Answers to Selected Computational Exercises

Chapter 1: Drawing Statistical Conclusions

1.17 The two-sided p-value is 3/35 = 0.0857.

1.19 Coin .ips will not divide the subjects in such a way that there is an exact age balance.

However, it is impossible to tell prior to the .ips which group will have a higher average age.

1.21 There is no computation involved. This is, however, a sobering exercise.

1.25 (a) No radiation median is 0; radiation median is 1.

(b) Both distributions are positively skewed. The radiation group has a larger spread.

(d) It is observational data, so a strict interpretation would say that causation cannot be inferred. But what else could it be?

Chapter 2: Inference Using t -Distributions

2.13 (a) (Fish, Regular): Averages are (6.571, .1.143); SDs are (5.855, 3.185)

(b) Pooled SD = 4.713

(d) d.f. = 12; t12(.975) = 2.179

(e) 95% CI from 2.225 to 13.203mm

(f) t-stat = 3.062

(g) one-sided p-value = .005.

2.15 t-statistic = 9.32, with 174 d.f. Very convincing, indeed.

2.19 (a) Average= .1.14; SD = 3.18; d.f. = 6.

(b) SE = 1.20

(d) t-statistic = .0.95; two-sided p-value = .38.

Chapter 3: A Closer Look at Assumptions

3.21 (a) One sample t-test on differences (observed-expected) for the subset of umpires whose lifetimes were not censored (Censored = 0): t-stat = .0.987, d.f. = 194,

1 2 Answer Section

p-value = 0.32 (one-sided p-value = .16). A 95 percent con.dence interval for mean life length minus expected life length: 1.6 years to 0.54 years.

3.23 (a) Yes. One should expect the rates to follow a time series where serial correlation is present.

(b) There is a problem: there is a steady increase, or “trend,” in the series. There is also a somewhat cyclic behavior.

3.25 Use the computer. Refer to Display 3.7.

3.27 (a) (i) Oil exporters are positively skewed. Industrialized are reasonably symmetric.

(ii) The log makes the oil exporters look OK, but the industrialized group gets squashed and has its outlier made more prominent. (iii) estimate = 1.703, SE = 0.280. (iv) The median per-capita income in the industrialized countries is 5.5 times that in the oil exporting companies. (v) From 1.128 to 2.278. (vi) The industrialized-to-oil exporting ratio of median per-capita incomes is estimated to be between 3.1 and 9.75 (95% con.dence interval).

(b) (i) Oil exporters group is a bit long-tailed, and it has greater spread than industrialized group. (ii) Uniform health standards; technology that insulates health conditions from environmental factors. (iii) The log, square root, and reciprocal will fail because the group with the higher average has the smaller spread. The square mightwork. (iv) These are the conditions where the t-tools fail—the group with the higher spread has the smaller sample size.

3.29 With all the data, the one-sided p-value is .0405; without the 0.659 value, the one-sided

p-value is .0900. This is a fair swing; the evidence goes from suggestive to faint.

Chapter 4: Alternatives to the t -Tools

4.15 One-sided p-value = 2/10 = .20.

4.17 .0396.

4.19 (a) 0.1718

(b) Normal approximation

(d) t-test gives p = .081; t-test with removal gives p = .180; rank sum gives p = .1718.

(e) The rank sum test is valid AND it uses all the data.

4.21 Two-sided p-value = .00314

4.23 Two-sided p-value = .00643, compared to .00537.

4.25 CI: (39.59, 165.81), based onWelch’s t with 97 d.f. two-sided p-value = .0016. No. It looks like something else is involved.

4.27 Two-sided p-value = .00452, from signed rank test on the log(ratio) values. On the straight difference scale, the signed rank gives .00208 . . . close. It is not particularly apparent.

Chapter 5: Comparisons Among Several Samples

5.14 (a) 6.914, with 39 d.f.

(b) t-statistic = 5.056, with 39 d.f., giving one-sided p-value .0001.

Answer Section 3

5.18 (a) CPFA50 CPFA150 CPFA300 CPFA450 CPFA600 Control 168.3 171.7 146.7 151.0 152.3 185.6 There is no suggestion that the size of residuals depends on the average protein level.

There is a suggestion that the mean level may change from one day to another.

(b) There is convincing evidence that the means are different (p-value = .0001). There is ample evidence to suggest that the means under the control treatment are different on different days (p-value = .0021).

5.21 (d) .0777.

Chapter 6: Linear Combinations and Multiple Comparisons of Means

6.13 The 95% con.dence intervals are: (Amputee . Crutches) between .3.009 and 0.025; (Amputee . Wheelchair) between .2.431 and 0.603; (Crutches . Wheelchair) between

.0.939 and 2.095.

6.15 (a) 4.484

(b) 1/3, .1/2, .1/2, 1/3, 1/3

6.18 g = 0.803

6.20 (a) Two-sided p-value = .005.

Chapter 7: Simple Linear Regression: A Model for the Mean

7.12 (a) 15.53

(b) 3.94

7.14 (a) Intercept estimate = 0.7648; slope estimate = 0.5369.

(d) 0.1357, with 8 d.f.

7.16 (b) Estimate of µ{life | income} = 68.87+0.00077 income. one-sided p-value = 0.0053

7.18 The left-hand side expressions require a second pass through the data, because one full pass must be made to get the averages of X and Y. Therefore, either the data must be entered twice or it must be stored the .rst time through.

7.20 (a) SE{pred} = 0.0875

(b) 5.6139 mean pH 6.0175.

7.22 About 109.

7.24 (a) H. nudus L. bellus C. productus

slope 0.4083 2.9737 2.0685 SE(slope) 0.5426 0.6125 0.4275

(b) C. productus vs. L. bellus: t-stat = 1.212. Two-sided p-value = 0.24

C. productus vs. H. nudus: t-stat = 2.403. Two-sided p-value = 0.025.

7.26 (b) t-statistics: .2.07, .5.71, .4.02, .5.78.

4 Answer Section

Chapter 8: A Closer Look at Assumptions for Simple Linear Regression

8.15 (b) Estimated mean number of species = 24.04928 + 0.00211 ¡¿ Area.

(c) The regression line does not come near hitting the center of the distribution of species numbers from islands with similar area. There is a pronounced curvature in the residual plot.

8.17 (b) Estimated mean log(Mass) = 3.797 . 0.262 ¡¿ sqrt(Load)

8.19 (a) A residual plot from a .t of all data shows possible curvature and possible outliers.

(b) Transformation of the scales does not accomplish much in clarifying the situation.

(c) Fits with and without the bees with duration over 30 seconds give quite different results for the regression. Examination of a residual plot from the .t without durations over 30 seconds shows no further problems.

(d) Conclusion: For visits under 30 seconds, a straight-line regression appears to give a reasonable summary. That description does not extend to visits over 30 seconds.

8.21 No transformation appears necessary. Case 33 and, possible, case 35 are potential outliers.

With 33 deleted: g = .21.81; se(g) = 50.62; t-statistic = .0.43. ANOVA p-value for group differences = .3744. The conclusions remain the same with and without case 33.

Chapter 9: Multiple Regression

9.13 (a) The two-sided p-value is .0132.

(b) The two-sided p-value is .50.

9.15 (b) Estimate of µ{yield | rain} = .5.0 + 6.0rain . .229rain2

SEs: (11.4) (2.0) (.089)

(d) Estimate of µ{yield | rain, year} = .263 + 5.7rain . .216rain2 + .136year

SEs: (98) (1.900) (0.082) (0.052)

(e) Estimate of µ{yield | rain, year} = .1909 + 159rain . 0.186rain2 + 1.00year . 0.081rain ¡¿year

SEs: (486) (45) (.072) .26 .023 The two-sided p-value for signi.cance of the interaction term is .0016. This indicates that the effect of rainfall on yield is smaller for years closer to 1927 than for years closer to 1890. One possible explanation is that in later years the yield became less dependent on rainfall.

9.17 (b) Let datej represent the indicator variable for date j, for j = 2, . . . , 8).

Estimate of µ{interval | dur, DATE} = 32.9 + 10.9dur + 1.3date2 + 0.8date3

(3.1) (0.66) (2.7) (2.7)

+ 0.2date4 + 0.2date5 + 2.0date6 . 0.2date7 . 0.7date8

(2.6) (2.6) (2.6) (2.7) (2.7)

Answer Section 5

9.19 (a) â0 + â1hiplus + â2hionly + â3age + â4hiplus.age + â5hionly.age; â5 measures divergence.

(b) â0 + â1hiplus + â2hionly + â3age + â4hionly.age; â4 is the parameter of interest.

â7hionly.age + â8hionly.age; no single parameter describes the divergence group.

Chapter 10: Inferential Tools for Multiple Regression

10.1 (a) 32.

(b) .0014

10.11 (a) The two-sided p-value = .2443; the one-sided p-value is .1222; yes. Failure to account for sampling effort could lead to concluding there was a relationship between species numbers and reserve size, when the full data would suggest that different effort may be a better explanation.

(b) The p-value is .0001.

(d) 88.59%

10.13 (b) The slope estimate is 0.8150 for all three groups. The intercept estimates are: (i) .1.5764 for non-echolocating bats; (ii) .1.4741 for birds; and (iii) .1.4977 for echolocating bats.

(d) Same as b.

(e) Two-sided p-value = .8828

10.15 The F-statistic for inclusion of the day indicators is F = 0.209. The p-value is .98.

10.17 (a) R2 = 92.64%; adjR2 = 91.16%

(b) R2 = 99.03%; adjR2 = 98.55%

(d) R2 = 99.98%; adjR2 = 99.95%

(e) R2 = 99.996%; adjR2 = 99.977%

(f) R2 = 100%; adjR2 = unde.ned

10.19 (c) The p-value = .89.

10.23 (a) F-statistic = 8.166 on 2 and 17 d.f.; 0.01 > p-value 0.001.

(b) (i) Yes: two-sided p-value = 0.0032.

(ii) No: two-sided p-value = 0.2914.

(iii) No: two-sided p-value = 0.7302.

(iv) No: two-sided p-value = 0.2618.

10.25 (a) De.ne site indicators: s2, s3 and s4, an irrigation indicator, irr and set N = nitrogen level.

(b) The full model is:

â0 + â1N + â2N2 + â3irr + â4irr.N + â5irr.N2 + â6s2 + â7s2.N + â8s2.N2 +

â9s2.irr +â10s2.irr.N +â11s2.irr.N2 +â12s3+â13s3.N +â14s3.N2 +â15s3.irr +

â16s3.irr.N+â17s3.irr.N2+â18s4+â19s4.N+â20s4.N2+â21s4.irr+â22s4.irr.N+

â23s4.irr.N2

6 Answer Section

Chapter 11: Model Checking and Re.nement

11.11 (a) Cook’s distance identi.es sample #17 as being highly in.uential. This is the result of a high leverage (0.755) in combination with a large Studentized residual (3.468).

(b) Sample #1 now becomes a slight problem.

11.13 We do, naturally.

11.15 (c) The externally Studentized residual for case 22 is 3.327.

(d) They differ because including case 22 increases the estimate of the residual standard deviation, which is a divisor of the raw residual.

(e) The long-tailed aspect of these data is apparent in a normal plot of the residuals.

(f) The externally Studentized residual for case 22 = 3.327. This is very large for a normal deviate. However, recall that predictions are reliant on the normal distribution assumption, which is in question from other cases besides 22 in the normal plot. Because the normal plot shows that case 22 belongs to the same long-tailed pattern exhibited by the remainder of the data, there is not convincing evidence that this election was fraudulent.

11.17 (d) The strongest relationship is between the response and Sacri.ce Time.

11.19 (a) Estimates for means of log tumor-to-liver ratio are (SEs in parentheses):

Sacri.ce time (hours)

0.5 3.0 24 72 BD .3.505 (0.195) .2.371 (0.205) +0.752 (0.209) +1.649 (0.209) NS .4.302 (0.205) .3.168 (0.195) .0.044 (0.209) +0.852 (0.209)

(b) Same results!

Chapter 12: Model Selection with Large Numbers of Explanatory Variables

12.11 Forward selection settles on the model B.

12.13 Here are the results of one simulation. Each solution will differ, however, in detail.

(a) 11.9%.

(b) One variable entered, using a 4.0 cutoff for the F-statistic.

(d) The BIC (correctly) chose a model with the constant term only.

12.15 (a) The model with all four design variables has the smallest Cp statistic, at 3.50.

(b) Forward selection proceeds directly to the model with the four design variables, adding Sac Time 72 (F = 24.83), then Sac Time 24 (F = 126.28), then Sac Time 3 (F = 10.86), and then BD (F = 18.89).

(d) Stepwise does not alter the selection outcome.

(e) The smallest BIC (= .25.18) is for the model with only the four design variables.

12.17 (a) The p-value = .0083.

Answer Section 7

(b) The p-value = .0308. The conclusion is about the same, although there is some disagreement about the strength of the evidence.

Chapter 13: The Analysis of Variance for Two-Way Classi.cation

13.11 (a) p-value for interaction = .72.

(b) p-value for treatment effect in the additive model, = .012.

13.15 (b) p-value = .92.

Chapter 14: Multifactor Studies

14.9 (b) The difference between slopes at H2O = ..05 and ..40 is .1.44713. A 95% con.- dence interval for difference between slopes is (.4.24109, +1.34683). Exponentiate to get an estimate of 0.235, with a 95% CI of (0.014, 3.845).

14.11 (c) There is a much more obvious treatment effect when the sex is accounted for. Although the estimate of the effect is the same in both cases, the standard error is much smaller when sex is accounted for (in the randomized block version of the experiment in part (b). It is smaller because SD{score | treat, sex} < SD{score | treat}.

14.13 (a) Two-sided p-value = .50.

Chapter 15: Adjustment for Serial Correlation

15.7 (b) Two-sided p-value, .0201.

(d) Two-sided p-value = .9856

15.9 (c) For this segment of the sunspot series, an AR(2) model for the square root transformed numbers appears satisfactory.

Chapter 16: Repeated Measures

16.5 (a) r = 0.1310

(b) (i) R-squared = 11.08% (ii) p-value = .7379 (iii) R-square = 0.49% (iv) p-value = .0153

16.7 (a) Neither 95% con.dence interval includes zero.

(b) p-value = .0643.

8 Answer Section

(d) This may also seem to be contradictory, but only to those who cling doggedly to the .05 level cutoff. The evidence registered against zero values—.0409, .0366, .0643—is relatively consistent.

16.9 Here are four examples representing different regions around the con.dence ellipse.

Short term Long term T-square F-statistic p-value 8 8 13.76 6.45 .0095 11 .7 7.80 3.65 .0509 25 0 5.61 2.63 .1049 0 0 26.30 12.53 .0007

16.11 (d) (i) A multivariate regression is the best way to view the situation, with the P2P ratio

and the %Indigenous as responses to %Catholic as the explanatory variable. (ii) The inference from multivariate regression can be approximated by separate inferences from univariate regression. (iii) Countries with high population percentages being Catholic generally have low numbers of priests per parishioner; these countries with lownumbers of priests per parishioners generally have higher percentages of the priests being nonindigenous.

Chapter 17: Exploratory Tools for Summarizing Multivariate Responses

17.9 The .rst three principal components account for about 95% of the total variation. (1) (M2+ M5 + M6 + M7 + M11)/5; M2; and (M5 + M11)/2 . (M6 + M7)/2; or (2) (M2 + M5 + M7)/3; M2; and M5 . M7.

17.11 (a) The largest canonical correlation between passionate responses and compassionate responses is 0.5433. The test that all four canonical correlations are zero has a p-value of .4375, indicating no evidence of any correlation between the two sets of responses.

(b) The largest canonical correlation between husbands’ responses and wives’ responses is .5717, with associated p-value = .4572. Again, there is no evidence that the two sets of responses are correlated in any way.

Chapter 18: Comparisons of Proportions or Odds

18.9 (a) (i) 0.01832 in obese group; 0.01537 in not obese group. (ii) 0.00505. (iii) .0.00696 to

+0.01285.

(b) z-statistic = 0.5825; one-sided p-value = .2801.

(d) The odds of CVD death in the obese group are estimated to be between 0.66 and 2.17 times the odds of CVD death in the not obese group (95% con.dence interval).

18.11 (a) (i) 0.00050 (ii) 0.000250 (iii) 0.6429

(b) (i) 0.00025 (ii) 0.00025 (iii) 0.4737

(c) (i) 0.00025 (ii) 0.00025 (iii) 0.1692 Retrospective samples (of equal size) do not estimate the population proportions.

Answer Section 9

18.13 (a) If, for example, ðu = .0010 and ðv = .0002, the relative risk is ñ = 5.0, while the odds ratio is ù = 5.004.

(b) 3.69

Chapter 19: More Tools for Tables of Counts

19.11 C503,0 ¡¿C317,5/C820,5 = (317¡¿316¡¿315¡¿314¡¿313)/(820¡¿819¡¿818¡¿817¡¿816) = .0085.

19.13 (a) Excess = 5.5; SE(Excess) = 1.963; z-statistic = 2.80; one-sided p-value = .0026.

(b) Fisher’s Exact Test: one-sided p-value = .0044.

19.15 (a) No. The expected number of used and not used houses of old wood are both smaller than 5.

(b) One-sided p-value = .0057.

Chapter 20: Logistic Regression for Binary Response Variables

20.9 (a) Estimated survival probabilities for males are .413 at age 25 and .091 at age 50. For females, the estimates are .777 at age 25 and .332 at age 50.

(b) For females, the age of 50% survival is 41.0 years; for males it is 20.5 years.

20.11 (a) logit( . ð) = 10.8753 . 0.1713 temperature (5.7031) (0.0834)

(b) One-sided p-value = .0200, from z-statistic = .2.054.

p-value for the coef.cient, so the approximate one-sided p-value based on the deviance would be .0074.

(d) The 95% con.dence interval extends from .0.3348 to .0.0078.

(e) logit = 5.5650; estimated probability of failure = .9962.

(f) It represents an extrapolation beyond the range of the available data.

20.13 p-value = .3008.

Chapter 21: Logistic Regression for Binomial Counts

21.9 (b) The estimate of the intercept is 0.2805, with standard error = 0.2309. The slope estimate is .0.0187, with standard error 0.0068.

21.11 The drop in deviance for interaction terms is 2.2391, with 5 d.f. (p-value = .8152). There is no evidence to suggest that the equal odds ratio assumption is inadequate.

21.13 (b) The .t: logit( . ð)= .1.2003 . 0.0346 dose (0.0617) (0.0711)

10 Answer Section

The goodness-of-.t p-value = .7674, from the deviance = 0.5296 with 2 d.f. Wald’s test for the coef.cient of dose is z = .0.4866, giving a two-sided p-value = .6265.

The drop in deviance test has a p-value = .6253, based on the drop = 0.2385 with 1 d.f.

(c) There is not evidence that the model is inadequate. Nor is there much evidence that the odds of a cold are associated with the daily dose of vitamin C.

Chapter 22: Log-Linear Regression for Poisson Counts

22.15 (a) Estimate of µ{matings1/2 | age} = .0.8122 + 0.0632age

(b) Estimate of µ{log (matings + 1) | age} = .0.6989 + 0.0509age

22.17 The two-sided p-value from the deviance test is .73.

22.19 (a) The deviance goodness-of-.t p-value is less than .0001, providing overwhelming evidence of lack of .t of the Poisson model.

(b) Using quasi-likelihood analysis and backward elimination—discarding a variable at each step if its t-statistic is smaller in magnitude than 2—leads to the model with log area and log of area of the nearest neighbor for describing the mean number of nonnative species.

(c) For each doubling of island area the mean number of nonnative species increases by 35%. For a given island area, the mean number of nonnative species decreases by 9% for each doubling of area of the nearest island.

Chapter 23: Elements of Research Design

23.13 Here the response is binary: Y = 1 for responding to the drug and = 0 for nonresponse. The control proportion is 0.60 for the standard treatment, and the desired alternative is 0.75 for taxol. The desired odds ratio is 2.0. The sample size in each group should be 304.

23.15 The nonvictim proportion is 0.20, corresponding to odds of 0.25. An odds ratio of 3.0 translates to odds of 0.75, or a victim proportion of 0.4286. Distinguishing the two can be done with a sample of 132 from each group.

Chapter 24: Factorial Treatment Arrangements and Blocking Designs

24.11 (b) There are indicator variables for A, for B, for the interaction of A and B, and for 11 of the 12 leaves (blocks). The resulting regression .t is: 0.715 + 0.616A . 0.272B + 0.119AB.0.433L2.0.841L3+0.009L4.1.039L5.1.200L6.2.013L7.1.144L8. 1.803L9 . 1.294L10 + 0.013L11 . 1.211L12. Adding 6 to both responses in leaf 8, 2 to both responses in leaf 11, and 15 to both responses in leaf 3 does not change coef.cients of A, B, and AB. The coef.cients of L3, L8, and L11 increase by 15, 6, and 2.