram for the ages of those who died is extremely left skewed – almost all the deaths are among older cases.

c) The death rate increases dramatically after age 50.

Chapter 1 – Page 10

EXERCISE 9 a) Median was 249 and mean 292.4.

The mean is larger than (to the right of) the median, indicating a distribution skewed to the right. Yes, both the stem and leaf plot and the box plot reveal the skewness of the distribution. b) The outliers 955 and 1160 may have resulted from younger patients or from patients diagnosed earlier or in whom the disease was less severe. c) 38 out of 51 patients were in remission for less than one year, so approximately 75%.

EXERCISE 10 a) For the combined group, scores are almost evenly distributed over the interval from 0 to 100.

17.5 15.0 12.5 P e r 10.0 c e n 7.5 t 5.0 2.5 0 0

10

20

30

40

50

60

70

80

90

100

score

Chapter 1 – Page 11

b)

30 25 P e 20 r 0 c 15 e n t 10 s t a t u s

5 0 30 25 P e 20 r 1 c 15 e n t 10 5 0 0

10

20

30

40

50

60

70

80

90

100

60

70

80

90

100

score

30 25 P e 20 r 2 c 15 e n t 10 s t a t u s

5 0 0

10

20

30

40

50 score

As expected, the placement scores among those who passed the class (Status=0) tend to be higher than the scores of those who failed the class (Status=1). Surprisingly, the scores for those who withdrew (Status=2) are as high as those for those who passed. Perhaps this is because those who withdrew do so for a variety of reasons frequently unrelated to their math ability.

NOTE TO INSTRUCTOR: The differences between the groups might best be displayed with boxplots, though boxplots have a tendency to hide multiple modes.

Chapter 1 – Page 12

100

75

s c o r e

50

25

0 0

1

2

status

Chapter 1 – Page 13

EXERCISE 11 a) Since the data is sequential by year, use a time series plot. b) Prices have been increasing, with an especially sharp jump in the period from 2007 to 2007.

EXERCISE 12 a) Reliability (36.7%) and Enthusiasm (25%) are the traits most frequently thought of as most important.

40

Relative Frequency

30

20

10

0 R E LB LE N T H US LF S TG G R M HE LO Q TP U S H Y

TRAIT

Chapter 1 – Page 14

EXERCISE 13 a) 4

Half-Life of drug

3

2

1

0 1

2

3

4

5

6

7

8

9

10

11

Dosage in mg per kg

The initial dosage seems to be higher for drug A but the half-lives are not much different. b) There is not any strong relationship between half-life and dosage, either overall or within the individual drugs. c) Drug A: M = 9.21, SD = 1.14; Drug B: M = 2.67, SD = 0.44;

Typical dosage is much higher for

Drug A. This supports the conclusion in part (a).

Chapter 1 – Page 15

EXERCISE 14 a) IQR = 119-18 = 101, upper fence = 119+1.5*101 = 270.5, upper whisker will stop at 253 and there is one outlier at 300. See boxplot below.

This distribution is very right (or positively) skewed. A small number of homes linger on the market for a long time. b) The mean will be greater than the median.

EXERCISE 15.

The boxplots show that typical wind speeds are higher at 4PM than at 4AM. There is not much difference in the dispersions (the IQR is slightly higher at 4AM but the range slightly higher at 4PM). Both distributions are very right skewed, with numerous outliers on the right side.

Chapter 1 – Page 16

CHAPTER 2 EXERCISE 1 a) Y

50,000

10,000

1,000

10

0

P(y)

1/150,000

5/150,000

25/150,000

1,000/150,000

148,969/150,000

b) mean= (1/150000) * 50000 + (5/150000) * 10000 + (25/150000) * 1000 + (1000/150000) * 10 + (148,969/150,000)*0 = $0.90 c) no, expected value is only $0.90 so there would be an expected (average over many tickets) loss of $0.10 per ticket. d) σ²= (50000-.9)²*(1/150000) + (10000-.9)²*(5/150000) + (1000-.9)²*(25/150000) + (10-.9)²*(1000/150000) + (0-.9)²*(148969/150000) = 20166.5233 σ = $142.01 This distribution is very skewed, so the standard deviation does not reflect the very large (but low probability) prizes, even if you use an interval of mean +3 standard deviations.

EXERCISE 2 a) 0.5, using area of rectangle with base from 0 to 1. b) using formulas for uniform distribution in section 2.4, µ = 1, Variance = 4/12 = 1/3.

EXERCISE 3 a) mean=(0*.3277)+ (1*.4096) + (2*.2048) + (3*.0512) + (4*.0064) + (5*.0003) = 0.9999 Variance= 0.7997 b) µ= np = (5)(.2) = 1

σ² = np(1-p) = (5)(.2)(.8)= .8

Yes, they agree except for small rounding error.

EXERCISE 4 The system will fail if at least one component fails (since they are in series). Hence, they will fail if they do not all work.

Probability of failing = 1 – probability all work = 1 – (0.999)10 = 0.00996

Chapter 2 – Page 1

EXERCISE 5 Arrangement 1: probability of failure = prob [(A1 fails and A2 fails) OR (B1 fails and B2 fails)] = (0.01)2 + (0.01)2 – (0.01)4 = 0.0001999 Arrangement 2: probability of failure = Prob[(A1 fails or B1 fails) AND (A2 fails or B2 fails)] = (0.01 + 0.01 – 0.012)2 = 0.000396 Arrangement 1 is most reliable.

EXERCISE 6 If you use a scientific calculator rather than table A.1, your values may differ slightly. a) 0.1587 b) 0.8413 c) 0.5 – 0.1587 = 0.3413 d) 1 – 0.9332 – 0.0668 e) 0.9808 – 0.1635 = 0.8173 f) look for 0.05 in the column for probability, A is between 1.64 and 1.65 (or use Table A.1A to find it is 1.645, or an inverse normal function on a scientific calculator or software). g) look for 0.025 in the column for probability, C is 1.96

EXERCISE 7 a) P(Y >15) = P(Z > 1)= 0.1587 on TI-84, normcdf(15,1E99,10,5) b) P(8 < Y < 12) = P( -2/5 < Z < 2/5) = .6554 - .3446 = 0.3108 on TI-84, normcdf(8,12,10,5) = 0.3108 c) P(Z < c) = .90 P( Z > c) = .1 c = 1.282 = (Y – 10)/5 Y = 16.41 On TI-84 calculator, invnorm(0.90,10,5) = 16.41

EXERCISE 8 Treating grades as a continuous random variable. a) Prob (Y < 60) = P(Z < (60-76)/14) = P(Z < -1.14) = 0.1265 b) (Y – 76)/14 = 1.28, so lowest score for an A is 93.92, or 94. c) Adding points changes the mean, not the standard deviation. If only 5% are to fall at or below 60, then the new mean must be such that Chapter 2 – Page 2

(60 - µ)/14 = -1.645, or µ = 60+1.645*14 = 83.03. Since the original mean was 76, we need to add 7 points.

EXERCISE 9 The sample proportion favoring the increase has mean = p = 0.20, and variance = p(1-p)/n = (0.2 * 0.8)/250, so std dev= 0.025298 Z= (0.24 - 0.2)/0.025298 = 1.581 P(Y > 1.581) = 0.0571 (or on TI-84, 0.0569) EXERCISE 10 a) µ = 0(0.368)+1(0.368)+2(0.184)+3(0.061)+4(0.015)+5(0.003)+6(0.001)= 1.00 Variance = (0-1)2(0.368)+(1-1)2(0.368)+ . . .+(6-1)2(0.001) = 1.004 b) the mean is correct, but the variance is off a little bit. Perhaps it is rounding error.

EXERCISE 11 Let O be the number of operators and Y = number of calls arriving. If number of calls exceeds number of operators, get a busy signal. Want to choose O so P(Y > O) = 0.05.

Z = 1.645 = (O-40)/(sqrt(40)) →

O = 50.4, so need 50 or 51 operators.

EXERCISE 12 We set µ = np = 200*0.015 = 3. The probability of 0 cases is e-3*30/0! = 0.0498.

EXERCISE 13 standard error of sample mean = 25/sqrt(100) = 2.5 Z1 = (138-140)/2.5 = -0.8

Z2 = (142-140)/2.5 = 0.8

P(138 < Y < 142 ) = P( -.8 < y < .8) = .7881 - .2119 = 0.5762 On TI-84 calculator, normcdf(138,142,140,2.5) = 0.5762 EXERCISE 14 We replace products with lifetime less than A, where P(X < A ) = 0.10. The z-score that corresponds is -1.28, so (A – 1000)/150 = -1.28, hence A = 808. So the manufacturer should guarantee the product for 808 days.

EXERCISE 15 P(Z < z1) = .1 and

P(Z > z2) = .1 Chapter 2 – Page 3

Z1= -1.282 Z1 = -1.282 = (60 - µ)/σ and

and

Z2= 1.282

Z2 = 1.282 = (90 - µ)/σ

Solving two equations for two unknowns yields: σ = 11.70

σ² = 136.9

µ = 75

EXERCISE 16 a) If treatment not effective, probability of survival still 0.3. Probability of at least 2 out of 3 using binomial is 0.216. b) Probability of at least 4 out of 6 using binomial is 0.070. c) In both cases, at least 2/3 of victims survived, but the chance of this happening in a sample of 6 if the treatment is NOT effective is much smaller, and less likely to be due to chance.

EXERCISE 17 For a normal distribution, the quartiles are about 0.67σ above and below the mean, so the IQR is about 1.34σ. The upper fence starts at 0.67σ + 1.5*(1.34σ) = 2.68σ. The probability beyond this is 0.0037. Counting the lower tail as well, there is about 0.0074 probability (0.74%) that a value will be an outlier.

EXERCISE 18 The upper limit for the control chart is 0.05 + 3 * .05 * .95 / 100 = 0.115 . The process was in control throughout the first shift. During the second half of the second shift an upward trend began in the fraction of nonconforming. No single shift was listed as ‘out of control’ until the third shift, but the persistent high nonconforming rates during the second shift could be viewed as a warning.

Chapter 2 – Page 4

Proportion Nonconforming

0.20

0.15

0.10

0.05

0.00 0

2

4

6

8

10

12

14

16

18

20

22

24

Sample Number

EXERCISE 19 The number of different tickets is 53*52*51*50*49*48/6! = 22,957,480 a) There is only one choice of numbers that will win the grand prize, probability = 1/22957480 = 4.356*10-8 b) 1.233*10-5 There are 6 choices for a number to get wrong, and 47 numbers to insert in its place. c) 0.01484

EXERCISE 20 The control limits are 4 ± 3 0.25 / 5 = (3.33,4.67) The process was initially producing windshields that were too thin, shifted to ones that were too thick, and then back to windshields that were too thin.

Perhaps the production control is over-compensating in its attempt to correct the process.

Chapter 2 – Page 5

6

Mean Thickness

5

4

3

2 0

2

4

6

8

10

12

14

16

18

Sample

EXERCISE 21 a) Use binomial with p = 0.05. A: 0.939, B: 0.921 b) Use binomial with p = 0.15. A: 0.322, B: 0.097 c) Plan B is more expensive but has a smaller probability of missing an increase in the true error rate.

EXERCISE 22 a) All people who have recently filed for unemployment benefits in your county. b) If the true proportion p is 10%, then the number in our sample should be a binomial with n = 68 and p = 0.10. Denote the number of those with these feelings in the sample as X. If you are using a calculator or program that has the binomial cdf, then Prob(X ≥ 12) = 1 – Prob(X ≤ 11) = 1 - 0.9638 = 0.0362 c) If the true proportion in the population is 10%, it would be unusual to find 12 or more in a sample of 68. I would conclude that the proportion of unemployed people who have had feelings of sadness recently is most likely more than 10%. d) The sample proportion would be nearly normally distributed with a mean of 0.10 and a standard deviation of

.1 * .9 = 0.0364 . Using the normal distribution, the probability that the sample 68

proportion will be less than 0.05 or more than 0.15 is 0.0848+0.0848 = 0.1696.

Chapter 2 – Page 6

EXERCISE 23 a) 119.2 b) 0.4972 (TI84: 0.4950) c) standard error = 15/sqrt(20) = 3.354, probability = 0.3734 (TI84: 0.3711) EXERCISE 24 a) If n = 50 and p = 0.03, then the expected number of birth defects is 50(0.03) = 1.5. Using the binomial distribution, the probability of X > 3 (note ‘more than twice’) is the same as 1 – Prob (X ≤ 3) = 1 – 0.9372 = 0.0628. b) If n = 150 and p = 0.03, then the expected number of birth defects is 150(0.03) = 4.5. Using the binomial distribution, the probability of X > 9 = 1 – prob (X ≤ 9) = 1 – 0.9845 = 0.0155.

Though the decision rules are very similar, our chances of a ‘false’ spike in the observed proportion of birth defects is much smaller when we have a large sample.

EXERCISE 25 Inspecting Table A.4 with 5 and 5 df, P(F > 5) = about 5%.

On TI84 calculator, get 0.051 = Fcdf(5,1E99,5,5).

Chapter 2 – Page 7

CHAPTER 3 Answers may differ slightly due to rounding.

EXERCISE 1 a) The most serious error would be to need the umbrella and not have it, so Ho: it will rain versus H1: it will not rain. Type I error is when it truly will rain but you say there is evidence it will not.

This is the situation where you would not have an umbrella when you needed it, and you will control the probability of this error at a low level. b) Most serious error would be to fire a good employee – null hypothesis is employee should not be fired. c) Usually consider that there needs to be evidence for a merit raise, hence null hypothesis is no merit and alternative hypothesis is that employee merits a raise. d) Must show evidence for claim, so null hypothesis is claim is not correct. e) Must show evidence for guilt, so null hypothesis is innocent. f) Must serious problem is to have battery run out during test, so null hypothesis is battery needs to be changed. g) Highly unpopular to restrict driving! Would need to justify with strong evidence it would improve achievement. Null hypothesis is that restricting driving does not improve achievement.

EXERCISE 2 a) Ho: µ = 8, H1:µ ≠ 8, reject Ho if x < 7.9 or x > 8.1.

Since the standard error is 0.15/sqrt(16) = 0.0375, the probability of rejecting Ho when the mean is truly 8 (that is, Ho true) is P(Z < -2.67 or Z > 2.67) = 2*0.00383 = 0.00766 = α b) The following type of calculation must be done for several choices of µ, then the resulting β graphed against the choices of µ.

When µ = 8.05, then probability of not rejecting Ho = = P(7.9 ≤ x ≤ 8.1) = P(-4 ≤ Z ≤ 1.33) = 0.9087 = β

EXERCISE 3 a) Do not reject Ho if -1.96 ≤ z ≤ 1.96  9.616 ≤ x ≤ 11.184. Assuming µ = 10.0 and standard error of mean is 2/sqrt(25) = 0.4, then the probability of this happening is P( -0.96 ≤ z ≤ 2.96) = β = 0.8300. On a scientific calculator, you may get 0.8299. b) Do not reject Ho if 9.3696 ≤ x ≤ 11.4304. β =0.9428 (may get 0.9423 on scientific calculator) c) when α=0.05, accept Ho if 10.416 ≤ x ≤ 11.984, β=0.1492; when α=0.01, β=.3372 Chapter 3 – Page 1

d) when α=0.05, accept Ho if 10.542 ≤ x , β=0.0877; when α=0.01, β=0.2514 e) As you reduce the chance of Type I error (reduce α) you increase the chance of Type II error (increase β). As the discrepancy between the true value of µ and the value hypothesized in the null hypothesis increases, your chance of failing to reject Ho decreases – that is, it is more likely that you will be able to ‘prove’ strong effects. One tailed tests have a smaller β than two-tailed tests.

EXERCISE 4

n = 100, µ = 10, σ = 2, standard error = 0.2

a) Ho: µ = 10.4 versus H1: µ ≠ 10.4 at α = 0.05, Do not reject Ho if 10.008 ≤ x ≤ 10.792. β = P(0.04 ≤ z ≤ 3.96) = 0.4840 b) at α = 0.01, do not reject Ho if 9.8848 ≤ x ≤ 10.9152. β = P(-0.576 ≤ z ≤ 4.576) = 0.7190 c) Ho: µ= 11.2 versus H1: µ ≠ 11.2.

For α = 5%, do not reject if 10.808 ≤ x ≤ 11.592; β = P(4.04 ≤ z ≤ 7.96) ≅ 0.000 For α = 1%, do not reject Ho if 10.685 ≤ x ≤ 11.715; β = P(3.4 ≤ z ≤ 8.58) ≅ 0.000 d) Ho: µ = 11.2 versus H1: µ < 11.2 at α = 0.05 Do not reject Ho if x > 10.871; β = P(Z > 4.36) ≅ 0.000 Ho: µ = 11.2 versus H1: µ < 11.2 at α = 0.01 Do not reject Ho if x > 10.735; β = P(Z > 3.67) = 0.0001 e) Comparing the answers here to those of Exercise 3, we see that increasing the sample sizes reduces β.

EXERCISE 5 a) Ho: µ = 100 versus H1: µ ≠ 100. z = (92-100) / (10/sqrt(30)) = -4.38, reject Ho, there is significant evidence that the mean in this class differs from 100. b) P(|z| > 4.38) is approximately 0

Chapter 3 – Page 2

c) Since the test shows this class has grades that are systematically less than is typical, it may be that the teaching methods need to be re-evaluated.

EXERCISE 6 a) Ho: µ = 48750 versus H1: µ ≠ 48750. Reject Ho if |z| > 1.96 z = 1.739 (p value = 0.082). There is no significant evidence that the mean income has changed. b) 51200 ± 1.645*12200/sqrt(75) = (48883, 53517) With confidence 90%, the true mean income is between $48,883 and $53,517. c) Accept Ho if 45988.88 ≤ x ≤ 51511,12. β = 0.1453

EXERCISE 7 a) Ho: p=.4 vs. H1: p=.6 α = probability of getting 4 red in 4 tries assuming Ho is true = (0.4)^4= .0256 b) β = probability of not rejecting Ho assuming H1 is true = P(y ≤ 3) = 1 – P(y = 4) = 1 – (.6)^4 = 0.8704 EXERCISE 8 a) P( x < 76) assuming µ = 80 = P(z < -5.56) ≅ 0.000. b) Reject Ho if x < 80-1.645(0.72) = 78.816 EXERCISE 9 a) E= (1.96)*(8)/12 = 1.31 b) 79.6 ± (1.96)*(8/12) (78.29, 80.91) With confidence 95%, the true mean reaction time is between 78.29 and 80.91s. c) n = (1.96)²*(8)²/(1)² = 245.8 so n = 246 EXERCISE 10 a) Ho: µ = 0.3, H1: µ > 0.3, reject Ho if z > 1.645. z = (0.35-0.3) / (0.4/sqrt(20)) = 0.56, p value = 0.288. There is no significant evidence that the drug produces a meaningful decrease. b) Use α = 1%. The lower the significance level, the stronger the required evidence.

Chapter 3 – Page 3

EXERCISE 11 a) Ho: µ = 100 versus H1: µ < 100 Reject Ho if z < -1.645. z = (91.6-100) / (20/sqrt(8)) = -1.188 There is no significant evidence that the true mean is less than 100.

Rejection region shown as shaded Area to right of -1.645, has area 0.05

b) p value = P(z < -1.188) = 0.1174

p value is area to right of -1.188, as shown as shaded area in graph. Note it is same shape as classic rejection region, but using observed -1.188 as dividing point.

EXERCISE 12 Ho: p = 0.2 versus H1: p < 0.2 where p is probability of exceeding threshold.

Using binomial with n = 8 and p = .2, p value = P(Y ≤ 1) = 0.503 There is no significant evidence that the probability is less than 20%.

EXERCISE 13 Ho: µ = 12.3 versus H1: µ < 12.3 (note, however, that many statisticians prefer the two-tailed test) z= (10.9 – 12.3)/(3.5/sqrt(100)) = -4.0 p value is .00003, much less than 0.05. There is significant evidence that the mean processing time has been reduced.

Chapter 3 – Page 4

EXERCISE 14 The population is all batches of loans that could have been produced by the simulation. In this population, with confidence 95%, the mean error rate would be between 4.57 and 4.63 per batch.

EXERCISE 15 n = (1.96)²*(0.5)²/(.01)²= 9604 EXERCISE 16 a) p = true probability of birth defect in this community Ho: p = 0.03 versus H1: p > 0.03 b) p value = P(Y ≥ 3) assuming p = 0.03, = 1 – P(Y ≤ 2 ) = 0.1178 c) Since the p value is not less than α=0.05, there is no significant evidence that the birth defect rate in this community is elevated.

EXERCISE 17 a) assuming Ho true (p = .03), the probability of rejecting Ho is P( Y ≥ 4 ) = 1 – P(Y ≤ 3) = 1 – 0.9686 = 0.0314

using the binomial distribution with n = 40

b) P(Y ≤ 3) assuming p = 0.10 = 0.4231 EXERCISE 18 a) µ = mean calorie consumption among teenage boys in your county Ho: µ = 2700 versus H1: µ ≠ 2700 Reject Ho if |z|>1.96. z = (2620-2700) / (450/sqrt(36))= -1.07 There is no significant evidence that the mean calorie consumption among teenage boys in this county differs from the national mean. (The p value is 0.286.) b) Reject Ho if x < 2553 or x > 2847. If µ = 2600, the probability of this happening is P(z < -0.63 or z > 3.29) = 0.2643 + 0.0005 = 0.2648.

EXERCISE 19 a) (2473, 2767) b) about 138 or 139

Chapter 3 – Page 5

EXERCISE 20 a) Use a one-sided confidence interval for µ, the mean overpayment among all claims from the chain of clinics.

Lower limit = 21.32 – 1.645(32.45/sqrt(100)) = $15.98 b) It is quite likely that the individual overpayments have a very positively skewed distribution (most overpayments near 0, but a few very large). The large size of the standard deviation versus the mean supports this notion. Hence, we need a large sample so that x will be normally distributed.

EXERCISE 21 a) P(Y ≥ 4) using Poisson distribution with µ = 1.8 = 0.1087 b) binomial with n = 12 and p = 0.1087, P(Y ≥ 1 ) = 0.7486 EXERCISE 22 a) Ho: p = .3 versus H1: p > 0.3, Here are some probabilities, by inspection, of the binomial distribution with n = 10 and p = .3 P(Y > 4) = 1 - .8497 = 0.1503, P(Y > 5) = 1 - .9527 = 0.0473. There is no value with α exactly .10, so we would go as close as we can without exceeding 0.10. So our rejection rule is ‘There is significant evidence of effectiveness if more than 5 (or at least 6) victims survive.’ b) There is no significant evidence that the treatment is effective. c) Power = P(Y > 5) assuming p = .6, power = 0.6331

Chapter 3 – Page 6

CHAPTER 4 EXERCISE 1

µ = mean weight loss among all if put on diet

Ho: µ = 0 versus H1: µ > 0 t = (2.4583-0) / [2.1339/sqrt(12)] = 3.9907 With 11 df, reject Ho if t > 1.7959 So reject Ho, yes a mean weight loss was achieved.

OR p value = 0.0011, which is less than 0.05, so reject Ho. (The peril of a one-tailed test is that if the subjects actually gained significant amounts of weight, we would not catch it!)

EXERCISE 2

x =2.458, s = 2.134 µ is the mean weight loss of all people if put on this diet program Ho: µ=1 versus Ho: µ>1. Reject Ho if t > 2.718 using 11 df. t = 2.367 (p value = 0.0187).

At α=1%, there is no significant evidence that the mean weight loss exceeds 1 pound.

EXERCISE 3

Without assuming the true mean is 0, SS = 0.9217 calculated about the sample mean, and

we have 14 df. Ho: σ² = 0.01 versus H1: σ² < 0.01 (range/4 = 0.1, so want std dev < 0.1) X²=0.9217 / 0.01= 92.17 Reject Ho if X² < 6.571. Do not reject Ho. There is no significant evidence that the watches have the claimed accuracy.

Note that it is also reasonable to take the SS calculated about the claimed value of 0, in which case there are 15 df.

EXERCISE 4

x =1.87, s = 1.719 With confidence 95%, the mean value among all claims is between 1.066 and 2.675 (in thousands of dollars). Since the distribution is quite skewed and the sample size small, this interval may not have the claimed confidence level. This interval is extremely wide, so it is not very helpful in establishing the true value of the population mean.

Chapter 4 – Page 1

EXERCISE 5 a) Ho: p = 0.2, H1: p < 0.2 npo = 6, n(1 − po )= 24 , so yes b) z = (0.167-0.2) / sqrt(0.2*0.8/30) = -0.46, p value = P( z <-0.46) = 0.324. No significant evidence the mill meets the requirement. c) z = -2.28, p value = 0.011 Now there is evidence the mill meets the requirement.

EXERCISE 6 Ho: µ = 500 versus H1: µ < 500

452, s = 56.471, t = y= −4.66, p value < 0.0001.

There is significant evidence that the mean is less than 500.

EXERCISE 7 With 29 df, the critical values from the Chi-squared distribution are 16.0471 and 45.722 (using Microsoft Excel function chisq.inv). The confidence interval for the variance is

(29)(56.4712 ) / 45.722 ≤ σ 2 ≤ (29)(56.4712 ) /16.0471 To get the confidence interval for the standard deviation, take the square roots of the limits, to get that the standard deviation is between 44.97 and 75.91, with confidence 95%.

EXERCISE 8 Ho: µ = 129 versus H1: µ > 129 We have selected a one-tailed alternative in order to detect high blood pressures.

Reject Ho if t > 2.7181. t = (133.0-129)/[13.941/sqrt(12)] = 0.994 (p value 0.1708) There is no significant evidence that the mean blood pressure in this community is higher than normal.

EXERCISE 9

p = proportion of all voters who prefer candidate X

Ho: p = 0.50 versus H1: p > 0.50 z = 0.65328 reject Ho if z > 1.645 Do not reject Ho, she should not reduce campaign funds.

Asymptotic requirement fulfilled: 150*0.5=150*(1-0.5) = 75 EXERCISE 10 ˆ = p

= 79 / 150

0.5267

With confidence 95%, the true proportion of voters favoring candidate X is between 0.4468 and 0.6066.

Chapter 4 – Page 2

ˆ = 79 ≥ 5 and n(1 − p ˆ) = 71 ≥ 5.

This data set is sufficiently large for the normal approximation, np

EXERCISE 11 µ = mean weight of all healthy newborns in this neighborhood Ho: µ=7.5 versus H1: µ < 7.5 t = -1.079 Reject Ho if t < -2.8214, so do not reject Ho. We cannot conclude that the babies from this neighborhood are, on average, underweight.

EXERCISE 12

= x

7.080, = s

1.231

With confidence 99%, the mean weight of all newborns in this neighborhood is between 5.815 and 8.345 pounds.

EXERCISE 13 We need the standard deviation to be less than 0.2/2.576 = 0.07764 Ho: σ = 0.07764 versus H1: σ < 0.07764, reject Ho if X² < 10.117 (lower tail with 19 df) SS=0.1445 X²=23.97. Do not reject Ho. There is no evidence that the units satisfy the regulation.

EXERCISE 14 s2 = 0.0076 , SS = 0.1445.

Upper limit is 0.1445/8.907 = 0.0162 and lower limit is 0.1445/32.852 = 0.0044. With confidence 95%, the variance in the weights of all units is between 0.0044 and 0.0162.

EXERCISE 15 Management will routinely sample washers and calculate the mean and standard deviation. If the null hypothesis that the standard deviation is at most 0.002 is rejected, then they will have to stop production and fix the machine. The choice of significance level depends on how difficult or costly the adjustment is. If it is a quick and cheap adjustment, then they may accept a high chance of a Type I error (unnecessary adjustments) in order to keep quality high. This would equate to a high choice for α. But if the adjustment is very difficult or costly, then they would want to avoid Type I errors by setting α low. A low α would mean more Type II errors, where the machine really does need adjustment but they do not take any action.

EXERCISE 16 Ho: σ = 0.002 versus H1:σ > 0.002

X2

[24 = * 0.00372] / (0.002)2

82.14 Chapter 4 – Page 3

Using 24 df and a TI-84 calculator or Microsoft Excel, the p value is < 0.0001, this would be significant even at a very small α. So the machine should be adjusted.

EXERCISE 17 normality assumption violated, very right (positively) skewed distribution, assumptions are not valid . Try a transformation such as logarithm of claim.

EXERCISE 18 Stem Leaf # 101 9 1 101 04 2 100 56 2 100 01244 5 99 577777 6 99 24 2 98 6 1 98 2 1 ----+----+----+----+ Multiply Stem.Leaf by 10**-1

Boxplot 0 | | +--+--+ *-----* | | 0

The distribution appears reasonably symmetric, but perhaps slightly long tailed (two moderate outliers). If the sample were larger, we might base our test on the proportion of observations more than 0.2 from the 10.0 target value.

EXERCISE 19 1.87546 ± (2.0796)(0.6299/sqrt(22)) = 1.87546 ± (2.0796)(0.134299) (1.596, 2.155) With confidence 95%, the true mean half-life of this drug is between 1.596 and 2.155.

Chapter 4 – Page 4

EXERCISE 20 SS = 8.33275. Using Microsoft Excel function CHISQ.INV to find the 5th and 95th percentile with 21 df, we get 11.591 and 32.67. The lower limit is 8.33275/32.671 and the upper limit is 8.33275/11.591.

With confidence 90%, the variance of the half-life is between 0.2551 and 0.7189.

EXERCISE 21 p = true proportion of customers who also drink another brand Ho: p = 0.10 versus H1: p > 0.10 (H1 contradicts claim) . Reject Ho if z >1.645. z = 2.667 Reject Ho, there is evidence that the proportion is more than 10%. OK to use z-test here because np= 100*0.10= 10, n(1 − po= ) 90, both ≥ 5. o

EXERCISE 22 Ho: p = 0.1 versus H1: p > 0.10 Using α = 5%, we will reject the null hypothesis if the sample proportion is more than 0.1+1.645*sqrt(.1*.9/100) = 0.1506. That is, we will reject the null hypothesis if X ≥15, where X is the number in the sample of 100 who drink another brand. Using the binomial distribution to calculate power = P(X ≥ 15) with n = 100 and p = several choices at or above 0.10 gives:

Note that this critical region is only approximately of significance level 5%. The actual probability that we will reject Ho if p = 0.1 is 7.3% rather than 5%. This is because we used the normal approximation to set the critical region. The binomial is a discrete distribution and there is no rejection region with probability exactly 5%.

Chapter 4 – Page 5

EXERCISE 23 Ho: σ² = 0.01 versus H1: σ² < 0.01 However, since the mean difference should be 0, we need to take the sums of squares around 0 (the uncorrected sums of squares). Hence, the χ² should be based on 15 degrees of freedom. Reject Ho if χ² < 7.261if we use α = 5%. The device does not meet the standard for accuracy. SS = 0.5032, X²= 50.32

EXERCISE 24 a) The 6.285 is an extreme outlier. There are three other values that are moderate outliers: 5.52 on the high side and

6.50

5.115 and 5.125 on the low side.

6.25

b) The extreme positive outlier causes the distribution to be typical pH.

PH

skewed. Hence, medians would be a better descriptor of

6.00

5.75

c) M = population median

5.50

Ho: M ≥ 5.40 versus H1: M < 5.40

5.25

The number of observations strictly larger than 5.40 is 4. If

5.00

the median was 5.4, the probability an observation would exceed 5.4 would be 0.50. Using the binomial distribution, the probability of 4 or fewer successes in 20 tries when p = 0.50 is 0.0059. This is the p value.

There is significant evidence that the median is less than 5.40. Note that if this had been a two-tailed alternative, we would have doubled this to get a p value of 0.0118.

This data set is large enough to use a z test for the proportion,Ho: p = 0.5, z = -2.68, p value = 0.0036.

Answers differ slightly because z test for proportion gets more accurate when samples really large.

EXERCISE 25 pˆ = 22 / 44 = 0.50,

0.50 ± 1.645 0.5*(1 − 0.5) / 44

= (0.376, 0.624)

With confidence 90%, the percentage of all Kintyre residents who would rate windfarms this way is between 37.6% and 62.4%.

EXERCISE 26 Ho: µ = 90 versus H1: µ > 90 Reject Ho if t > 1.383 a) t = -2.68, p value 0.987. Do not reject Ho. There is no significant evidence the mean sound level exceeds 90dBa. Chapter 4 – Page 6

b) t = 2.19, p value 0.028. Reject Ho. There is significant evidence that the mean sound level exceeds 90 dBa. (Pay attention to negative signs!) c) Given the potential for hearing damage, we want to require protection if there is even modest evidence the sound level exceeds the threshold.

EXERCISE= 27 H 0 : p 0.10, H1 : p > 0.10 Since np0= 70*.1= 7 ≥ 5 and

n(1 − p0 ) =70*.9 =63 ≥ 5 we can use the z = 1.195, p value = 0.116. There is no significant evidence that the proportion of times the noise is greater than the threshold exceeds 10%. The company does not need to begin requiring ear protection.

EXERCISE 28 a) p = proportion defective among all parts from this vendor Ho: p = 0.05 versus H1: p > 0.05 b) npo = 20*0.05 = 1 which is less than 5. Hence, cannot use z test for proportions. c) Using the binomial with n = 20, the probability of 4 or more defects in 20 parts when p = 0.05 is 0.0159. d) Since the p value is less than α = 0.05, there is significant evidence that the proportion of defectives exceeds 5%.

EXERCISE 29 a) X 2 = 2.18, 11 df . Critical value from table A.3 = 3.053. There is evidence variance is less than 0.25. b)

boxplot appears symmetric, normality is reasonable

Chapter 4 – Page 7

normal probability plot – points cluster along straight line so that normality is reasonable. b)

EXERCISE 30 p = probability a survey respondent will be Hispanic Ho: p = 0.59 versus H1: p ≠ 0.59 We can use the z test for proportions because 454*0.59 = 268 and 454*(1-0.59)=186 which are both at least 5.

Reject Ho if z < -2.576 or z > 2.576. z =

(0.36 − 0.59) 0.59 * 0.41 / 454

=

−9.96

The proportion of Hispanics in the survey differs from the census figure by more than can be attributed by chance. It is apparently much lower than it should be.

EXERCISE 31 a) p = probability of improving. Ho: p = 0.5 versus H1: p ≠ 0.5 z = 2.49, p value = 0.0127. At α = 5%, there is significant evidence that the proportion improving differs from chance. b) Using the binomial distribution with n = 9 and p = 0.5, the probability of 6 or more out of 9 improving is 0.2539. The two-tailed p value is 0.5078. There is no significant evidence that the proportion improving in Putnam differs from that due to chance.

Chapter 4 – Page 8

CHAPTER 5 EXERCISE 1

Ho: µA = µB versus H1: µA ≠ µB.

Using the pooled t test, there are 23 df, so reject Ho if |t| > 2.0687.

Sp²= 77.885; t = 4.321/3.5329 = 1.223 with 23 df, p value = 0.2337.

Do not reject Ho. There is no significant evidence that there is a difference between methods.

Alternatively, unequal variance test had t’ = 1.219 with 22.5 df, p value = 0.235. Same conclusion.

EXERCISE 2 Using Class B – Class A, the Confidence interval is 4.321 ± 2.0687 77.885 (1 / 12 + 1 / 13) =4.321 ± 7.309

With confidence 95%, the difference in the mean scores when taught by the two methods is between -2.99 and 11.63. This answer uses the pooled version of the test statistic. If you used the unequal variance version on a scientific calculator or statistical software you would get -3.02 to 11.66. However, if you did the unequal variance version and guessed the df as 12-1 = 11, then you would get a wider interval.

EXERCISE 3

Ho: µA = µB versus H1: µA ≠ µB.

Pooled t test: There are 14 df. Reject Ho if |t| > 2.1448.

Sp²= 26.2201/14 = 1.873 t = 1.479, p value 0.1616.

Unequal variance t test: t’ =1.478, df = 14, p value 0.1616.

Do not reject Ho. There is no significant evidence the mean pollution indexes differ for the two areas.

EXERCISE 4 It is now a paired t test. By using pairing, we can eliminate day-to-day variability in the indices, and focus just on the difference in the areas. Taking differences as Area B – Area A, Ho: µ d = 0 versus H1: µ d ≠ 0 d = -1.011, sd = 0.1974, t = -14.49 with 7 df.

Reject Ho if |t| > 2.3646.

There is significant evidence the areas differ in their mean index.

EXERCISE 5

Ho: µNew = µReg versus H1: µNew > µReg or µNew-µReg > 0

Pooled t test has 16 df.

Reject Ho if t > 1.7549 Sp²= 1578.393;

t = 3.136, p value 0.0032 Chapter 5 – Page 1

Unequal variance t test has 13.8 df, t’ = 3.08, p value 0.0042.

Reject Ho. The new diet does seem to result in higher mean weights. (Pooled t test is reasonable here, Ftest comparing variances had p value 0.18.)

EXERCISE 6 Ho: µNew = µReg versus H1: µNew > µNew+25 or µNew - µReg > 25

= yold 845.5, = ynew 904.6, = snew 43.28, = sold 36.73 Pooled t test has 16 df.

Reject Ho if t > 1.7459 Sp²= 1578.393;

t = 1.81, p value = 0.0446,

Unequal variance t test has 13.8 df, t’ = 1.775, p value = 0.0488 Reject Ho. There is evidence the new diet increases mean weight by more than 25 pounds.

EXERCISE 7

p1, p2 the true probabilities that the two machines will produce a defective.

Ho: p1 = p2 versus H1: p1 ≠ p2 ˆ1 = p

ˆ2 .0786, = p

= 0.05, p

.0618,, so then z =1.077, p value 0.281.

There is no significant difference in the reliability of the two machines, at any reasonable significance level.

The asymptotic requirement is satisfied because in the smallest sample,

np=

8.65 ≥ 5

and n(1 − p) =

131.3 ≥ 5.

EXERCISE 8 Paired t test, differences are mpg with device – mpg without device.

Ho: µD = 0 versus H1: µD > 0. Assume α=5%, reject Ho if t > 1.7959. = d

0.7667, = sd

= 1.3647, t

1.95 , p value = 0.0388

At α=5%, there is significant evidence that the device improves mean mileage. [If the device is expensive, it is worth noting that there is NOT evidence if α = 2.5%.]

EXERCISE 9

Ho: σ(new) = σ(standard) versus H1: σ(new) < σ(standard),

or H1: σ(standard)/σ(new) > 1 2 2 F= sstan = (3.9049)²/(3.1862)² = 1.502 with 15, 15 df. / snew

P value = 0.2200, or Reject Ho if F > 2.40.

Do not reject Ho. No significant evidence that the methods differ in their consistency.

Chapter 5 – Page 2

EXERCISE 10 a) This is a paired t test. If d = µafter-µbefore, then Ho: µd = 0 versus H1: µd > 0. = d

= 2.167, sd

= 1.722, t

3.08, = p value

0.0137.

There is significant evidence that the participants feel the seminar improved their knowledge. b) However, it is very possible that only the participants who had the most positive attitude about the seminar were the ones returning the follow-up survey. That is, those who did not feel it was helpful are not represented in the sample. This would cause the sample to appear better than the group as a whole.

EXERCISE 11 Ho: µ1 = µ2 versus H1: µ1 ≠ µ2 If α = 5% and with 8 df, reject Ho if |t| > 2.306 Sp²= 3755.227

t = -1.1798

p value 0.2720

Do not reject Ho. No significant difference in mean level of hydrocarbons.

An F-test for the null hypothesis of equal variances had F = 2.21 and p value 0.41, so the equal variance assumption is reasonable.

EXERCISE 12 a) Independent Sample t test.

Ho: µped can=µobst versus H1: µped can ≠ µobst Unequal variance t’ = 1.887, df=11.4, p value = 0.0848 Pooled t = 1.860, df = 13, p value = 0.0857 There is no significant evidence the groups differ in their mean burnout value. b) F test. Ho: σ2ped can/σ2obst = 1 versus H1: σ2ped can/σ2obst ≠ 1 F = 1.148, p value = 0.820 There is no significant evidence that the variances differ c) Samples are very small, and the F test is particularly sensitive to the normality assumption.

EXERCISE 13 Ho: p1 = p2 versus H1: p1 ≠ p2 where p is probability of improving z = -1.27, p value 0.2038 There is no significant evidence that counties differ in their probability of school improvement.

Asymptotic requirement satisfied, 89*0.261=23.3 ≥ 5, 89*(1-0.261)= 65.7 ≥ 5.

EXERCISE 14 Paired t test. d = µtreated - µuntreated. If treatment works, expect d to be positive.

Ho: µd = 0 versus H1: µd > 0.

Chapter 5 – Page 3

= d

1.425, = sd

= 1.722, t

2.341, = p value

0.0259.

There is significant evidence (at α = 5%) that the treatment improves the mean number of surviving fish.

(Note, however, that if we had a two-tailed alternative, there would not be significant evidence at 5%!)

EXERCISE 15 Ho: µa = µg versus H1: µa ≠ µg The F test shows moderate evidence that the variances differ F(21,20)=2.51, p value = 0.0441). Use the unequal variance version of the t test. Df are at least 20, at which we would reject if |t| > 1.7247, but may be as much as 41, at which we would reject if |t| > 1.6839 t = -0.89 with 35.7 df. Scientific calculators, Excel, or stat software will give p value of 0.3814.

There is

no significant evidence that the drugs differ in the mean of their half-lives.

EXERCISE 16 (Refers to data for Exercise 15)

The half-lives for Aminikacin (“A”) is nearly symmetric, while that for Gentamicin (“G”) has a very slight positive skew. Fortunately, the t test is robust against mild skew.

EXERCISE 17 a) p1 = probability manager in industry 1 will rate reliability as most important.

Ho: p1 = p2 versus H1: p1 ≠ p2 Reject Ho if |z| > 1.96 p = (44 +60)/(120+150) = 0.385

z = -0.0333/0.0596 = -0.56

Do not reject Ho, proportions rating reliability as most important are not significantly different for the two industries.

OR p value = 2*P(|z| > 0.56) = 0.575, no evidence proportions differ.

Chapter 5 – Page 4

Asymptotic requirement fulfilled: 120*0.385=46.2, 120*(1-0.385)=73.8. b) Difference in sample proportions is normal with mean 0.1 and true standard deviation will be sqrt( 0.3*0.7/120+0.4*0.6/150)=0.0579.

Following instructions in problem, assume estimated std error very close to this.

P(z < −1.96

or z > 1.96) =

0.0001 + 0.4085 = 0.4086.

ˆ1 − p ˆ2 | > 0.1134) P(| p =

EXERCISE 18 a) paired t test (final-initial) t = 3.06, one-tailed p value = 0.0189 There is significant evidence of an increase in mean BUN in the control group. b) paired t test (final-initial) t = 0.56, one-tailed p value = 0.3036 There is no significant evidence of an increase in mean BUN in the intervention group. c) independent samples t test comparing the mean change in the two groups. Variances do not differ significantly. Equal variance t = 2.49 with 8 df, p value = 0.0377. There is significant evidence that the groups differ in their mean change. The mean change was apparently less in the intervention group.

EXERCISE 19 a) p = probability of requiring ICU or death Ho: pHCQ = pnoHCQ z = -0.399, p value = 0.69, There is no significant evidence that patients on HCQ differ in their probability of a worsening condition.

Asymptotic requirement fulfilled, 84*.215=18, 84*(1-.215)=66. b) With confidence 95%, the probability of developing a heart problem is between 3.2% and 15.8%.

Asymptotic requirement fulfilled, y=8, n-y=76. c) No clear benefit (as shown in part a) and potential side effect (part b) so this data does not support use of the drug.

EXERCISE 20 p = probability of making the choice that detracts from the imbalance Ho: ppeople = pcontrol versus Ho: ppeople ≠ pcontrol z = -2.73 with p value 0.0063 There is significant evidence the groups differ in their probability of making the choice that detracts from the imbalance.

EXERCISE 21 p = probability of spending more time looking at the new goal events Chapter 5 – Page 5

Ho: preach = pfirst versus H1: preach ≠ pfirst z = 2.556, p value = 0.0106 At α = 5%, there is significant evidence that the proportions differ by more than would be expected by chance, but not at α = 1%.

Check for sufficient sample size: p = 0.50 , np = n(1 − p) = 7.5 ≥ 5 in each group.

EXERCISE 22 Independent sample t tests (unequal variance version) are used to compare the means in each group.

Exhaustion: t’ = -8.06, p value = 0 Cynicism: t’ = -2.30, p value = 0.0216 Efficacy: t’ = 0.39, p value = 0.6966 This is consistent with the authors’ statement. Police were significantly different (apparently lower) for both exhaustion and cynicism. The difference does seem to be greater for the exhaustion scale.

EXERCISE 23 McNemar’s test. There are 36 discordant pairs, split 11 to 25.

Testing Ho: p = .5 using the z test for proportions, z = -2.33, p value = 0.0196. There is significant evidence of a difference, with the wife apparently more likely to favor background checks than the husband.

Asymptotic requirement: 36*.5 = 18 ≥ 5

Chapter 5 – Page 6

CHAPTER 6 EXERCISE 1 There are 40 observations and T = 5 groups, so DfB between is 5-1=4 and DfW = 40-5=35 Source

Df

SS

MS

F

Between

4

25

6.25

1.411

Within

35

155

4.429

Total

39

180

F(4,35)=1.411, p value = 0.2506. There is no significant evidence that the mean number of words memorized differs for any group.

EXERCISE 2 Since there are 4 groups, there are dfB = 4-1 = 3 and dfW=60-4 = 56.

Source

Df

SS

MS

F

Between

3

155

51.67

3.23

Within

56

896

16.0

Total

59

1051

F(3,56) = 3.23 with p value = 0.029 (using TI 84 calculator or Excel). There is significant evidence that at least one type of insulation gives a different mean change in the energy consumption.

EXERCISE 3 TSS = 29*(3.424)2 = 339.99, SSW = 9*(2.52+32+32) = 218.25 Source

Df

SS

MS

Between

2

121.74 60.87

Within

27

218.25 8.08

Total

29

339.99

F 7.53

F(2,27)=7.53, p value = 0.0025. There is significant evidence that the mean number of words memorized differs for at least one group.

EXERCISE 4

SSW= 5(4.22 + 5.32 + 5.02 + 4.92 )= 473.7, overall mean = 23.55, 2 ] 6*135.53 = SSB 6 [(15.5 − 23.55) 2 + .. + (31.8 − 23.55) = = 813.18

Source

Df

SS

MS

Between

3

813.18 271.06

Within

20

473.7

Total

23

F 11.44,

23.685

Chapter 6 – Page 1

p value = 0.0001 There is significant evidence that at least one type of insulation has a different mean change in energy consumption.

EXERCISE 5 Using Tukey’s Procedure with 3 group and family-wise significance level of 5% gives q between 3.52 (25 df) and 3.49 (30 df). Will use 3.52 for calculations here. Declare two groups different if means differ by more than | t | > 3.52 / 2= 2.489 ⇒ | ya − yb | > 2.489 8.08(2 /10)= 3.164.

Examining the table of means, we see that Group 2 is higher than 1 and 3. Groups 1 and 3 do not differ significantly.

Group

1

3

2

Mean

8.5 a 9.0 a 13.0 b

EXERCISE 6 The researchers should use Dunnett’s test, since they are comparing to a control. Reject Ho if |t| > 2.54 using Table A.5.

All t will have denominator = σˆ L

= 23.685(1/ 6 + 1/ 6) 2.81.

Insul. 2 vs 1: t = (22.3-15.5)/2.81= 2.42 Insul 3 vs 1: t = (24.6 – 15.5) / 2.81 = 3.24 Insul 4 vs 1: t = (31.8 – 15.5) / 2.81 = 5.80 Insulations 3 and 4 are significantly better than the standard insulation #1.

EXERCISE 7

a) Ho: µ length of stay is the same in all groups, H1: not all means are equal

F= 3.5791667/0.2335526316 = 15.3249 with 3 and 76 df At α = 0.05 the critical value is 2.73, so reject Ho (p value < 0.0001).

There is significant evidence that the mean length of stay is different for at least one group. b) Since the variable is discrete, it is unlikely to follow a normal distribution. Moderate sample size (20 per group) may make up for this if there are no extreme outliers. c) Normal probability plot of residuals and boxplots indicate a serious failure of the normality assumption. Levene test shows unequal variances (p value = 0.0276).

Chapter 6 – Page 2

1.5

1.0

r e s i d s

0.5

0

-0.5

-1.0 -3

-2

-1

0

1

2

3

N orm al Q uanti l es

d) Tukey’s HSD indicates groups 1 and 2 have shorter mean stays than groups 3 and 4. However, patients may not be suitable to give treatments 1 and 2.

EXERCISE 8 a) Ho: all the colors have the same mean time (there is no effect due to door color) F = 20.01 with 2 and 12 df, p value = 0.0002 There is significant evidence of an effect due to door color. b) All pairwise comparisons should be made, so it is appropriate to use Tukey’s HSD test. Green has a significantly different (apparently higher) mean than either Red or Black. Red and Black are not significantly different from each other. c) This could be interpreted as meaning that we want to test the contrast Ho: µred – 2µgreen + µblack = 0, which has F=38.64 (t = 6.22). If the contrast was decided after-the-fact, its critical value should be determined using Scheffe’s method (critical value for F = 2*3.89 = 7.78, for t = 2.79). There is significant evidence that the mean for green differs from the combined mean of the other two colors.

EXERCISE 9

a) Ho: variance among suppliers is 0, H1: variance among suppliers is not 0

Note that supplier is a random effect, however, the test statistic is computed the same way as for the usual one-way ANOVA. F = 19.04 with 3 and 12 df, p value < 0.0001.

There is significant evidence that the variance in mean tensile strength among suppliers is greater than 0. b) est(σ²)=139.65

est(σs²)= (2659.4 – 139.65) / 4 = 629.94

Chapter 6 – Page 3

EXERCISE 10 a) Ho: mean compression resistance is the same for all percent sand F = 14.87 with 4 and 20 df, p value < 0.0001 There is significant evidence mean compression resistance is different for at least one sand percentage. b) Since the researchers wish to explore ideas, they need to use Scheffe’s test. Contrast will be significant if F > (5 − 1)(2.87) = 3.39, where 2.87 comes from the table of the F distribution using α = 5% and 4 and 20 df.

For Medium versus Low, F = 33.91.

For Medium versus High, F = 25.10.

Medium sand groups appear to do better than either very low or very high sand groups.

EXERCISE 11 a) F = 53.55 with 4 and 20 df, p value < 0.0001.

There is significant evidence that at least one group has a different mean, so there is evidence that one or more of these instructors should be rewarded. Tukey’s HSD shows that instructors 5 and 3 are significantly higher than all other instructors, and 2 is lower than all other instructors.

Instrct

2

1

4

3

5

Mean

7.82 c

10.68 b

12.12 b

14.3 a

14.58 a

b) CONTROL L1= µ1 + µ3 + µ4 + µ5 - 4 µ2

F= 104.04/0.7266= 143.19 , p value < 0.0001

There is significant evidence that the average of the means with some type of additive differs from the mean with no additive.

MFG

L2= µ1 + µ3 - µ4 - µ5

F= 5.09, p value = 0.0354.

There is significant evidence that the average of the means for manufacturer I differs from that for manufacturer II. ADD

L3= µ1 + µ4 - µ3 - µ5

F = 63.59, p < 0.0001

There is significant evidence that the average of the means for Additive type A differs from that for Additive Type B. EXERCISE 12 Using the pooled version of the independent sample t test in Exercise 3 of Chapter 5, t = 1.48 with 14 df, and the p value was 0.1616.

Chapter 6 – Page 4

Using an ANOVA, F = 2.18 with 1 and 14 df, and the p value was 0.1616. The value of t2 = 1.482 = 2.19 ≅ F, with the difference attributable to round-off. The p values are identical.

EXERCISE 13 Ho: all treatments have same mean hours of sleep H1: not all treatments have the same mean hours of sleep MSB= 16.58222334

MSE=1.102445556

F = 15.0413 with 2 and 15 df, p value = 0.0003

There is significant evidence that at least one treatment has a different mean number hours of sleep.

Since one group is a placebo, it is reasonable to compare both drugs to the placebo (and not make the comparison between the 2 drugs). Using Dunnett’s test, both the standard drug and experimental drug showed significantly greater mean sleep than the placebo. Dunnett’s test will not compare the standard and experimental drug, for that you need to use Tukey’s test throughout for all comparisons. Tukey’s shows that the two drugs do better than the placebo, but don’t differ significantly from each other.

EXERCISE 14 Source SS

df

MS

F

Between

452.6853

150.9

1.69

Within

1784.0020

89.2

Reject Ho if F > 3.10. There is no significant evidence that the paints differ in their mean time to peel.

EXERCISE 15 MSB = 162.9305556

MSW= 9.0583333

F = 17.99 , p value < 0.0001.

There is significant evidence that at least one insecticide has a different mean number of dead. Using Tukey’s HSD, insecticides D and C have a different (apparently higher) mean number of dead than A and B. Boxplots of the data show that the groups with the small means tend to have much larger variability. To stabilize the variances, the Arcsine (sqrt (proportion)) transformation was applied. The variances are more similar, though one group still shows an outlier. F = 23.48, again there is significant evidence that at least on insecticide has a different median number of dead. Using Tukey’s HSD, insecticide D now appears to have a significantly higher median number of dead than C, which is in turn greater than B or A. Insecticides B and A do not differ significantly from each other.

EXERCISE 16 There are 30 – 5 = 25 dfW. a) ordinary p value from t distribution (2 tailed)=2*(0.0255)=0.0510 On a TI84 calculator, right tail is TCDF(2.05,1E99,25) Chapter 6 – Page 5

b) Adjust p value using Bonferroni’s Method = 4*.0510 = 0.204 c) Scheffe’s method = F * (2.05) 2 = / (5 − 1) 1.051 p value is area to right of 1.051 for F distribution with 4 and 25 df p value = 0.4011 d) Tukey’s method (not available in Excel or most scientific calculators) Recall that the scale for the studentized range statistic is for = | t | 2 2.05 = 2 2.899 Using Probmc(‘RANGE’,2.899,.,25,5) in SAS we get the right tail probability is 1-0.7274 = 0.2726 If you tried to use Bonferroni for all pairwise comparisons, there are 5*4/2 = 10, so the p value would be at most 10*.0510 which is an extremely conservative estimate.

EXERCISE 17 Cadmium: MSB = 0.192845

MSW= 0.00332467

F = 58, p value < 0.0001

There is significant evidence that at least one soil type has a different mean absorption of cadmium. Using Tukey HSD, soil type D seems to have the lowest mean absorption, followed by C and E (C and E do not differ significantly from each other).

Levene’s test shows no significant evidence that the variances differ (F = 2.02, p value = 0.1227). The QQplot of the residuals shows one or two points below the line at the upper end, indicating some mild departures from normality. These are not outliers, but rather evidence of a ‘short tail’ at the upper end.

Lead (PB) : MSB = 27.3667

MSW = 1.98

F = 13.82, p value < 0.0001

There is significant evidence that at least one soil type has a different mean absorption of lead. Using Tukey’s HSD, soil types B, C and E do not differ significantly from each other, but are lower than A and D. Levene’s test shows significant evidence that the variances differ (F = 3.31 with p value = 0.0263) and the QQ plot shows a possible moderate outlier. Hence, this data might benefit from some type of transformation. A logarithmic transform shows somewhat more similar variances (except for soil C). The conclusions are similar to those from the original lead contents.

EXERCISE 18 a) There is extremely strong evidence that the mean fungus diameter is different for at least one of the mediums (F(6,21) = 168.61, p value < 0.0001). (Note the style in which the df are given inside parentheses for the F statistic.) b) Using Tukey’s HSD with a family-wise significance level of 5% Chapter 6 – Page 6

medium

WA

TWA

PCA

CMA

NA

RDA

PDA

Mean

4.25 a

5.175 b

6.125 c

6.275 c

6.80 d

7.00 d

7.250 e

diameter The fungus grows best in medium PDA, and next best in RDA/NA (statistically tied).

EXERCISE 19 UDL= 53.775 +2.68 * 9.445 *sqrt(3/24) = 62.72 LDL= 53.775 – 2.68 * 9.445 * sqrt(3/24)= 44.83 All means fall in this range. This is consistent with the results of Exercise 14.

EXERCISE 20 The mean number of defects across all lines is c = 150 and the estimated standard deviation is = s

= 150

12.25. If α=5%, the value of hα = 2.56 and the control limits are

UDL = 150+2.56*12.25*sqrt(4/5)= 178.05 LDL = 150-2.56*12.25*sqrt(4/5)= 121.95 Line 4 is significantly worse than the average, the manufacturer should focus on improving that line. Line 5 is substantially better than average, try to find out what is being done right on that line!

EXERCISE 21 a) 28*0.05 = 1.4 b) Using Bonferroni’s Method, they could use 0.05/28 = 0.0018 as a comparison-wise α for each test. c) No, neither of the tests would have been significant.

EXERCISE 22 b) Grand mean = 64.77/125 = 0.5182 SSB = 0.2881, SSW = 8.5062, F(2,122) = 2.07 with p value = 0.13 (using Excel) There is no significant evidence that any of the means differ, that is, there is no significant evidence that the instructions on Focus affected the mean IAT. c) The p values are the same as those that would be cited for an ordinary independent samples t test with 122 df. They do not appear to have been adjusted in any way for the multiple comparison problem, not even by requiring the overall F test to be significant.

Chapter 6 – Page 7

CHAPTER 7 EXERCISE 1 a) F(1,13) = 1.8148, p value = 0.2009 No significant evidence of an association. b) t = 0.445 , ± 1.8148 = −1.347 = −0.6 / std .err. → std .err. = Confidence interval for expected change is 4*(-0.6 ± 2.160*0.445)=(-6.2,1.4) Note this interval contains 0, consistent with test of no association. c) 1.6 d) 1.6 ± 2.16 1.22 (1 + 1/15 + 4 / (14*32 ) = 1.6 ± 2.72 EXERCISE 2 a)

µ(sugar|days)= 12.04189 – 0.82162*Days

0.1587. There is no significant evidence of a linear relationship between days after picking and sugar content.

Residuals

The t-test for the slope has t = -1.61 with p value =

b) The plot of the residuals versus the predicted values suggests a nonlinear model might be better.

EXERCISE 3

Predicted value

a) µ(y|x)= 14.39557 + 0.76509x F(1,13) = 19.41, with p value 0.0007, so there is significant evidence of a linear relationship between the scores on the midterm and on the final. b) Obs

final

predict

resi

lower lim

upper lim

1

76

77.1328

-1.1328

51.2937

102.972

1.9211

60.1782

113.980

2

83

70.2470

3

89

87.0789

4

76

64.8914

12.7530 11.1086

44.6196 39.1221

95.874 90.661

Prediction intervals (limits in last 2 columns) are very wide, so an individual student’s final exam can vary wildly around the expected value. A final is necessary. c)

Plot of actual observations and predicted equation. The points are widely scattered about the line.

Chapter 7 – Page 1

90 80

Final

70 60 50 40 30 20 30

40

50

60

70

80

90

100

Midterm

d) Predicted final is 77.1, which matches height of line if you draw vertical up from 82. e) r = .7739 R-Square= 0.5989, F = (15-2)(0.5989)/(1-0.5989) = 19.41, which agrees with (a) EXERCISE 4 a) βˆ1 =

1.5, t =

2.65 with 12 df, p value = 0.0213. Note that as X increases by one unit, we expect

an increase in the mean of Y of size 1.5. As X increases two units from -1 to 1, we expect an increase of size 3. This is the same as the difference between the sample means for the two groups. b) That X takes on just two values suggests an independent samples t test comparing Y when X=-1 to Y when X=1. The equal variance version of the test has t = 2.65 with 12 df and p value 0.0213, that is, is exactly the same as the regression result!

Chapter 7 – Page 2

EXERCISE 5

10

a) The fitted equation is estimated Range = -6.48 + 0.75*Latitude. increases as one moves further north. b) The residual plot shows 5 large outliers on the negative side (less range than expected). Inspecting the listing of the

Residuals

That is, the discrepancy between the maximum and minimum

0

-10

city names, all of these outliers correspond to cities on the West Coast near the Pacific Ocean. Perhaps being right on the ocean keeps the temperatures less variable, by keeping it cooler in summer and/or warmer in winter than would be

-20 20

30

40

Latitude

expected.

Chapter 7 – Page 3

50

EXERCISE 6 An initial regression of FOWL on WATER showed strong evidence that the expected number of birds increased as open water increased: FOWL = 14.74 + 8.54Water, F(1,50)=181.26, p value < .0001.

However, the residual plot showed strong evidence of curvilinearity, and one point with exceptionally large water and fowl that was ‘controlling’ the regression.

After some experimentation, a better model is obtained by using LNFOWL=LN(Fowl+1) and LNWATER=LN(Water+1). Adding 1 is necessary because some observations have values of 0.

LN(fowl+1) = 1.262+0.952*LN(Water+1), F(1,50)=15.34, p value 0.0003.

Chapter 7 – Page 4

EXERCISE 7 a) Apparently, electricity usage increases as

80

average temperature increases.

70

b) predicted Kwh= -97.92389 + 2.00101(Tavg),

σˆ 2 = 70.355

KWH

60 50 40 30

F value = 110.2, p value < 0.0001, there is

20

significant evidence of a linear relationship

10

between daily mean temperature and electricity

60

65

70

75

80

Avg Temperature

consumption. For every increase of 1 degree in the mean temperature, the expected electricity consumption increases by 2 Kwh.

30

c) residual plot: no evidence of nonlinearity,

20

nonconstant variance or extreme outliers.

Residuals

10

0

-10

-20

-30 20

30

40

50

60

Chapter 7 – Page 5

70

EXERCISE 8 a) Using LNPE (as in Chapter 5, to stabilize variances) we get LNPE = 3.3173 + 0.352X, t = 1.84, p value = 0.0744.

The intercept is the expected LNPE when x=0, that is, for NYSE exchange.

The slope (estimated at 0.352) is the increase in expected PE as x goes from 0 to 1, that is, the difference between the NYSE and NASDAQ exchanges. b) sample mean for NYSE is 3.3173, consistent with interpretation in (A), and difference between 2 means is 0.352.

Pooled t test has t = 1.84, p = 0.0744 same as regression result.

Unequal variance t is a little different, t’ = 1.83 with 32 df, p value 0.0766.

EXERCISE 9

N = 217 and r = 0.36, so F (1, 215) =(217 − 2)(0.362 ) / (1 − 0.362 ) =32.01,

p value < 0.0001 There is strong evidence of an association. The large sample size makes it possible to detect this association, even though it is too weak to make accurate prediction possible.

EXERCISE 10 a) Amikacin: predicted half-life = 1.75241 + 0.01336Dose_mg_kg Gentimacin: predicted half-life = 2.11150 + -0.03537Dose_mg_kg While one slope is positive and the other is negative, both are so close to 0 that these are essentially parallel (horizontal) lines. b) For the null hypothesis that β1 = 0 Amikacin: t = 0.11, p value = 0.9148, there is no significant evidence of a relationship Gentamicin: t = -0.17, p value = 0.8661, there is no significant evidence of a relationship As related to part (a), neither predicted line has a slope which differs significantly from 0.

Chapter 7 – Page 6

c)

4

The degree of scatter about the two lines makes it obvious why neither relationship appeared significant.

A

Half Life

3

G G G G

A

G G G G G G G G G G G G G G

2

A

AA

A A A A A A

AA A A

1

A A A A A A

0

0

2

4

6

8

10

12

Dose in mg/kg

EXERCISE 11 Index = 0.1357 + 1.3672*Future.

F(1,44) = 76.37, p value < 0.0001 showing strong evidence of an association, though sqrt(MSE) = 0.814 indicates very far from perfect prediction!

The residual plot shows no evidence of curvilinearity or heteroscedasticity. There are no obvious outliers.

EXERCISE 12 a) F =

(n − 2)r 2 = 1 − r2

176 * 0.18 = 0.82

38.63 with 1 and 176 df. There is significant evidence of a linear

relationship between the two, so the authors are justified in their claim. b) t =− F =−6.215 =−.62 / std. err. Hence, the standard error is approximately 0.10, and a confidence interval would be −0.62 ± 1.96 * 0.10 = (−0.82, −0.42). Since this interval is entirely negative, it allows the authors to describe the relationship as a decline.

Chapter 7 – Page 7

EXERCISE 13 200

The residual plot shown to the right

150

shows a failure of the regression of the residuals is greater for the houses of larger size. That is, there is more

100

Residuals

assumptions. Apparently, the magnitude

variability in house price for the very

50 0 -50

large homes than for the smaller homes.

-100 0

The Q-Q plot shows an apparent failure

1

2

3

4

5

Size in 1000 sf

of the normality assumption. However,

this same pattern can also be caused by a set of residuals coming from a much larger variance.

A plot of the price per square foot versus size shows both that the price per

200

sf is generally rising as size increases,

150

and becoming more variable.

Residuals

100 50 0 -50 -100 -3

-2

-1

0

1

2

3

Normal Quantiles

Chapter 7 – Page 8

EXERCISE 14 a) For X1 and X2, F(1,13) = 6.26, which exceeds the critical value from the F table for 1 and 13 df with α=5% (4.67). So this correlation coefficient is significantly different from 0.

For X1 and X3, F(1,13) = 0.44, which does not exceed the critical value. So this correlation coefficient is not significantly different from 0. b) Yes, these statements are consistent with the hypothesis tests in (a).

EXERCISE 15 a) increases in reading times are associated with decreases in response times b) Since the t statistic is the ratio of the estimate to its standard error, we can solve for the standard error as -0.03/-2.11=0.0142. 20 *(−.03 ± 2.1009 * 0.0142) = -1.196 to -.034 With 95% confidence, if reading time increases by 20 seconds, the mean response time will decrease between .034 and 1.196 ms.

18r 2

c) t 2 = F= 4.45 =

1−r

2

⇒ 4.45 − 4.45r 2 = 18r 2 ⇒ 4.45 = 22.45r 2 so r

2

= 0.198 , no

Chapter 7 – Page 9

EXERCISE 16 (a & b) The relationship for the original variables is clearly not linear. After logarithms, the relationship is more linear for the majority of the time period, but at the end levels off.

2.4

12

2.3

10

2.2 2.1

Ln(Water)

6 4

2.0 1.9 1.8 1.7

2

c)

0 0

1

2

3

4

5

6

1.6 1.5 -2

-1

0

1

2

Ln(Soaking Time)

Soaking Time

LN(water)=1.979+0.260LN(Stime). However, the residual plot clearly indicates the remaining nonlinearity in the model.

0.15 0.12 0.09 0.06

Residuals

Water

8

0.03 0.00 -0.03 -0.06 -0.09 -0.12 -0.15 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5

Predicted

Chapter 7 – Page 10

CHAPTER 8 EXERCISE 1 For the X’X matrix, the sum of the squares of the X1 values yields 378, that is the entry on the X1/X1 entry. For the X’Y matrix, the sum of the products of the X1 times Y entries is 544.9, that is the entry on the X1 /Y cell of the table.

To verify the inverse, note that the first row of the X’X inverse matrix times the first column of the X’X matrix is 12.76103*8 – 0.762244*48 – 1.89706*34 = 1.0005 (should be 1, but there is some roundoff).

The third row of the X’X Inverse matrix times the first column of the X’X matrix is -1.89706*8 + 0.1078431*48 + 0.2941176*34 = -0.00001 (should be 0, but there is some roundoff).

EXERCISE 2 a) F(10,489)=(500-10-1)*(0.07)/[10*(1-0.07)]=3.681, p value = 0.00009 Yes, it is ‘significant’. b) There is evidence that at least one of the variables is linearly related to socioeconomic status, because the F test in part (a) tests the null hypothesis that none are related. c) No, individual prediction is very poor, R-square is only 0.07, meaning that the independent variables only explain 7% of the variability in socoeconomic Status. d) F(10,39)=(50-10-1)*0.07/[10*(1-0.07)]=0.294, p value = 0.979, no.

EXERCISE 3

Result for Stress: µ(y1|x) =

700.61805 - 1.52568X1 + 175.98394X2 - 6.69714X3

The fitted regression shows that the stress at

300

failure apparently decreases as percent binder increases, however this particular effect is not weaker as the ambient temperature increases (holding other variables constant.) However, it becomes significantly stronger as the loading

Residual stress

significant. The material becomes significantly

200

rate increases (also holding other variables

100

0

-100

constant).

The residual plot does not show any indication of specification error. There are no features of

-200 -200

0

200

400

600

800

1000

1200

Predicted stress

curvature or flare, nor are there any obvious outliers.

Chapter 8 – Page 1

Result for Strain µ(y2|x)= -5.61130 + 0.66754X1 - 1.23535X2 + 0.07319X3 This shows that the strain at failure significantly

6

increases as the percent binder increases (that is,

5

this type of strength increases with binder, weakens as the loading rate increases, unlike the stress at failure. It becomes significantly stronger as the ambient temperature increases.

Residual strain

holding other variables constant). It significantly

4 3 2 1 0 -1

However, the residual plot shows clear

-2

indication of a nonlinear relationship, and

-3

possibly an increase in variances. A transformation of the dependent variable might

-4 -3 -2 -1 0

1

2

3

4

5

6

7

8

9 10

Predicted strain

improve the model.

Chapter 8 – Page 2

EXERCISE 4

90

a) The residuals from the regression of PASS on

80 70

the quantitative variables shows clear grow as the predicted size grows.

50

Residual

misspecification error. The variances seem to

60 40 30 20 10 0 -10 -20 -30 -40 -50 -20

b) The logarithmic transformation makes sense,

0

20

as it often helps when variances grow with predicted size. The residual plot is enormously reasonably stable variance.

The fitted equation is Predicted ln(passengers) =

60

80

100

120

140

1

Residual

improved, showing reasonable linearity and

40

Predicted value

2

0

-1

-6.47 - 0.44Lmiles + 2.91Linm +.78Lins+.43Lpopm +.39Lpops+.71Lairl It makes sense that the number of passengers should

-2 0

1

2

3

4

5

Predicted value

increase with the population size, but less sense that it decreases with distance. One might think that there would be more incentive to fly than drive, but perhaps the more the distance, the less strong the ties between the cities and the fewer the people who need to make the trip. c) Using airlines as an independent variable and passengers as dependent (as in parts A and B) is treating capacity as the given variable. But treating Airlines as the dependent variable is thinking of it as if airlines are reacting to demand by moving into the market.

Chapter 8 – Page 3

EXERCISE 5

a) µ(y|x) = -379.248 + 170.220(DBH) + 1.900(height) + 8.156(age) – 1192.868(grav)

Notice residuals, this model is clearly mis-specified and not very useful. A transformation would be helpful. Since both nonlinearity and nonconstant variance appear, perhaps a multiplicative model.

400

Residual logweight

300 200 100 0 -100 -200 -300 -200

0

200 400 600 800 1000 1200 1400 1600

Predicted logweight

b) All variables were transformed by taking logarithms. The regression showed that

0.4

log(grav) was not significant, it was dropped. The µ(y|x) = -1.727+2.147Ldbh + 0.996Lht 0.152Lage.

The residual plot is much improved with respect to linearity, but still shows some residuals with greater magnitude on the left.

Residual logweight

fitted model was

0.3 0.2 0.1 0.0 -0.1 -0.2 -0.3 -0.4 -0.5 4

Log(age) does not contribute significantly to the

5

6

7

8

Predicted logweight

regression. Since it is harder to estimate than Height or DBH, it could be dropped and still have R-square of 98%.

Chapter 8 – Page 4

EXERCISE 6 a)

Analysis of Variance Sum of Mean DF Squares Square 4 409928367 102482092 16 6781702 423856 20 416710068

Source Model Error Corrected Total

Variable Intercept value doct nurse vn

F Value 241.78

Parameter Estimates Parameter Standard Estimate Error t Value 530.26514 263.57743 2.01 1.29583 0.51457 2.52 0.46388 1.71375 0.27 -0.88013 0.90711 -0.97 2.14219 0.68198 3.14

DF 1 1 1 1 1

Pr > F <.0001

Pr > |t| 0.0614 0.0228 0.7901 0.3464 0.0063

Variance Inflation 0 56.71 121.31 77.55 31.67

The number of deaths appears to increase as the number of doctors and vocational nurses increases, but decreases when there are more nurses! The former relationships don’t make sense. However, since large population regions have more deaths, but also tend to have more doctors and nurses, the results are probably the result of population. That is also the cause of the multicollinearity problem. b) Converting all variables to per capita basis by dividing by population should help, as it should put large population regions and small ones on the same footing. c)

Converting to per capita values Analysis of Variance Sum of Mean Squares Square 24.29002 6.07251 50.16061 3.13504 74.45063

Source Model Error Corrected Total

DF 4 16 20

Variable Intercept pcvalue pcdoct pcnurse pcvn

Parameter Estimate 7.11622 -0.84243 0.17416 -0.28438 1.49954

DF 1 1 1 1 1

Parameter Estimates Standard Error t Value 2.56261 2.78 1.51528 -0.56 2.95725 0.06 1.00699 -0.28 0.65146 2.30

F Value 1.94

Pr > F 0.1533

Pr > |t| 0.0135 0.5859 0.9538 0.7813 0.0351

Variance Inflation 0 2.87 3.16 1.56 1.20

Chapter 8 – Page 5

The multicollinearity is greatly reduced, but now there is no significant evidence that any of the variables are related to death rates. Most likely, the relationships seen in part A were just due to the relationship of all variables to population. You cannot compare the R-squared since you have changed the dependent variable.

EXERCISE 7 a) µ(y/x) = 219.27461 + 77.72454(time) Distance significantly increases with time, but a

100

linear model is probably not satisfactory.

The residual plot shows nonlinearity. It also misrecorded.

0

Residual

suggests that the 18th observation may have been

-100

-200 200 400 600 800 1000 1200 1400 1600 1800 2000

Predicted

b) µ(y/x)= 178.07841 + 93.10596(time) 0.72874(time²). The addition of the square term does improve the model. The size of shows curvature (see the residual plot for c).

Negative coefficient for quadratic term shows that the effect of increase in time is less at higher times than at lower times (rate of increase is slowing). c) Residual plot for quadratic model: A model using time and square of time was better but makes it clear that the 18th observation was probably incorrect.

Residual

the residuals is somewhat reduced, but still

90 80 70 60 50 40 30 20 10 0 -10 -20 -30 -40 -50 -60 -70 -80 -90 -100 -110 0

200 400 600 800 100012001400160018002000

Predicted

Note: replacing square of time with log(time) or square root of time are even better models.

Chapter 8 – Page 6

Chapter 8 – Page 7

EXERCISE 8 a) Fitted goalmade = -67.365 – 0.142*dash100 + 1.203*height – 0.011*weight, MSE = 0.57508, F(3,21)=127.10 b) The largest VIF is 1.011, so there is no multicollinearity (surprising since both height and weight are present). c) The third observation (height = 70, weight=170,

d) The residual plot suggests some mild nonlinearity, which can be improved by fitting a full quadratic model for height, weight, their squares, and the interaction height*weight. (Dash100 is not

Jackknifed residual

goalmade=11) is an outlier, using the jackknifed residuals.

helpful, and was dropped.) However, the outlier remains, and one suspects the outlier is partially responsible for the apparent

Predicted value

nonlinearity.

EXERCISE 9

a) COOL SEASON

5

coolweight= -2.63755 +

4

0.43937(coolwidth) +

3

0.11038(coolheight) Residual plot shows nonconstant variance and nonlinearity, suggesting a

2 1

transformation of the dependent variable

0

and possibly the independent variables.

-1 -2 -1

0

1

2

3

4

5

6

7

8

9

P redi cted val ue

Chapter 8 – Page 8

WARM SEASON Warmweight= -2.11678 + 0.20691(warmwidth) + 0.11846(warmheight) This residual plot also shows nonconstant variance and possible nonlinearity.

5 4 3 2 1 0 -1 -2 -3 -4 -5 -1

b) transforming all variables to logarithms.

COOL Season logcoolweight= -4.59650 + 1.57149(logcoolwidth) + 0.74670(logcoolheight) The residual plot is greatly improved.

Weight increases as width and height increase, which makes sense.

WARM SEASON logwarmweight= -4.42125 + 1.66939(logwarmwidth) +

0

1

2

3

4

5

6

7

8

9

10 11

P redi cted val ue

2

1

0

-1

-2 -2

-1

1

0

2

3

P redi cted l og w ei ght

0.20887(logwarmheight) The residual plot is greatly improved.

Weight increases as width and height increase, which makes sense. Note that the coefficient with respect to height seems somewhat less in warm season rather than cool.

2

1

0

-1

-2 -3

-2

-1

0

1

2

3

P redi cted l og w ei ght

Chapter 8 – Page 9

EXERCISE 10 a) The fitted model suggests increasing survival time with increasing CLOT, PROG, ENZ and LIV (though the latter is not significant. However, as the residual plot shows, the relationship is both nonlinear and has variance that grows with magnitude of predicted value, suggesting a logarithmic transformation. b) The best residual plot is from a model using log(time), but leaving the other variables unchanged (see below). All variables except LIV have significant positive impacts on log survival time. You cannot compare R-square or MSE to the results of part (A) because the dependent variable has been changed, but the residual plot is now a featureless blob, indicating we have reduced curvilinearity and heteroscedasticity.

Predicted ln(Surv) =

1.1254+0.1578*clot+0.213*Prog+0.0218*Enz + 0.0044*Liv

F(4,49) = 430.98, MSE = 0.01188 0.4 0.3

Residuals

0.2 0.1 0.0 -0.1 -0.2 -0.3 3

4

5

6

7

Predicted

Chapter 8 – Page 10

EXERCISE 11 crimes= -10.30475 + 0.37775(age) + 2.29391(sex) + 0.17933(college) + 0.29339(income) The residual plot shows some evidence of nonconstant 7

variance.

F

M

6 5

M

4

F

Residual

3

F

1F 0 -1 -2

F F

F FF M M F F

F F

M

F

2 F

M F

F

M

M M

M F

F F

F

M MM F F

M M F

F F

-3

M F

-4

M

-5

M

5 6 7 8 9 1011121314151617181920212223

Predicted value sex

A transformation of all variables except SEX using logarithms stabilized the variances but suggested some

significant, probably due to multicollinearity with

0.1

Residual

0.2

0.0

F

F

FM

F F M M F F M F

F

F

F F

+0.597Log(age)+.1514Sex+1.083Log(Income)

-0.4 1.6

M

M M M MM

F F M F F FM

F M

F F

-0.2 -0.3

F

F

F

-0.1

Log(crimes_=-3.5074 Holding other variables constant, the perception of

M

0.3

model was obtained by) dropping LCOLLEGE (not

is still some suggestion of nonlinearity.

MMM1

0.4

nonlinearity. After exploration, a somewhat better

LAGE). The residual plot is shown to the right. There

FFF0

F M M

F

M

F 1.8

2.0

2.2

2.4

2.6

2.8

3.0

3.2

Predicted value

crimes increases with age and income. Men tend to perceive more crimes than women of the same age and

sex

0

1

income.

EXERCISE 12 a) Fitted model predicted ln(Water) = 2.055+0.301X-0.071X2 with R2=0.9895 b) Fitted model predicted ln(Water) = 2.047+0.304T-0.061T2 with R2=0.9914

Chapter 8 – Page 11

c) The second model fits slightly better, since it does a better job of mimicking the ‘flattening’ of the curve on the far right.

Chapter 8 – Page 12

EXERCISE 13 a) HT = 15.994 + 2.331*DFOOT + 0.309*HCRN, F(2,61)=24.41, p value < 0.0001 Plot of residuals versus predicted value appears featureless, so curvilinearity and heteroscedasticity are minimal.

A model using sqrt(DFOOT) is also sensible, but is very similar. b) However, R-square only 0.4445, this model does not do a good job of accurately predicting HT. EXERCISE 14 a)Note: price is in 1000s of dollars and size is in 1000s of sq ft. The selected model is Price = 36.93 -0.319(AGE) – 11.868(BED) + 61.774(SIZE) The coefficients are similar to those in the full model discussed in Example 8.2. The increase of price with size and the decrease of price with age, other values being held constant, makes sense. The negative coefficient with Bed was explained in Example 8.2. aii) The full data set (table 1.2). Expressing price in 1000s of dollars and house size and lot size in 1000s of sq ft, the selected model is Price = -42.466 – 1.05(AGE) +90.623(SIZE) + 0.303(LOT) This model makes sense regarding the signs of the coefficients, but is much different from the model fit on the lower priced homes. Its residual plot shows evidence of nonlinearity and increased variance on the right side.

If I was looking for only lower priced homes, I would use the first model. If I was looking for higher priced homes, I would re-fit a model using only more expensive homes. aiii) A large positive residual may signal a home that is overpriced, while a very large negative residual may signal a bargain (or a home with a problem!) However, until a suitable model without curvilinearity is built, then the residuals don’t mean very much. As this problem stands, the only exceptionally large residuals are for the large homes that are showing higher prices than we would expect. b) Price=-42.14 + 79.9(SIZE)-1.01(AGE)+0.29LOT+13.0GARAGE Garage slightly improved prediction, but model still shows nonlinearity. Log price showed slightly better residual plots with less curvilinearity, and some possible ‘bargains’ at the low end of the price range.

β= 0 , H1: at least one coefficient is nonzero EXERCISE 15 Ho: β= 2 3 F (2,46) = [(932-901)/2]÷(901/46) = 0.79, p value 0.4599, not significant. Model 2 does not fit significantly better.

Chapter 8 – Page 13

EXERCISE 16 = a) F

(256 − 194) / 1 = 194 / 46

14.7 with 1 and 46 df. Since this is well beyond the critical value of the F

table even using α = 1%, there is significant evidence that X3 is related to Y, after controlling for X1 and X2. b) t = ± F with the sign matching the sign for the parameter estimate. So t = 3.83 with 46 df. Once again, there is significant evidence that X3 is related to Y, after controlling for X1 and X2. c) t =

βˆ1 / std.error =

3.83 =

2.1 / std.error so the standard error must be 0.548. A 95%

confidence interval would be 2.1 ± 2.013 * 0.548 using the value from the t-table with 46 df (you can use TI-84 or Excel to get this t critical value). With confidence 95%, the expected increase in Y if X3 increases by 1 and other variables remain constant, is between 0.997 and 3.203.

EXERCISE 17 a) 16

B

14

G

12 10

G

8 6 4

B 0

1

2

3

4

5

x2

b) For low values of X2, girls tend to score higher than boys. But for high values of X2, boys tend to score higher than girls. c) girls-boys: = ( β 0 + β1 + 3β 2 + 3β3 ) − ( β 0 + 3β 2 ) = β1 + 3β3 d) 5-1.5*3= 0.5

Chapter 8 – Page 14

EXERCISE 18 a) b) When X2 is very low, the value of Y

9

will decrease slightly as X1 increases. However, when X2 is

8

high, the value of Y will increase steeply as X1 increases.

Predicted Y

10

7

c) No, βˆ1 = 1.5 is the expected increase in Y if X1 increase

6 5

by 1, provided that X2=0.

4 3 0

1

2

X1 x2

-1

1

d) Now the value of Y tends to increase as X1 increases, no matter what the value of X2.

8

Predicted Y

7

6

5

4

3 0

1

2

X1

x2

-1

1

Chapter 8 – Page 15

EXERCISE 19 a) Model 1: R-squared = 0.07 = SSR/45.778, therefore SSR = 3.2, SSE=42.578, F(2,97) = 3.65, p value 0.0298. There is significant evidence that at least one of the two independent variables is related to RO Model 2: R-squared=0.19 = SSR/45.778, therefore SSR = 8.70, SSE=37.08, F(4,95) = 5.57, p value 0.0005. There is significant evidence that at least one of the 4 independent variables is related to RO b) t(95) = -3.29, significant. Since the coefficient is negative, workers in diversion have higher expected RO than those in non-diversion, all other independent variables held constant. c) CI for coefficient for type of work = -0.56 ±1.985*0.17 = (-0.897, -0.222) d) F(2,95) = [42.574-37.08)/2]÷(37.08/95) = 7.04 p value 0.0014. There is significant evidence that at least one of type of work or employment length are related to mean RO, after controlling for social support and cultural competency.

EXERCISE 20 Model 1 must have SSE = (1-0.05)*33.7 = 32.015 with 217 df.

Model 2 must have SSE = (1-0.34)*33.7 = 22.242 with 214 df.

F = [(32.015-22.242) / (217-214)] ÷ (22.242/214) = 31.34 with 3 and 214 df. There is significant evidence that at least one of the burnout scores is related to frequency of complaints, after controlling for gender and age.

Chapter 8 – Page 16

CHAPTER 9 EXERCISE 1 The two-way ANOVA indicates significant evidence for each type of effect.

Source

DF

Type III SS

Mean Square

F Value 85.88

<.0001

4.41

0.0414

a

1

56.25000000

56.25000000

t

3

8.66750000

2.88916667

a*t

3

18.62000000

6.20666667

Pr > F

9.48

0.0052

The contrast of the Control treatment versus the average of the other three treatments (coefficients 3, -1 , -1, -1) shows that the average of the other three treatments do differ significantly from the control. The p value here has not been adjusted for the multiple comparison problem.

Contrast

DF

Contrast SS

Mean Square

F Value

Pr > F

vs control

1

8.16750000

8.16750000

12.47

0.0077

Using Tukey’s HSD to compare the Treatment main effects shows that M, N, and Q do not differ significantly from each other. While Q is different from C, M and N are intermediate and not significantly different from the control.

Inspection of a profile plot suggests that Factor A has very little effect in the control treatment, but for the other treatments level A yields a higher value on average than level B. This is the source of the interaction.

EXERCISE 2 Sum of Source

DF

11

Squares

79.54833333

Mean Square

Error

12

8.37000000

0.69750000

Corrected Total

23

87.91833333

Source

DF

Type III SS

c

3

a*c

6

Model

a

2

F Value

Pr > F

Mean Square

F Value

Pr > F

33.54833333

11.18277778

16.03

0.0002

7.79666667

1.29944444

1.86

0.1691

38.20333333

7.23166667

19.10166667

10.37

27.39

0.0002

<.000

Chapter 9 – Page 1

Since we do not have significant interactions, we can compare main effects only using Tukey’s HSD. For Factor A, Level M is significantly lower than R or P (which do not differ from each other). For Factor C, Level C is significantly higher than all the others. B appears next highest, but B and D do not differ, though B is higher than A.

EXERECISE 3 An ANOVA (using Proc GLM) with dependent variable log(time) showed a significant main effect for LIVCAT (two categories).

A cell mean plot shows that those with low liver function tended to have shorter log(time).

The plot suggests that the effect of liver function is greater among those with better prognosis, but this interaction was not strong enough to be significant.

To test whether LIV has any kind of effect (either main or interaction), construct F test based on full and restricted model.

Full model SSE=10.753, df=50; Restricted Model (w/o LIV or interact.) SSE=15.579, df=52, F(2,50)=11.2, p value < 0.0001. Yes, knowing LIV does improve prediction of survival time, after controlling for PROG.

Chapter 9 – Page 2

EXERCISE 4.

Source

DF

Type III SS

Mean Square F Value

Pr > F

period

1

13.22500000

13.22500000 26.02

<.0001

day

4

13.65000000

3.41250000

6.71

0.0006

day*period

4

9.65000000

2.41250000

4.75

0.0044

There was a change due to period (before and after the rules), but according to the significant interaction, the change was greater on some days than on others.

Examining the cell mean plot, the reduction in addons was primarily on Mondays.

Chapter 9 – Page 3

EXERCISE 5 Since the dependent variable was in the form of percentages, they were transformed to Y=arcsin(sqrt(survive/100)).

Sum of Source

DF 5

Squares

4.42484933

Mean Square

Error

24

2.62772434

0.10948851

Source

DF

Type III SS

fung

2

1.28497456

Model

Corrected Total

concen

concen*fung

29

1 2

0.88496987

Pr > F

8.08

0.0001

Mean Square

F Value 20.41

Pr > F

0.0001

0.64248728

5.87

0.0084

7.05257367

2.23508411

F Value

2.23508411

0.90479067

0.45239534

4.13

0.0287

The profile plot (still using the transformed values of the dependent variable), help us interpret the ANOVA. The high concentration results in significantly higher proportions of seeds that

1.2

survive. This is the main effect for concentration.

1.1 1.0

The effect of concentration is much greater for

0.9

fungicide A than for the other fungicides. This main effect for Fungicide. At low concentrations, the fungicide don’t differ very much, but at high

0.8

Mean Y

explains the interactions. It is also the source of the

0.7 0.6 0.5 0.4

concentrations, fungicide A is much better.

0.3

Tukey’s HSD was used to compare the 6

0.2

combinations of fungicide and concentration.

Fungicide A, at concentration 1000, has a significantly higher mean (transformed) survival than any other combination.

H

0.1

H

H L

L L

0.0 a

b

c

Fungicide

Chapter 9 – Page 4

EXERCISE 6 The scatter plot suggest a curvilinear

50

relationship of growth rate to nutrient 40

concentration. A quadratic regression Growth

yielded Predicted growth = 7.5238 + 1.2964(NC) -0.0143(NC2) with SSE = 1040.54 and 33 df.

30

20

10

A one-way ANOVA using 9 groups had SSE

0 0

= 959.32 with 27 df.

10

20

30

40

50

Nutrient Concen.

The lack of fit test gave F = [(1040.54-959.32)/6] / (959.32/27) =

0.38 with 6 and 27 df.

There is no significant evidence that the quadratic model does not fit. We may take the quadratic as a model.

NOTE: Comparing the one-way ANOVA model to a simpler linear model had lack-of-fit F(7,27) = 1.81, p value = 0.126. Not significantly worse than the ‘full’ model, but quadratic model is significantly better than linear model.

EXERCISE 7 The profile plot, where plotting symbol is the number of the level for TEMPR, is is greater for higher levels of CLEAN than it is at low levels of CLEAN. At higher levels of CLEAN, the higher values of ELAST seem to be found at intermediate levels of TEMPR (levels #2=1.65 or #3 =2.38) rather than the highest or lowest levels.

The ANOVA confirms both the main effects and interactions.

Mean Elast

shown to the right. Clearly, the effect of TEMPR

150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0

3 2

4 3 2

1

4 3 4 1 2 1 0 3 4 2 0

3 2 4 1 0

0 1

1

0

0

2

CLEAN

SSE = 872.69, df = 50

Chapter 9 – Page 5

Source

DF

Type I SS

Mean Square

F Value

Pr > F

clean

4

86093.77441

21523.44360

1233.17

<.0001

clean*tempr

16

24042.93605

1502.68350

86.09

<.0001

tempr

4

22554.07993

5638.51998

323.05

<.0001

Given the strength of the interactions, to further investigate the comparisons, we would want to treat this as a very large one-way ANOVA with 25 groups. The precise list of comparisons would depend on the goal. If we wish to identify the combination of TEMPR and CLEAN which gives the highest ELAST, we should make the pairwise comparisons among all 25 groups (using Tukey’s HSD on the 25 groups, CLEAN=2.0 and TEMPR =1.65 or 2.38 are significantly better than any other combination). If, however, we wish to describe the best value of TEMPR within each selection of CLEAN, then we have a different set of comparisons to make.

Alternatively, we can treat this as a multiple regression. µ(ELAST|x) = -8.03515 -30.95195(clean) + 36.27533(tempr) + 23.78724(clean²) -12.46042(tempr²) + 17.65392(clean*tempr) However, the SSE of this model is SSE=13313.57 with 69 df. Lack of fit test has F(19,50)=[(13313.57872.69)/(69-50)] / (872.69/50) = 37.5 By comparison to the two-way ANOVA, we see this regression does NOT fit the data. Higher order polynomial and cross product terms are needed.

Chapter 9 – Page 6

EXERCISE 8.

Since we do not have the original replicated data, we in essence only have one observation per cell.

Hence, we must use the SS for the highest order interaction to estimate the error variance. Also, the interactions with the nitrogen application rate seem weak, so we will delete them also. The ANOVA for the final model is shown below.

The effect of nitrogen is relatively simple to describe: the more the better. Tukey’s HSD applied to the main effects shows the 150 rate gave significantly higher yields than the 60 rate. The 120 and 90 rates are intermediate and not significantly different from any other rate.

A profile plot is helpful in understanding the effect of variety and location. At most locations, the varieties are fairly similar. However, at location E, variety L is much better.

We can fit a linear trend in Nitrogen application: t = 3.92 with 33 df. There is a significant linear trend in nitrogen.

Source Model Error Corrected Total

DF 14 33 47

Squares 68117055.67 7281838.33 75398894.00

Source variety nitapp location variety*location

DF 2 3 3 6

Type III SS 12448648.62 3439716.67 34391297.50 17837392.87

Sum of Mean Square 4865503.98 220661.77

F Value 22.05

Pr > F <.0001

Mean Square 6224324.31 1146572.22 11463765.83 2972898.81

F Value 28.21 5.20 51.95 13.47

Pr > F <.0001 0.0047 <.0001 <.0001

Chapter 9 – Page 7

EXERCISE 9 The sums of squares can be obtained from the means and standard deviations given in the table. For example TSS=119*(4.902)2 = 2=2859.5, SSE=29*(3.72 + 4.12 + 4.42 + 4.82) = 2114.1, some small rounding errors will occur when compared to the values obtained from squaring the cell means.

Source

df

SS

MS

F

P value

Model

3

745.4

248.5

13.43 < 0.0001

Political

1

270

270

14.81 0.0002

CommDisSt 1

367.5

367.5

20.16 < 0.0001

Interaction

1

107.9

107.9

5.87

Error

116 2114.1 18.225

0.0169

Examining the cell means, we see that those living in Hot Spots will have significantly lower scores than those in less affected communities. There is an interaction, with the effect of community disease status being smaller among those who identify as liberal. In general, liberals tend to score lower than conservatives.

0.7

using weights as the dependent variable showed a

0.6

tendency towards greater variance at higher predicted

0.5

values. Two transforms, log(1+weight) and sqrt(weight), were tried and the square root transform seemed slightly better. The profile plot for the

Mean sqrt weight

EXERCISE 10. The residuals from the first ANOVA

R

R B N

K N

K R

0.4 B

0.3 0.2

sqrt(weight) is shown to the right. Clearly, all the

0.1

varieties are better at the lower temperatures, and this

0.0

is reflected in the significant evidence for a main effect

K N

B

N R K B

15

of temperature. Given the lack of significant evidence

20

25

30

Temperature

for an interaction, primary interest is on the varieties that have the highest main effect. Using Tukey’s HSD, variety BUR has overall lower mean weight than NOR or KEN, but not significantly lower than RLS. The other varieties do not differ significantly among themselves.

Chapter 9 – Page 8

Sum of

Source

DF

Squares

Mean Square

F Value

Pr > F

Model

15

3.42854201

0.22856947

5.15

<.0001

Corrected Total

95

6.97575156

Source

DF

Type III SS

Mean Square

F Value

Pr > F

temp

3

2.42938098

0.80979366

18.26

<.0001

temp*var

9

0.44016902

0.04890767

1.10

0.3704

Error

80

var

3

3.54720955

0.55899201

0.04434012

0.18633067

4.20

0.0082

EXERCISE 11 Using only treatments 1 through 9, we can perform a two-way ANOVA. There are significant effects due to PREP but not for the interaction with GRAIN. Source

DF

prep

grain

grain*prep

2

Type I SS

621.262344

Mean Square

F Value

Pr > F

2

1551.674144

775.837072

7.33

0.0018

4

784.949511

310.631172 196.237378

2.93 1.85

0.0635 0.1353

Pairwise comparisons of the main effects for PREP show that BSB has a significantly higher expected value than either WHOLE or DECORT (which do not differ significantly from each other).

To compare to the control group, we used a one-way ANOVA with all 10 treatment groups. The contrast compared the average of the means in the first 9 groups to the control group, coefficients for TRT were -1, -1, -1, -1, -1, -1, -1, -1, -1, 9. F(1,50)=38.51, p < 0.0001. There is significant evidence that the average of the mean values in the treatment groups differs from that in the Control group. Inspecting the sample averages, the Control group has a HIGHER mean biovalue. Note that this has a different MSE than the first analysis.

Chapter 9 – Page 9

EXERCISE 12 There are no significant interaction effects, so we can focus on the significant main effects for program.

Using Tukey’s HSD to compare the main effects, we find that the mean Risk for those trained by Program A is significantly worse than either Program B or C. Programs B and C do not differ significantly, you may choose either of those.

Program

A

B

C

Mean Risk

24.90 a 15.30 b 16.20 b

Means with same letter Are not significs. diff.

EXERCISE 13 a) Source

df

SS

MS

F

Model

5

92

18.4

4.94

Test

2

15

7.5

2.01

Distraction

1

69

69

18.5

Test*Dist.

2

17

8.5

2.28

Error

22

82

Corrected

27

174

3.727

b) At α=5%, the only significant effect is the main effect for distraction. Scores in one Distraction condition were consistently higher than in the other condition, over all versions of the test. But without the cell means, we don’t know which distraction condition was higher. c) the SS for the main and interaction effects don’t sum to the Model SS.

Chapter 9 – Page 10

EXERCISE 14 Source

df

SS

MS

F

Model

7

1840

262.857

3.613

Medication

3

910

303.333

4.170

Dose

1

500

500.000

6.873

Med*Dose

3

430

143.333

1.970

Error

40

2910

Corrected

47

4750

72.750

Using EXCEL to assign p values to the F statistics, there is significant evidence of main effects for Medication and Dose. However, there is no significant evidence of an interaction.

EXERCISE 15 a)

consistent in showing a very strong main effect of Outcome, with participants showing the highest

mean out. satis.

b) The plot and test statistics are

7 6

L H

5 4

H

3

L

2

H L

1 E q

satisfaction when Outcome is perceived

B tr

W rs

Outcome

as Equal. There is a weak main effect for Cog. Busyness, with a tendency for those with the Low Busyness to have less satisfaction. However, this tendency seems strongest in the Better Outcome category, leading to a significant interaction. c) 6*5/2 = 15 d) Yes, as noted in (b), the difference between the High and Low Busyness categories is most pronounced in the Better outcome category. There is less of a difference in the Equal and Worse categories. Means when outcomes Equal significantly better than in any other groups.

Chapter 9 – Page 11

EXERCISE 16 5

means, and 2 to mark the Minority source cell means.

The plot shows strong interactions. Among those who did not change their attitudes, those who were told their source was a

Mean Beh. Intent.

a and b) The plot uses 1 to mark the Majority source cell

Majority opinion had slightly higher mean Behavioral

2

4

3

1

1

2

2

Intention. Among those who did change their attitude, those

No

Yes

Attitude Change

who were told their source was a Minority opinion had much

higher mean Behavioral Intention. The significant interaction shows that there is significant evidence that the effect of source status differs among those who did and did not change their intention. While the main effects are significant, the interactions are so strong that the main effects have little meaning. c) Labeling the cell means using subscripts that correspond to the table, the contrasts are: #1: Ho : µ11 − µ21 = 0, ˆ = L 3 − 2.29 = 0.71, t = 0.71 /

2.248 * (1 / 16 + /17) = 1.36

ˆ = #2: Ho : µ12 − µ22 = 2.40 − 5.00 = −2.60, 0, L

t= −2.60 / 2.248 * (1 / 10 + /10) = −3.88 Each t-statistic has 53-4=49 df.

To control the experimentwise error rate at 5%, apply Bonferroni’s method and set the comparisonwise rate for each hypothesis at 2.5%. Reject Ho if |t| > 2.31 There is no significant difference in the means by source status for those who did not change their attitude, but there is a significant difference among those who did change their attitude. Yes, these results are consistent with the authors’ statements.

EXERCISE 17 Full model with all factors and their interactions had SSE = 447.50 and df = 72.

Reduced model dropping all terms involving TIME had SSE=542.822, df=88, = F(16,72)

(542.833 − 447.5) / (88 − 72) = 447.5 / 72

0.96

p value is 0.509. There is no significant evidence that TIME has any kind of effect.

Chapter 9 – Page 12

EXERCISE 18 Applying the coefficients for the quadratic trend to each sample cell mean, ˆ = L

= 267.7900, t

267.7900 /

20.2241 = * (60 / 3) 13.3

Applying the coefficients for the quadratic trend to each sample cell mean, ˆ = L −72.8399, t = −72.8399 /

20.2241 * (12 / 3) = − 8.1

The t statistic has 24 df. Using α = 2.5% (to control for multiple comparisons using Bonferroni) reject Ho if |t| > 2.39. Reject Ho for each contrast.

The significant linear trend has a positive coefficient, apparently the expected yield increases with increasing phosphorous. However, here is significant evidence of a quadratic trend and its coefficient is negative, meaning that the increase in yield per unit increase of P is not constant and tends to fall off as P increases. Both these statements are consistent with the profile plot.

EXERCISE 19 a) For each contrast, the coefficients sum to 0 so these are legitimate contrasts. For each pair of contrasts, the products of the corresponding coefficients sum to 0, so they are orthogonal. For example, L1 and L3 have dot-product 1*0 + 1*1 +1*(-1) +(-1)*0+(-1)*1+(-1)*(-1) = 0. b) L4: the difference between Standard and the average of Multi and GasMiser is the same for 6-cylinder cars as it is for 4-cylinder cars.

L5: the difference between Multi and GasMiser is the same for 6-cylinder cars as it is for 4-cylinder cars. c) The regression gives the same SSE (1.08433) as the two-way ANOVA. The regression gives a t = 0.43 for the Multi vs GasMiser comparison (L3) where the hand calculation gave 0.44, this is probably rounding error. d) The regression gives t = 4.25. Doing it using the cell means gives ˆ = L

3.96, = t

3.96 / 1.08433 *= (4 / 5)

4.25 , so they do correspond.

Chapter 9 – Page 13

CHAPTER 10 EXERCISE 1

This is a randomized block design with sampling, where Replicate is block. The proper

denominator is the interaction. There is no significant evidence of a difference in mean oxygen consumption by treatment. Hence, this data does not support either of the hypotheses.

Source

DF

Type III SS

Mean Square

F Value

Pr > F

rep

1

1.57597920

1.57597920

7.74

0.0103

trt*rep

2

2.12804180

1.06402090

5.23

0.0130

Source

DF

Type III SS

Mean Square

F Value

Pr > F

trt

2

2.76847407

1.38423703

6.80

0.0046

Tests of Hypotheses Using the Type III MS for trt*rep as an Error Term

trt

2

2.76847407

1.38423703

1.30

0.4346

EXERCISE 2 Each replication is a block, and we have a 3x4 factorial (unreplicated) design within each block. The denominator for the test of SOLN, VAR, and SOLN*VAR uses the pooled SS for all the interactions with REP. Source

DF

Type III SS

Mean Square

F Value

Pr > F

*

soln

2 3

92764

9177.000000

46382

3059.000000

68.86

<.0001

soln*var

6

1776.000000

296.000000

0.44

0.8445

rep *

var

Error: MS(Error)

2

107.722222

22

53.861111

14819

0.0127

200 175

Mean Weight

strong effect of the solution amount. Variety #1 (7) seems to analysis confirms a significant main effect due to Variety.

4.54

0.9234

673.588384

The profile plot is consistent with the significant evidence for a have a higher mean weight, while variety #3 the lowest. The

0.08

150 125 100 75 50 25 0 5

6

7

8

9 10 11 12 13 14 15

Solution amount

Chapter 10 - Page1

EXERCISE 3. a)

11.0

It appears that Variety A may have slightly lower Yield, when averaged over Nitrogen judging from the way that Variety B has higher observed yield under the low

10.6

Mean YIELD

levels. There may be an interaction,

L

10.8

H

10.4 10.2 10.0 9.8 9.6

nitrogen level (marked with an L on the

9.4

plot).

L

H

9.2

b)

H L

A

B

C

VARIETY

Source

DF

Mean Square

F Value 2

29.83474444

14.91737222

9.65

0.0018

VAR

2

6.23078611

3.11539306

2.02

0.1657

NIT

1 4

0.00027222

33.53125556

0.00027222

0.00

0.9896

YEAR*NIT

2

22.92551111

11.46275556

7.42

0.0053

REP

3

1.98078333

0.66026111

0.43

0.7363

REP*YEAR

6

2.03080000

0.33846667

0.22

0.9650

REP*NIT

REP*YEAR*VAR

3

12

0.68818333

0.22939444

0.15

0.9292

REP*YEAR*NIT

6

3.08036667

0.51339444

0.33

0.9101

REP*VAR*NIT

6

3.17314167

0.52885694

0.34

0.9043

YEAR

YEAR*VAR VAR*NIT

REP*VAR

2

6

Type III SS Pr > F

15.66710278

8.38281389

1.01085833 4.81083333

7.83355139

0.16847639 0.40090278

5.42 5.07

0.11 0.26

0.0059 0.0197

0.9941 0.9887

Main effect for VARIETY: F(2,6) = [6.2308/2] / [1.0109/6] = 18.49, p value = 0.0027 There is significant evidence of a main effect for variety.

Main effect for NITROGEN: F(1,3) = [0.0003/1] / [0.6882/3] = 0, p value = 0.975 There is no significant main effect for Nitrogen.

Interaction effect for VAR*NIT: F(2,6) = [15.667/2] / [3.173/6] = 14.81, p value = 0.0048 There is a significant interaction, that is the effect of NIT is different in at least one variety.

All these results are consistent with the profile plot. c) These results are specifically for random blocks (REPs) in these years, and not for years in general.

Chapter 10 - Page2

EXERCISE 4 In a classical split plot design, the experiment for the main plot (VARiety) is treated as a randomized block where block is the REPlication. Hence, the test for variety will use the REP*VAR interaction (sometimes called error(a)). The tests for TRT and VAR*TRT will used the pooled SS for the interactions with REP, excluding REP*VAR. Dependent Variable: brate

Source

DF

Type III SS

Mean Square

F Value

Pr > F

Trt

1

1.071019

1.071019

21.61

0.0002

trt*var

5

0.589569

0.117914

2.38

0.0799

rep(var)

18

2.938962

0.163276

3.29

0.0076

Error: MS(Error)

18

0.891963

0.049553

*

* This test assumes one or more other fixed effects are zero.

Source *

Var Error: MS(rep(var))

D F 5 1 8

Type III SS

Mean Square

4.142785

0.828557

2.938963

0.163276

F Value

Pr > F

5.07

0.0045

Chapter 10 - Page3

The profile plot is useful to understanding the results. The

4.0 3.8

varieties 4 and 5, particularly with TRT=2 (the higher

3.6

Mean BRate

varieties with the highest blooming rate appear to be planting rate). Tukey’s HSD applied to the main effects for VAR, using the MS(REP(VAR) interaction as error,

3.4 3.2 3.0 2.8

showed that varieties 4 and 5 were significantly higher

2.6

than 1 and 6, but could not confirm a difference with 2 and

2.4 1

3.

2

3

4

5

6

Variety

Dependent Variable: mrate Source

DF

Type III SS

Mean Square

F Value

Pr > F

Trt

1

4.687500

4.687500

0.09

0.7691

trt*var

5

481.937500

96.387500

1.83

0.1583

rep(var)

18

1362.875000

75.715278

1.43

0.2256

Error: MS(Error)

18

949.875000

52.770833

*

* This test assumes one or more other fixed effects are zero.

Source *

Var Error: MS(rep(var))

D F 5 1 8

Type III SS

Mean Square

160.937500

32.187500

1362.875000

75.715278

F Value

Pr > F

0.43

0.8252

MRATE does not seem to differ by any of the factors.

To choose the best combination, it would seem sufficient to focus on the best BRATE.

Chapter 10 - Page4

EXERCISE 5 a) The profile plot uses symbol to mark the number of the level for SALT, 1 = lowest and 4 = highest.

The highest level of SALT (4) tends to have a

90

lower mean percent of seeds emerging at every day. days, where an increasing amount of salt is clearly associated with a decreasing percentage of seeds emerging. The effect fades over time (a possible interaction). By DAY 14, only the highest level of salt seems different.

Mean Emergence

The effect of SALT is most noticeable early, at 5

80 70 60

2 1 3

1 2 3

1 2

1 3 2 4

4

4

50 40

3

30 20 10

4

5

6

7

8

9 10 11 12 13 14

DAY

b) Salt is a between subjects factor, since each plot (REP, which is subject) was entirely treated with the same level of Salt. DAY is a within subjects factor.

Source

DF

Type III SS

Mean Square

F Value

Pr > F

SALT

3

4343.395833

1447.798611

185.48 49.13

<.0001

<.0001

26.69

<.0001

DAY

3

8129.395833

2709.798611

DAY*SALT

9

3451.520833

383.502315

REP(SALT)

8

1666.666667

208.333333

ERROR

24

187.3333

7.80556

347.16

<.0001

Main effect for SALT: F(3,8) = [4343.396/3] / [1666.667/8] = 6.95, p value = 0.0128. There is significant evidence of a main effect for SALT. Interaction for Day*SALT: F(9,24) = 49.13, p value < 0.0001. There is significant evidence for a SALT*DAY interaction. c) The hypothesis tests confirm the main effect for SALT, which is apparently of the form that the higher the SALT level, the less the percentage of emerging grass. However, as confirmed by the interaction, this effect varies by DAY, and seems to be most important at days 5 and 8.

Chapter 10 - Page5

EXERCISE 6.

From the profile plot (averaging together data from the two 7

panels) there seems to be a strong effect due to temperature.

Meat stored at the higher temperature (level 4, 38 degrees) While all the meat tends to taste worse as time progresses, the effect is particularly noticeable among meat stored at the highest temperature.

Mean Rating

seems considerably worse than at the other temperatures.

6

5

4

3

2 1 2 3 4 5 6 7 8 9 10 11 12 13

TIME

For the formal analysis, I treated this as a two-way factorial design (TIME and TEMP as factors) embedded in a randomized block where BLOCKS were the panels of judges. Normally, it is important to include a panel or judge effect, as panels can differ in their preferences or scoring attributes. PANEL is a random effect. The interaction of PANEL with the other factors are pooled used to estimate Error.

Source

DF

Type III SS

Mean Square

F Value

Pr > F

TEMP

3

20.54393000

6.84797667

48.65

<.0001

TIME

TEMP*TIME

4

12

15.27168500

3.81792125

27.12

<.0001

PANEL*TEMP

3

0.23781000

0.07927000

0.56

0.6496

PANEL*TIME

4

0.39696500

0.09924125

0.70

0.6036

PANEL

Error

1

12

0.77841000

4.88609500

1.68921500

0.77841000

0.40717458

0.14076792

5.53

2.89

0.0366

0.0390

Pooled estimate of error is (0.23781+0.39697+1.68922)/19 = 0.1223.

Test for TIME: F(4,19) = 3.818/0.1223 = 31.2, p value < 0.0001. There is a strong main effect for time.

Test for TEMP: F(3,19) = 6.848/0.1223 = 55.99, p value < 0.0001. There is a strong main effect for temperature.

Test for TIME*TEMP: F(12,19) = 0.4072/0.1223 = 3.33, p value = 0.0095. There is evidence of an interaction.

All these results are consistent with the profile plot.

Chapter 10 - Page6

EXERCISE 7 a) If Rep is nested within Light, as in Example 10.7, and the Rep(Light) effect is random, we get an analysis as shown here.

There are significant main effects due to Light and Leaf but no interaction.

b) If both Light and Leaf are treated as quantitative variables, we might try a quadratic in both variables with a crossproduct. The predicted efficiency is given by 1.781 + 0.013LIGHT – 0.000 LIGHT2 +0.267LEAF -0.047LEAF2 + 0.0005LIGHT*LEAF This model has SSR = 2.463 with 5 df. c) By comparison, the model fit in part (a) has SSR=5.431 with 26 df (it is a much more complicated model). The lack of fit test is F(21,48) = [(5.431-2.463)/21] / 0.02236 = 6.32, p value < 0.0001, showing that the quadratic model does not fit this data very well.

EXERCISE 8 The profile plot suggests that Additive 1 is overall the best, 23

and Additive 3 the worst, but with such large variability among Treating this as a randomized block with subsampling then the denominator of the F-test for a main effect for Additive would be the Car*Additive interaction.

This F(2,4) = 3.39 has p value = 0.1379 which is not significant, that is, there is no significant evidence of a

22

Mean MPG

cars it isn’t clear whether the effect is significant.

21 20 19 18 17 16 1

2

3

ADDITIVE

difference in the additives. This may seem surprising given the strength of the effect in the graph and in the ordinary (incorrect) two-way ANOVA. In essence, this analysis says that the observed difference Chapter 10 - Page7

could be explained as a chance effect due to the particular characteristics of the three cars in the experiment. Note that more repetitions per car/additive combination will not increase the small number of degrees of freedom in the denominator of the F-test for Additive. Perhaps another car would be needed.

Results of Ordinary two-way ANOVA Source

DF

Type III SS

Mean Square

F Value

Pr > F

car

2

65.15166667

32.57583333

48.57

<.0001

car*additive

4

18.46666667

4.61666667

6.88

0.0006

additive Error

2 27

31.26166667 18.1100000

15.63083333 0.6707407

23.30

<.0001

Table of Expected mean squares.

Source

Type III Expected Mean Square

car

Var(Error) + 4 Var(car*additive) + 12 Var(car)

car*additive

Var(Error) + 4 Var(car*additive)

additive

Var(Error) + 4 Var(car*additive) + Q(additive)

Tests based on table of Expected Mean Squares.

Source

DF

Type III SS

Mean Square

F Value

Pr > F

additive

2

31.261667

15.630833

3.39

0.1379

Error

4

18.466667

4.616667

Source

DF

Type III SS

Mean Square

F Value

Pr > F

car*additive

4

18.466667

4.616667

6.88

0.0006

car

2

Error: MS(car*additive)

Error: MS(Error)

65.151667

27

32.575833

18.110000

7.06

0.0488

0.670741

Chapter 10 - Page8

EXERCISE 9 These analyses were run as a split-plot design with COWID nested within COLOR as a random effect.

40

Surface: The profile plot suggests that there are very

B

Mean Surf. Temp

strong treatment effects, with Fan and Mist doing the best

39

job of keeping the cows at a lower surface temperature.

There is not much of an effect due to color (used as the plotting symbol), nor is there a very strong interaction.

W

38

W

37

B

36

The ANOVA is in agreement, there are strong main

B W

B W

fan

mist

35

effects for Treatment, but no significant evidence of a main effect for color or a Treatment*Color interaction.

Source

DF 3

73.54093750

cowid(color)

2

trt*color Source

trt

Type I SS

shade

sun

Treatment

F Value

Pr>F

16.61

<.0001

1.7856

0.60

0.5550

3

3.17593750

0.72

0.5523

DF

Anova SS

Mean Square

Tests of Hypotheses Using the Anova MS for COWID(Color) as an Error Term color

1

0.00281250

0.00281250

temperatures. There may be a tendency for white cows to have lower temperature, though there is an exception in the shade group. The ANOVA does not find any significant effects, either for main effects or

0.00

0.9603

B

40.1

Mean Rect. Temp

cows with the Fan treatment to have lower

Pr > F

40.2

Rectal: The profile plot for the rectal temperatures shows less clarity. There is possibly a tendency for

F Value

B

40.0 39.9

W

39.8 39.7

W W

B

39.6 39.5

B W

39.4

fan

interactions.

mist

shade

sun

Treatment

Source

DF

Type III SS

Mean Square

F Value

Pr > F

trt

3

1.25250000

0.41750000

1.79

1.23

0.1783

0.3130

trt*color

3

0.52125000

0.17375000

0.75

0.5366

cowid(color)

2

0.57125

0.285625

Tests of Hypotheses Using the Type III MS for cowid(color) as an Error Term Source

DF

Type III SS

Mean Square

F Value

Pr > F

color

1

0.06125000

0.06125000

0.21

0.6888

Chapter 10 - Page9

EXERCISE 10 An analysis using Route as block (a random effect) and year as a qualitative variable showed no significant main effect for year. However, a linear contrast for year is significant (F=6.46, p = 0.0143) showing a trend upward at a rate of about 11.70 per year.

Source

DF

Type III SS

Mean Square

F Value

Pr > F

year

24

2693219

112217

1.36

0.1822

Contrast

DF

Contrast SS

Mean Square

F Value

Pr > F

route

Error: MS(Error)

linyear

2

48

1

4172531 3973337

534713.7323

2086265 82778

534713.7323

25.20

6.46

<.0001

0.0143

Fortunately, the residuals within each route/year are not strongly correlated with those of the previous year. However, the residuals show positive skew with larger variability for routes A and B (construct a boxplot).

An analysis using ln(BIRD) as the dependent variable produced more nearly symmetric and equally distributed boxplots of residuals, but also more clearly correlated errors. It also shows strong evidence of a linear trend in ln(BIRD).

EXERCISE 11 a) Averaging within each lab / material combination and submitting the resulting means to a randomized block design where Labs are the blocks yields exactly the same F values for LAB (F(12,72) = 7.64) and MATERIAL (F(6,72) = 135.48). We can no longer test the LAB*Material interaction. b) As we average together the n values within each cell, the resulting mean has variance σ δ2 / n + σ 2 . So the leading two terms in the EMS for treatments has decreased by a factor of 1/n. However, the third term now has n=1, so the third term will also be decreased by a factor of 1/n. Hence, all terms in the numerator and denominator of the F-test will decrease by the same factor, and the F-value will not change.

Chapter 10 - Page10

EXERCISE 12 This is a repeated measures design, with a two-way factorial experiment embedded within each subject.

The effects of interest are SHOCK, NOISE and the SHOCK*NOISE interaction, all within-subjects factors. The denominator for the test of each effect is its interaction with SUBJECT. The three-way interaction is used to estimate error.

Both Shock and Noise have significant main effects but the interaction is not significant.

Focusing on the main effects, mean response at Noise=80 is significantly higher than at 40. To compare main effects for Shock, we use Tukey’s HSD with error term set to the same Shock*Subject interaction used to construct the F-test. The mean response at Shock=0.75 was significantly higher than that at either Shock=1 or 0.5. The mean response at Shock=0.25 was significantly lower than at any other level.

Source

DF

Type III SS

Mean Square

F Value

Pr > F

noise

1

65.0250000

65.0250000

42.29

<.0001

shock

3

219.2750000

73.0916667

47.54

<.0001

subj

4

361.8500000

90.4625000

58.84

<.0001

shock*noise

3

12.6750000

4.2250000

2.75

0.0890

shock*subj

12

39.3500000

3.2791667

2.13

0.1020

Error

12

18.4500000

1.5375000

Source

DF

Type III SS

Mean Square

F Value

Pr > F

Error

4

6.350000

1.587500

noise*subj

*

noise

4

1

6.3500000

65.025000

1.5875000

65.025000

1.03

40.96

0.4300

0.0031

Error: MS(shock*subj)

* This test assumes one or more other fixed effects are zero.

Source

DF

Type III SS

Mean Square

F Value

Pr > F

*

3

219.275000

73.091667

22.29

<.0001

12

39.350000

3.279167

shock

Error

Error: MS(noise*subj)

* This test assumes one or more other fixed effects are zero.

Source

DF

Type III SS

Mean Square

F Value

Pr > F

subj

4

361.850000

90.462500

27.17

0.0004

Error

6.4322

21.413805

3.329167

Error: MS(shock*subj) + MS(noise*subj) - MS(Error)

Chapter 10 - Page11

Source

DF

Type III SS

Mean Square

Error: MS(Error)

12

18.450000

1.537500

shock*noise

3

12.675000

4.225000

F Value 2.75

Pr > F 0.0890

If you analyze this data as a two-way factorial embedded in a block (subjects as blocks, as in Section 10.4) the model will not usually include the subject*treatment interactions. These will be pooled with the three way interaction to form the estimate of error, yielding 28 df for SHOCK, NOISE and SHOCK*NOISE tests. This may be reasonable if all the Subject interactions appear weak, but in repeated measures designs, they frequently are not weak, as subjects vary in their reactions to the treatments. This is the primary difference between a repeated measures analysis with two within-subjects factors, and a factorial embedded in a block.

EXERCISE 13 The individual fish within each bucket are not independent of each other, because there is possibly an effect for each bucket’s individual characteristics. Including an effect of bucket in the analysis should reduce the dependence between the individual fish. This bucket effect should be a random effect nested within Treatment level. The resulting F-statistics for the TREAT effect are WEIGHT: F(4,15) = 16.54, p < 0.0001 LENGTH: F(4,15) = 8.31, p = 0.001 RELWT: F(4,15) = 27.91, p < 0.0001 That is, feeding level does have a significant effect on each of these size factors. Profile plots suggest that increasing TREAT leads to larger size, until about TREAT=1 or 2. Thereafter, further increases do not lead to larger size.

Note that the individual buckets are independent of each other, so we can analyze the total, or average, weight of the fish in the buckets using a one-way ANOVA. The resulting F statistics are exactly like those given above.

Chapter 10 - Page12

EXERCISE 14 a) Now the design is a factorial experiment (SALT*DAY) embedded in a randomized block (REP). The interactions with REP are pooled to estimate experimental error.

Source

DF

Type III SS

Mean Square

F Value

Pr > F

REP

2

251.375000

125.687500

2.35

0.1124

SALT

3

4343.395833

1447.798611

27.10

<.0001

DAY*SALT

9

3451.520833

383.502315

7.18

<.0001

* *

DAY

Error: MS(Error)

3

8129.395833 30

2709.798611

1602.625000

50.73 53.420833

<.0001

* This test assumes one or more other fixed effects are zero.

b) The difference between the two problems is that in the description of problem 5, we actually have 12 plots (three at each level of SALT) that are observed repeatedly on 4 different days. This makes it a repeated measures model with DAY as a within-subjects factor and SALT as a between subjects factor (a subject is a plot, identified as a combination of REP and SALT).

In problem 14, however, there are actually three locations (identified by REP), assumed randomly selected. Rather than repeatedly observing the same plot multiple times, we select 16 separate plots. Some are observed at day 5, others at day 8, etc. We might do this if the actual observation process of measuring emergence damaged the plot in some way (e.g. by walking on it or digging in it), making it impossible to measure the same plot repeatedly.

Chapter 10 - Page13

CHAPTER 11 EXERCISE 1 Source

DF

Type III SS

Mean Square

F Value

Pr > F

wwt

1

394.0805861

394.0805861

34.24

<.0001

weaned

2

289.8186081

144.9093040

12.59

0.0004

There is significant evidence that weaning time is related to FWT, even after adjusting for WWT. Pairwise comparisons using Bonferroni’s to control the family-wise error rate shows that the LATE group is significantly less than either the EARLY or MEDIUM, after adjusting for weaning weight.

Check for unequal slopes: F= (195.6337-179.2549)/(17-15) / 11.9503259 = 0.6853 (p value 0.519) From Table A.4, critical value F(2,15)= 3.68, so we do not reject Ho that slopes are equal.

However, piglets from within the same litter are probably not independent because of maternal and genetic effects. Piglets were not randomly assigned to weaning time (needed to keep all piglets of same litter together).

EXERCISE 2. a) Unequal variance t’(12.1)=0.94, p value = 0.3675. Pooled t test has similar conclusion, t(14)=0.94, p value = 0.3651. Though the Calcium plus exercise group had a slightly higher change in bone density, the difference was not significant. b) overall mean BMI = 25.9625 Among Calcium group, expected change = 3.8325 - 0.0956*25.9625 = 1.35 Among Ca+Ex group, expected change = 3.8325 + 0.4233 - 0.0956*25.9625 = 1.77 Difference is significant (t(13)=2.77, p value = 0.0158) c) Interaction formed by multiplying dummy for group times BMI did not significantly improve model (t (12)=-0.94, p value = 0.3666) d) ANCOVA is valid because treatment was assigned randomly. Source of difference is that BMI explained some of the variation in the response, reducing the size of the error variance.

EXERCISE 3 a) People in Class 1 have expected value β0 + β3PRE and those in Class 3 have expected value β0 + β2 + β3PRE . The expected difference is β2 with confidence interval 4.058 ± 2.0066 2.664 = (0.78,7.33)

Chapter 11 – Page 1

b) People in Class 2 have expected value β0 + β1 + β3PRE . The difference between a person in Class 3 and one in Class 2 (having same PRE) is β2 − β1 . Confidence interval = 5.015 ± 3.463 = (1.55,8.48) (4.058 − (−0.957)) ± 2.0066 2.664 − 2(1.095) + 2.504

c) Person in Class 3 with PRE = 6 has expected value β0 + β2 + β36 and the person in Class 1 with PRE=10 has expected value β0 + β310 . The difference is β2 − 4β3 . The confidence interval is

(4.058 − 4 * 0.773) ± 2.0066 2.664 − 2 * 4 * 0.026 + 16 * 0.029 = 0.966 ± 3.429 = (−2.46,4.40)

EXERCISE 4 a) Using an ANCOVA, trees on the south side have the fitted equation Circum = -17.084 + 2.701*Rings While those on the north side have the fitted equation Circum = 40.366+2.701*Rings.

The expected difference between two trees with the same number of rings, one on the north side and one on the south, is 57.45, which is significantly different from 0 (t = 8.13, p value = 0.0001).

The overall mean number of rings was 27.714. Given this number of rings, the adjusted circumference on the north side is 115.2 and that on the south side is 57.8.

Trees were not randomly assigned to sides of the mountain. While the number of rings tended to be higher on the south side, the difference was not significant. b) An interaction between Side and Rings was not significant (t = 1.68, p value = 0.1003). c) Growth is the growth rate per year, and is a direct measure of the relation of tree size to age. An independent samples t test (unequal variance version) can be used to compare growth rates on the two sides. The north side had substantially higher growth rates (t = 6.8 with 31 df, p value = 0.0001).

EXERCISE 5 a) Expected difference = E(width – length) = E(width) – E(length) = 12 – 12 = 0 b) Variance = 0.0025 – 2*(0.0026) + 0.0064 = 0.0037 and the standard deviation is 0.0608 EXERCISE 6 a) The cell mean plot shows that for most risk groups, the mean hospital time is less in the Antiviral group than in the Placebo group. However, in 3 of the 5 older-age risk groups, there is very little difference.

Risk Group digits are in the order of Diabetes, Hypertension, Imm. Suppression, Age.

Chapter 11 – Page 2

b) F(21,203)=12.12, p value < .0001. At least one group has a different mean hospital stay. c) The antiviral has a significant main effect (t(215)=-2.75, p value =0 .0064) showing it tends to reduce hospital time on average by 2.79 days. However, there is a significant interaction with AGE with a regression coefficient of 2.48 days that lessens the impact of the antiviral among older patients. This is consistent with the cell mean plot. This is the only significant interaction. Diabetes, Immune Suppression and Age are all risk factors that tend to increase hospitalization time. d) full model (part B) had SSE = 1794.79 and df = 203 Restricted model (part C) had SSE = 1907.525 with df=215 F(12,203)=[(1907.525-1794.79) / (215-203)] / (1794.79/203) = 1.06, p value = 0.396. No significant evidence that the model in part C does not fit, we can accept the model in part C.

Chapter 11 – Page 3

EXERCISE 7

a)

Proc ANOVA: MSE = 2.4819 Source

DF 2

Anova SS

216.7736862

Mean Square

F Value

Pr > F

tread

2

203.6755235

101.8377618

41.03

<.0001

pave*tread

4

22.1545082

5.5386271

2.23

0.1086

pave

108.3868431

43.67

<.0001

DUMMY VARIABLE using proc GLM with class statement for PAVE and TREAD:

Source

DF

Type III SS

Mean Square

F Value

Pr > F

pave

2

233.5838333

116.7919167

47.06

<.0001

tread

2

212.4633531

106.2316765

42.80

<.0001

pave*tread

4

6.6989814

1.6747453

0.67

0.6186

Proc ANOVA is NOT NOT NOT for unbalanced multi-way ANOVA. Use PROC GLM. b) Performing an ANCOVA with TREAD as the covariate, Source

DF

Type III SS

Mean Square

F Value

Pr > F

pave

2

232.8180697

116.4090348

52.31

<.0001

tread

1

219.0624849

219.0624849

98.44

<.0001

This model had MSE 2.2254, which is even better than the MSE from the full two-way ANOVA model (MSE = 2.482). The Tread*Pave interaction was not significant and was dropped. c)

µ(y|x)= 26.19377022 + 28.65978927(friction) + 1.37431143(tread) with MSE = 2.474. This

model does not fit as well as the model in (B). The relationship is most likely non-linear in Friction.

EXERCISE 8 a) The overall mean dosage (X) is 6.014. At that dosage, the fitted mean half_life for drug A is 1.85 and that for drug G is 2.04. The fitted difference is 0.19 hours greater for Gentamicin. However, this difference is so small that there is no significant evidence that the drugs differ in their expected half-life, after controlling for dosage (t = 0.29, p value = 0.7714). b) There is no significant evidence that the slopes differ (t = -0.17, p value = 0.8694.) However, note that the dosage (the covariate) differs greatly for the two groups (that is, dosage is not independent of drug type), so that it is not reasonable to compare half-life for the two drugs at the same dose.

Chapter 11 – Page 4

EXERCISE 9

a)

Source

DF

Type III SS

Mean Square

F Value

Pr > F

medium

2

3137.391892

1568.695946

50.97

<.0001

medium*minutes

6

514.574392

85.762399

2.79

0.0812

minutes

3

1514.467662

504.822554

16.40

0.0005

There is no significant evidence of an interaction. There is significant evidence that the mean number of words differs by Medium and by number of Minutes. Using LSMEANs and Tukey’s HSD to compare main effects, it appears that mean number of words is higher at 10 and 20 minutes than at 5 and 15. Each of the mediums is significantly different, written material seems to be the highest. b) There seems to be a tendency for children in the longer time periods to drop out. This may produce a bias, if it is the weaker learners, who might be bored, who are falling asleep. There is also a tendency for students in the TV group to be less likely to drop out. Again, this may produce a bias.

EXERCISE 10 Note, all variables were transformed using natural logarithms giving a log-linear model.

A dummy variable was created for season, IS = 0 for cool, 1 for warm.

Two interactions were formed by multiplying IS times ln(WEIGHT) and IS times ln(HEIGHT).

The full model including independent variables ln(WIDTH), ln(HEIGHT), IS, IS*ln(WIDTH) and IS*ln(HEIGHT) had SSE = 23.2702 and 53 df.

The restricted model using only ln(WIDTH) AND ln(HEIGHT) had SSE = 31.33024 and df = 56.

The F test for the hypothesis that the 3 parameters related to season are 0 is F(3,53) = [(31.33024 – 23.2702)/3] / (23.2702/53) = 6.12, which has p value = 0.0012.

There is significant evidence that the relation between ln(weight) and ln(width) and ln(height) differs by season. Inspecting the fitted regression coefficients in the full model, it appears that the major difference is with respect to ln(height). In the cool season, the partial regression coefficient for ln(height) is +0.747.

However, in the warm season, the partial regression coefficient is +0.747-0.538=0.209.

Chapter 11 – Page 5

EXERCISE 11 a) Gas price (in cents) = 112.727 + 2.258(oil price in dollars per

60 50

barrel)

40

b) The residuals are plotted against the previous positive association. Residuals do not appear to be independent.

30

Residual

month’s residuals shows a moderately strong

20 10 0

c) The Durbin-Watson statistic was 0.744 and

-10

there was significant evidence of a positive

-20

correlation in the residuals. This is consistent

-30 -40

with the plot in (b).

-40 -30 -20 -10

d) Since the residuals are positively correlated,

0

10

20

30

40

50

60

Previous residual

we would expect gas prices to be lower than predicted next month also.

EXERCISE 12 a) Here is the covariance matrix of the parameter estimates.

Intercept

ISeason

LogHt

LogWt

LogHT*IS

LogWT*IS

Intercept

0.423

-0.423

-0.051

-0.155

0.051

0.155

Iseason

-0.423

0.647

0.051

0.155

-0.080

-0.211

LogHT

-0.051

0.051

0.041

-0.013

-0.041

0.013

LogWT

-0.144

0.155

-0.013

0.089

0.013

-0.089

LogHT*IS

0.051

-0.080

-0.041

0.013

0.061

-0.021

LOGWT*IS 0.155

-0.211

0.013

-0.089

-0.021

0.120

b) Mean ln(height)= 2.185 , mean ln(weight)= 2.287 Mean difference warm season – cool season using average values for ln(height) and ln(weight) is

βiseason + βislhd * 2.185 + βislwd * 2.287 Point estimate = 0.175 – 0.538*2.184+0.098*2.287 = -.776 (warm less than cool for given height, weight) In the calculation of the variance, use a ' = ( 0 1 0

0 2.185 2.287 )

Variance =0.042, std deviation = 0.203, t0.025(53) = 2.006, Confidence interval for difference in ln(weight) = -0.776 ± (2.006)*0.203 = (-1.18 ,-0.37). This is a confidence interval for log(wt) in warm – log(wt) in cool.

Chapter 11 – Page 6

EXERCISE 13 a) Among Gizzard Shad, a Two-way ANOVA with dependent variable LENGTH and factors NET_TYPE and mesh SIZE showed no significant evidence of an interaction.

Source

DF

Type III SS

Mean Square

F Value

Pr > F

SIZE

1

913381.3159

913381.3159

3881.24

<.0001

NET_TYPE

1

85.5505

85.5505

0.36

0.5472

SIZE*NET_TYPE Error

1

461.1210

461.1210

220

51773.107

235.332

1.96

0.1630

Dropping the interaction and fitting a model with only main effects showed significant evidence of an effect due to mesh SIZE (F(1,221) = 4968.49, p value < 0.0001) and due to NET_TYPE (F(1,221) = 4.42, p value = 0.0367). Mesh size seems to be the most important factor, with Mesh Size = 2 associated with much larger fish. b) Among Other fish, a two-way ANOVA showed a significant interaction of NET_TYPE and Mesh SIZE (F(1,33)=4.72, p value 0.0371).

Size had a strong main effect(F(1,33)=17.92, p value = 0.0002). Examining the means in the 2x2 table, it appears that for Mesh SIZE = 1, fish were larger with NET_TYPE = Mult, but when Mesh SIZE = 2, fish were larger with NET_TYPE = Mono. A profile plot is useful in understanding the interaction, the plotting symbol ‘1’ is for Mono and ‘2’ is for Mult type nets.

EXERCISE 14 Dropping the effect of P = Planting Rate (and all its interactions) and replacing it with NO = actual number of plants treated as a quantitative independent variable yields the following tests: Dependent Variable: TDM Source

DF

Type III SS

Mean Square

F Value

Pr > F

REP

3

4.730618

1.576873

1.11

0.4661

2.9966

4.246301

1.417055

Error

Error: 1.0007*MS(REP*WTR) - 0.0007*MS(Error)

Chapter 11 – Page 7

Source

DF

Type III SS

Mean Square

F Value

Pr > F

Error

3.0031

4.254776

1.416796

Source

DF

Type III SS

Mean Square

F Value

Pr > F

NRATE

2

87.780296

43.890148

36.18

<.0001

no

1

30.817525

30.817525

25.40

<.0001

Error: MS(Error) 83

100.694635

1.213188

WTR

1

45.102474

45.102474

Error: 0.9994*MS(REP*WTR) + 0.0006*MS(Error)

REP*WTR NRATE*WTR

3

4.250757

2

50.612990

1.416919

25.306495

31.83

1.17

20.86

0.0110

0.3270 <.0001

Note that the MS(ERROR) is much larger here (1.213) than it was in Table 10.22 (0.899) suggesting that dropping the interactions of Planting Rate with NRATE and REP was not a good idea. That is, it may be that the lines are not parallel.

Repeating the analysis using NO*WTR, NO*NRATE and NO*WTR*NRATE reduces the MSE (though only to 1.08). Given the strong interactions of NO with NRATE, an ANCOVA is not appropriate.

Source

DF

Type III SS

Mean Square

F Value

Pr > F

WTR

1

8.687273

8.687273

8.04

0.0089

Error

25.1

27.111166

1.080138

Error: 0.2721*MS(REP*WTR) + 0.7279*MS(Error) Source

DF

Type III SS

Mean Square

F Value

Pr > F

REP*WTR

3

3.797345

1.265782

1.25

0.2966

NRATE*WTR

2

11.551206

5.775603

5.71

0.0048

no

1

35.342156

35.342156

34.97

<.0001

no*WTR

1

0.285493

0.285493

0.28

0.5966

NRATE

no*NRATE no*NRATE*WTR Error: MS(Error)

2

2 2

2.965979

16.910387 4.575578 78

1.482990

8.455193 2.287789

78.837742

1.47

8.37 2.26

0.2369

0.0005 0.1108

1.010740

Chapter 11 – Page 8

EXERCISE 15 a) Source

DF

Type III SS

Mean Square

F Value

Pr > F

zip

3 1

617174205

30252281277

205724735

30252281277

0.58

85.55

0.6297

bed

1

2382102087

2382102087

6.74

0.0123

bath

1

65353263

65353263

0.18

0.6691

size

<.0001

Using reference cell coding with zip=4 as the baseline class, the fitted model was Price (in 1000’s) = 39.54 +63.86(size in 1000’s) -12.46(BED) -2.95(BATH) -10.097(ZIP1) -5.825(ZIP2) -4.143(ZIP3) Zip does not seem to make a difference, AFTER CONTROLLING for SIZE. But if you do a one-way ANOVA using only ZIP, ZIP might be significant if bigger houses fall in one of the zip codes. b) Source

DF

Type III SS

Mean Square

F Value

Pr > F

zip

3 1

13233407

541740770

4411136

541740770

0.02

0.9968

size

1

13676600860

13676600860

54.40

<.0001

bath

1

31275485

31275485

0.12

0.7263

lot

1

587574347

587574347

2.34

0.1346

garage

1

2338586052

2338586052

9.30

0.0042

age bed

exter fp

1

2 1

1532827804

1532827804

549059509 43352447

274529755 43352447

2.15 6.10

1.09 0.17

0.1503 0.0181

0.3458 0.6803

EXERCISE 16 a) The covariance matrix for Y = (output, demand) is  49 0  V =    0 100 

The expected difference is 500-500=0.

The variance is 49 + 2*(-1)*0 + 100 = 149, so the standard deviation of the difference is 12.21. b) Now the covariance is 0.7 49 * 100 = 49 so the covariance matrix is  49 49  V =   . The variance is 49 + 2*(-1)*49 + 100 = 51 and the standard deviation is 7.14. (Since  49 100 

the standard deviation is smaller than in part (A), there is less chance that the factory will have a large discrepancy between their output and demand.)

Chapter 11 – Page 9

EXERCISE 17 20

a) The plot shows no strong difference between the weight gains in the Girls (sex = 0) and in the Boys (sex = 1). This is consistent with the results of the independent samples t test (pooled t(22)=0.75, p value = 0.4692). There is no significant evidence that the mean weight change differs

10 c h a n g e

for the boys and girls.

0

-10

-20 0

1 sex

b) The ANCOVA shows that Girls are 20

expected to gain 10.8 kg more than a Boy of

B G

the same initial weight. The difference is significant (t(21)=-2.69, p value = 0.0138).

(A). Here we are comparing a Girl and Boy of the same initial weight. That would have to be a heavier than average girl or a lighter than

Change

Notice that this is a different statement than in

10

G G

B

G G G

0

G G

B BB

G

G

B B

B G B B

-10

average boy.

G

B B

-20 40

c) Since the students are not randomly

50

60

70

80

90

Weight Before

assigned a Sex independently of the initial weight, and the mean weights differ so greatly for the two groups, then the ANCOVA is not appropriate.

Chapter 11 – Page 10

EXERCISE 18 a) Since boys have SEX=0, a ' = (1 0 1.4 0) . The point estimate is  −75.75   20.73  1 0 1.4 0 ( )  78.48  = 34.12    −12.45 

The estimated variance of this estimate is  126.56 −126.56 −82.59 82.59   1   10.93   −126.56 288.07 82.59 −187.14   0   −10.93     = = (1 0 1.4 0)  (1 0 1.4 0)  82.59 54.06 −54.06   1.4   −82.59  −6.91        82.59 −187.14 −54.06 121.87   9   6.91 

1.26

The standard error is 1.26 = 1.12 Approximating the t with 192 df as a standard normal, a 95% confidence interval would be 34.12 ± 1.96*1.12 = 32.92 to 36.32 kg. b) The point estimate for an actual boy would also be 34.42, but the estimated variance of this estimate is 1.26 + 34.91 (using the MSE as an estimate for σ 2 ). The confidence interval for an actual boy’s weight would be 34.12 ± 1.96*6.01 = (22.34, 45.90).

EXERCISE 19 a) From the fitted regression parameters, we estimate that a southern state (SOUTH=1) will have a homicide rate that is 0.257 higher than a non-southern state (SOUTH=0) that has the same values for the other independent variables. This difference is significant (p value = 0.01). b) Since the states were not randomly assigned to levels of SOUTH, it may be that the southern and nonsouthern states differ substantially with respect to the values of the other variables. Comparing a southern and non-southern state at some mid-level of the other variables may be a way of extrapolating outside the range of the data. c) Oddly, Divorce Rate (DIV) is more strongly associated with homicide rate, as it has a standardized regression coefficient with a much larger absolute value. Other variables, such as RD and YOUNG, are of about the same influence as SOUTH.

Chapter 11 – Page 11

EXERCISE 20 a) F(3,162) =

(166 − 4)(0.34)2 = (1 − 0.342)

21.18 . Yes, there is significant evidence that at least one of the

independent variables is linearly related to child’s effort rating. b) The t test statistics for the null hypothesis that the coefficient is 0 are: AGE t = 0.12/0.03 = 4; SEX t = -0.24/0.14 = -1.71; CONDITION t = -0.06/0.14 = -0.43 so AGE is significant. c) We estimate that girls will rate effort, on average, 0.24 points lower than boys of the same age and condition. However, the difference between girls and boys of the same age and condition is not significantly different from 0 at α = 5%.

For each additional year of age, children are expected to have a rating that increases by 0.12, compared to other children of the same sex and condition. This increase is significantly different from 0 at α = 5%.

Chapter 11 – Page 12

CHAPTER 12 EXERCISE 1

a) The sample proportion is 524/737 = 0.71099. With confidence 99%, the probability of

a success for this set of dogs is between 0.668 and 0.754.

Calculation: 0.71099 ± (2.576)*(0.0166976) b) Ho: all 4 dogs have the same probability of success.

X²= 3.2268 with 3 df. Reject Ho if X² greater than 7.815 from table of Chi-squared. Do not reject Ho. Or from SAS, note that the p value is 0.3580 which is greater than 0.05. There is no significant evidence that the dogs differ in their probability of success.

EXERCISE 2 X2=8.46 with 1 df, p value = 0.0036. There is a significant difference in the crime patterns in the two cities. While both cities show more auto thefts than robberies, in City C the proportion of auto thefts is even higher than in City B. EXERCISE 3

Ho: pool and nonpool apts have the same probability of being leased by single occupants.

X²= (22-20.7)²/20.7 + (23-24.3)²/24.3 + (24-25.3)²/25.3 + (31-29.7)²/29.7= .275, p value = 0.6001.

Do not reject Ho. There is no significant difference in the proportion of single occupants in the pool and non-pool apartments.

EXERCISE 4 X 2 =156.8 with 3 df. There is significant evidence of a change in the distribution of the return times. Apparently many more surveys are being returned, and returned earlier.

EXERCISE 5 mean = 4.302, std dev. = 0.602. If use 7 categories of width 0.3 centered at the mean, dfoot ≤3.552 3.552 3.852 4.152 4.452 4.752 >5.152 3.852 4.152 4.452 4.752 5.152 N

8

7

7

15

E

6.81

7.74

11.15 12.59 11.15 7.74

6.81

χ2

0.21

0.07

1.54

0.48

0.46

9 0.41

13 3.57

5

χ 2 = 6.74 with 7 – 1 – 2 = 4 df, p value = 0.150, no significant evidence that the distribution is nonnormal.

CAUTION: YOUR VALUE OF CHI-SQUARED MAY DIFFER SOMEWHAT DEPENDING ON HOW YOU FORM YOUR CLASSES. Keep E values of at least 5.0.

Chapter 12 – Page 1

EXERCISE 6 Fisher’s exact test had two-sided p value = 0.0022. There is significant evidence that those who took the prerequisite had a different probability of passing the course compared to those who did not take the prereq.

EXERCISE 7 Ho: distribution of output quality has not changed.

E1 = (200 * .8) = 160

E2 = (200*.18) = 36

E3 = (200*.02) = 4

X²= 3.306, which is less than the critical value from the Chi-squared table with 2 df, 5.991 (p value 0.191). Do not reject Ho, no significant change in output of machine.

We should repeat this analysis combining the last unacceptable class with the good class, since the expected value for this class is less than 5. Then X²= 0.281 with 1 df. Again, there is no significant change in output quality for the machine.

EXERCISE 8 X 2 =79.35 with 2 df, p value < .0001, there is strong evidence of a difference in job distribution by gender. Examining the standardized residuals, women are underrepresented in accounting and overrepresented in secretarial posts, and the opposite pattern for men.

EXERCISE 9 X²= 7.38 with 4 df, p value = 0.1173. There is no significant evidence that the distribution of rating differs by city. SAS also reports the likelihood ratio statistic X²= 7.66 with p value of 0.1051, with a similar conclusion.

EXERCISE 10 Ho: each category has probability 1/3. Under Ho, each category has expected value 40.

X 2 = 3.8 with 2 df. Reject Ho if X 2 > 5.991. If using p values, p value = 0.15. There is no significant

evidence that any of the traits is more commonly valued than any other.

EXERCISE 11 a) X²= (50-45.22)²/45.22 + (44-36.17)²/36.17 + (10-22.61)²/22.61 + …..+ (2-.87)²/.87= 26.254 which has 6 df (p value from SAS was 0.0002). Critical value from Table A.4 is X²(6)= 12.592, reject Ho. The school types differ significantly with respect to the distribution of the type of person the students feel most affected their life. b)

problem: some cell frequencies are less than 5. Combine the Other group with the Politician

group. Adjust cells gives X²=23.39, p value 0.0001, still reject Ho.

Chapter 12 – Page 2

EXERCISE 12. a) X 2 = 13.62 with 4 df, p value = 0.0086. There is significant evidence of a relationship between the traits the managers consider most important in a salesperson and the traits they consider most important in a sales manager. b) The contingency coefficient is 0.319. While there is significant evidence the variables are not independent, the actual strength of the dependence is not very great.

EXERCISE 13 a) Forming four groups (for the four combinations of Size and Net_type) and testing that the proportion of gizzard shad is the same in all groups, X 2 (3) = 63.29, the proportion is different in at least one group. Using the Chi-squared test (or Fisher’s exact test) to make the six pairwise comparisons, using α = .05/6 = .0083 on each, we get that the MULT-2 group is the only one that is different. In the table, groups with the same letter were not significantly different.

Group

Mult-2

Mono-2

Mult-1

Mono-1

% G. Shad 37.9% a 86.1% b 91.7% b 94.6% b b) The probability of exceeding 193 is significantly different for some group, X 2 (3) = 96.95. This indicates that the median length is different for at least one group. Using Fisher’s exact test for pairwise comparisons at α = .0083, we see that nets with size 2 tend to catch longer fish than those with size 1.

Group

Mono-1

Mult-1

Mult-2

Mono-2

% exceeding 193 25.81% a 36.46% a 96.55% b 100.0% b EXERCISE 14 Ho: age distribution is the same for both those reporting and not reporting AI-driving. X 2 = 96.94 with 2 df, p value < 0.0001. There is significant evidence that the age distributions in the two groups do differ. Those reporting AI driving are under-represented in the 18-20 age group (where alcohol sales are illegal), and over-represented in the older age groups.

EXERCISE 15 Since the counts are small for the number developing fevers, use Fishers Exact Test, twotailed p value = 0.6827. This difference could easily arise by chance, there is no significant evidence that the groups differ in their probability of developing a fever.

EXERCISE 16 Ho: the distribution of ratings is the same in both communities. Combining rating categories 1 through 3 into a single category, X 2 = 6.84 with 2 df and p value = 0.0327. There is significant evidence of a difference in the distributions.

Chapter 12 – Page 3

CHAPTER 13 EXERCISE 1 a) The plot shows that

2

the ln(odds) of favoring the death In both races, the ln(odds) increases quickly in the 1970’s but the rate of increase slowed by the mid 1980’s. The size of the gap between whites and

Ln(ODDS)

penalty are much higher among whites.

1

W

B B

0 B BB

blacks seems stable across the years.

-1

b) We fit a model that replaced YEAR

W W WW WW

W W W W W W WW W W W W W W

BB B B

1970

B

B B B B BBB

1980

with YEARS=YEAR-1972, to avoid

BB B

B

1990

2000

YEAR

very large numbers in the quadratic term. We coded RACE = 0 for white and 1 for blacks.

Ln(odds) = 0.4291 -1.224Race + 0.0813YEARS -0.00219 YEARS2, with -2lnL = 38,448.182.

Yes, there is significant evidence that the quadratic term improves the model, it had Wald Chisquared = 68.05 with p < 0.0001. Race dummy variable had Chi-square 2104.14 with p < 0.0001, significantly lower approval of death penalty among Blacks. c) Adding an interaction of RACE with YEARS and RACE with YEARS2 gave -2lnL = 38,443.496. The likelihood ratio test has X 2 = 4.686 with 2 df, which is not significant at α = 5% (it has p value just below 0.10). Hence, there is no significant evidence of an interaction. d) Yes, it is reasonable to refer to the gap as enduring. The first model showed extremely strong evidence for a gap. Since the interactions were not significant, the gap in the ln(odds) does not appear to be changing much with time. Note, however, that the plot of the probabilities might show a slight change in the size of the gap, but it would be small.

Chapter 13 –Page 1

EXERCISE 2 a) The Chi-squared test of homogeneity has X2 = 8.2736 with p value 0.0160. There is significant evidence that the types of neighborhood differ with respect to the probability a robbery will involve a gun. b) Using 2 dummy variables I2 = 1 if faith is medium, 0 otherwise I3 = 1 if faith is high, 0 otherwise The Chi-squared test based on the likelihood ratio has X2 = 8.3468 with p value 0.0154, which shows significant evidence of a relationship between the probability a robbery will involve a gun and the level of Faith in Police. The medium faith neighborhoods had a probability that differed significantly from the low faith neighborhoods (estimated OR = 0.381, p value = 0.0076).

However, the high faith neighborhoods did not differ significantly, even though the estimated OR (0.456, p value = 0.0655) wasn’t much different from that in medium faith neighborhoods. This is possibly because the sample was smallest in the high faith neighborhoods.

you subdivided the categories. You may need to combine several categories in the lower ‘ventrical’ groups. There seem to be increasing probabilities of an abnormal EEG as ventrical size increases. However, it may be that the ln(odds) are not linear in ventricle size but somewhat curvilinear.

Ln(odds) abnormal EEG

EXERCISE 3 a) Your plot may vary depending on how

1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 -0.1 -0.2 -0.3 -0.4 -0.5 -0.6 -0.7 -0.8 -0.9 -1.0 30

40

50

60

70

80

90

Midpoint of Ventrical category

b) fitted model: Ln(odds) = -4.0478 + 0.0569V For each additional unit of ventricle size, the ln(odds) of an abnormal EEG increases by 0.0569.

This increase is significant (likelihood ratio Chi-square 6.836 with p value 0.0089, Wald ChiSquare p value 0.0149).

Chapter 13 –Page 2

EXERCISE 4 a) The first dummy variable was TIME (0 before intervention and 1 afterwards). Then there were 26 dummy variables for cities 2 through 27 (city 1 was the baseline city). The actual data set was rearranged to have 54 lines, two for each city (one with the Crashes and Years before intervention and one with the Crashes and Years after intervention). Ln(Years) was used as an offset variable.

The Poisson regression had a fitted coefficient of -1.1248 for TIME with Chi-square=118.01 and p value < 0.0001. There is significant evidence that the crash rate was reduced significantly.

With confidence 95%, the logarithm of the crash rate was reduced between -1.3277 and -0.9219.

That is, the actual crash rate after intervention was between 26.5% and 39.77% of what it was before the intervention. b) However, the title makes it sound like this might not have been a complete tabulation of all sites or even a random sample. The authors may have picked the sites with the greatest crash reductions to give in their table, thus biasing the results.

Chapter 13 –Page 3

EXERCISE 5 a) The plot shows that accidents involving male drivers are much more likely to be alcohol related. There is some indication of a mild

-1

women is somewhat larger in the later age categories.

Estimated ln(odds)

interaction, in that the gap between odds for men and

M

-2

M

M

-3

F

M

F F

-4 F

-5 1

2

3

4

AGE Category

b) The plot of the estimated probabilities still shows that crashes among men are more likely to be alcohol suggestion of an interaction is much stronger. c) Dummy variable for Sex was coded as 0 for men and 1 for women.

M

11 10

M

9 8 7 6 5 4

M

F

F F

3 2

There were 3 age dummy variables.

M

12

Estimated probability %

related than crashes among women. However, now the

13

1

A2 was 1 for age 18-20 and 0 otherwise.

F 1

2

3

4

AGE Category

A3 was 1 age 21-24 and 0 otherwise.

A4 was 1 for age 25+ and 0 otherwise.

Fitted ln(odds) = -3.2067 – 1.2144Sex + 1.051A2 +1.268A3 + 1.118A4 All the dummy variables were significant. Females have significantly lower ln(odds) of a crash being alcohol-related. All the older age groups have significantly higher ln(odds) of a crash being alcohol-related compared to the youngest age group. d) Fitted ln(odds) = -3.234 - 1.0503Sex + 1.0246A2 + 1.303A3 + 1.1699A4 + 0.1561Sex*A2 - 0.2128Sex*A3 - 0.3104Sex*A4 The interaction affects the size of the gap between the ln(odds) of the men and women. In the youngest age group, the ln(odds) for women is -1.0503 less than the ln(odds) for men. In the oldest age group, the ln(odds) for women are -1.0503-.3104 = -1.36 less than the ln(odds) for men. This is consistent with the plot in (a). e) In the reduced model without the interactions, -2lnL = 84011.029

Chapter 13 –Page 4

In the full model that includes the interactions, -2lnL = 84063.981.

X 2(3) = (84063.981 − 84011.029) = 52.952 , p value < 0.001

There is significant evidence that at least one of the interactions is non-zero, that is, that the gap in the ln(odds) for men and women is different in at least one age group.

EXERCISE 6 a) It looks like the maximum is about 10.5, so initial

11

estimate of θ1 =10.5. At time 0, it looks like the curve

10

would intercept the Y axis at about 4, so ln(4 / 10.5) =

−0.965 . It also looks like the

curve would achieve 90% of maximum sometime around hour 2, giving an initial estimate of θ3 =-1.1.

Water

θ2 =

9 8 7 6 5 4 0

1

b) fitted value of Water =

2

3

4

5

6

TIME

10.3986*exp(-1.1276*exp(-1.3686STIME)) c) We need 0.75 = exp(-1.1276*exp(-1.3686STIME)) -> ln(.75) = -1.1276*exp(-1.3686STIME) -> ln(0.25513) = -1.3686 STIME -> 0.998 = Time at which peas absorb 75% of their maximum.

This makes sense when we look at the plot. At time one hour, the peas have absorbed about 8 for water, which is about 75% of the approximate maximum.

EXERCISE 7 This uses Poisson regression where the dependent variable is the number of crashes and the independent variable of interest is a dummy variable for design (coded as 0 for design A and 1 for design B). ln(AWTL) is used as an offset variable. (Expected number of crashes = Rate*AWTL.) Fitted value ln(crash rate per unit of AWTL) is -0.8166 + 0.7784DVDesign. This corresponds to an increase in the ln(crash rate) of 0.7784 for Design B versus Design A. This difference is significant, Wald Chi-squared = 22.96 with p < 0.0001.

Chapter 13 –Page 5

EXERCISE 5

The null hypothesis is that the differences are symmetrically distributed about 0.

a) Using Table A.8, Reject Ho if T ≤ 3. T(+) = 2, T(-) = 34, so T = 2. There is significant evidence that the differences are not symmetrically distributed about 0. Since the test is insensitive to lack of symmetry, we interpret this is meaning that the typical value is not 0, that is, that the treatment teeth had significantly different (apparently LOWER) mineral content than the control teeth. b) t = -2.281, p value 0.0565. There is no significant evidence that the treatment differs from the control with respect to mean mineral content. This test is suspect since there are two teeth where the differences are much larger than for the other 6 teeth. This inflates the standard deviation.

EXERCISE 6 T(1) = 84, T(2) = 55.5, T(3) = 31.5. S2 = 28.471. Using SAS, Kruskal-Wallis H = 6.673.

Compared to a Chi-squared table with 2 df, the p value is 0.0356. Reject Ho, there is a significant difference in at least one teaching method. We declare two methods significantly different if the sample means in their ranks differs by more than 2.1314 28.4706 *

method

1

2

3

n

6

7

5

(

18 − 6.673 15

)

1 ni

1 += nj

9.883

1 ni

+

1 nj

Mean rank 14.0 a 7.93 b 6.3 b Teaching method 1 is significantly different from 2 or 3. Apparently, it has higher typical values.

EXERCISE 7

Counties are the treatment, because you are trying to find whether counties differ. Year is

the blocking variable. Sums of ranks for each county: 7, 6.5, 10, 6.5 A=89.5

B=77.83

T*

Chapter 14 – Page 2

CHAPTER 14 EXERCISE 1

Assuming a two-tailed alternative hypothesis with α=5%.

a) T(+)= 9 ; T(-)=66-9, so T=9. From table A.8, Reject Ho if T ≤ 10. (SAS proc univariate gives p value = 0.032.) Reject Ho. There is significant evidence that the location differs from 12.5. Note, however, that the significant value is in the wrong direction – it appears that the average is LESS than what the manufacturer claimed! b) t = -2.178, p value = 0.0544. There is no significant evidence that the mean differs from 12.5. While it is difficult to judge normality in small samples, the boxplot for the data is somewhat right-skewed. A nonparametric test may be safer.

EXERCISE 2 Using the Wilcoxon signed rank test, eliminating subject with change of 0, T(+) = 38.5 and T(-) = 27.5, so T = 27.5. From Table A.8, reject Ho if T ≤ 10 (using α = 5%). Do not reject Ho. There is no significant evidence that the differences are not symmetrically distributed about 0. (No significant evidence of a change in weight.)

Examining a boxplot of the data shows no particular reason a paired t test wouldn’t work. Student’s t = 0.73 with 11df, again, no significant evidence the mean loss differs from 0.

EXERCISE 3

Ho: groups have the same location. Using Table A.9 and α=5%, we reject Ho if T ≤ 78.

Since T= 81, we do not reject Ho. There is no significant evidence that the scores for the two professors differ with respect to location. These observations do not show any serious skewness or outliers, so normality is not seriously in doubt. Exam scores have the potential for a small proportion of very low or very high values, so in general medians might be better than means.

EXERCISE 4 Mann Whitney / Wilcoxon rank sum test. T(1)=15, T(2)=40. Using Table A.9 with n1=4 and n2=6, reject Ho if T ≤ 12 at α = 5%. SAS will provide an exact p value of 0.17. No significant evidence that the distributions differ for the two species. Pooled t test had t = -1.19 with p value 0.27, no significant evidence that the means differ.

Chapter 14 – Page 1