Solutions to Assignment 3Stat 504, Spring 20031. The row totals are fixed by design. Assuming that the handwriting expert knows that five of thesamples are from men and five are from women, the column totals are also fixed. Therefore, weshould use Fisher’s exact test. Under the null hypothesis that the expert is merely guessing atrandom, the probability of observing the table4 11 4is5451105=5 × 5252= .0992.The probability of observing the more extreme outcome5 00 5is5550105=1 × 1252= .0040.The Fisher exact p-value is 26/252 = .1032. The evidence that the expert has some ability ismoderately strong but not overwhelming.2. The usual test for independence in the table19 29 24497 560 269yields X2=7.88 with 2 degrees of freedom and a p-value of .0194, so there is clearly a relationship.To describe the relationship, we need at least two relative risks or odds ratios. But there are manyways to define these two effects.In this example, we may think of “carrier” as the risk factor and tonsil size as the outcome. First,let’s consider the risk of having enlarged tonsils versus normal tonsils:Enlarged NormalCarrier 53 19Non-carrier 829 497estimate 95% intervalodds ratio 1.67 (0.98, 2.86)relative risk 1.18 (1.02, 1.36)Then we may consider the risk of having very enlarged versus normal or slightly enlarged:Very enlarged Normal/slightlyCarrier 24 48Non-carrier 269 1057estimate 95% intervalodds ratio 1.96 (1.18, 3.27)relative risk 1.64 (1.17, 2.32)Or the risk of slightly enlarged versus normal:Slightly enlarged NormalCarrier 29 19Non-carrier 560 497estimate 95% intervalodds ratio 1.35 (0.75, 2.45)relative risk 1.14 (0.90, 1.44)Or the risk of very enlarged versus normal:Very enlarged NormalCarrier 24 19Non-carrier 269 497estimate 95% intervalodds ratio 2.33 (1.26, 4.34)relative risk 1.59 (1.20, 2.11)Or the risk of very enlarged versus slightly enlarged:Very enlarged SlightlyCarrier 24 29Non-carrier 269 560estimate 95% intervalodds ratio 1.72 (0.98, 3.02)relative risk 1.40 (1.02, 1.91)All of the estimated ratios are greater than 1.00, and many (but not all) are significantly greaterthan one at the .05 level. The overall pattern is clear: being a carrier of Streptococcus pyogenes isassociated with enlargement of the tonsils.3. The overall test of relatedness (i) yields X2=7.87 with 4 degrees of freedom, p-value = .0965;this gives moderate evidence of a relationship. To test (ii), we collapse the second and third rows,012+None 15 14 10Low/high 88 41 21which gives X2=5.66, 2 degrees of freedom, p-value = .0591. To test (iii), we eliminate the firstrow,012+Low 46 26 14High 42 15 7which gives X2=2.29 2 degrees of freedom, p-value = .3184. There is moderately strong evidencethat training does have an effect on the number of errors, but a high level of training does notseem to be any more effective than a low level of training.4. (a) McNemar’s test gives Z2=(67− 33)2/(67 + 33) = 11.56, p-value = .0007, so there isstrong evidence for a rate change. The estimated rate of Catholic affiliation at age 16 is.2743, and the estimated current rate is .2520, so the estimated difference in proportions is−.0223 = 2.23%. The standard error for this estimate is 115246715241 −671524+3315241 −331524+2(67)(33)15242= .0065,and a 95% confidence interval is (.0095,.0351) or from 0.95% to 3.51%.(b) An equivalent result from two independent, cross-sectional samples would have looked likethis:Catholic Non-CatholicAge 16 418 1106Current 384 1140This gives the same estimated difference in proportions, but the standard error is now.2743(1 − .2743)1524+.2520(1 − .2520)1524= .0159which is nearly 2.5 times as large as the actual standard error found in part (a). Using alongitudinal sample apparently produced a much more precise estimate of the rate changethan cross-sectional samples would have