A.9 Answers to Tutorial 9 tutorial

A.9.1 Answers for Sect. 9.6

The difference in odds is statistically significant, so either 2. or 4. is consistent with this conclusion… and we cannot tell which. The size of the OR is not the issue here: if the sample OR is statistically different from 1, it does not tell us much about the population OR except that it is (most likely) not one.

2. The odds are quite different, but not statistically significant, so chance cannot be ruled out as the reason why the OR is different from one. (This likely has a smaller sample size.) 3. The odds are a little different, but not statistically significant, so chance cannot be ruled out as the reason why the OR is different from one.

A.9.2 Answers for Sect. 9.7

1. Incorrect is the one about means. 2. \(113/626 = 18.05\%\). 4. 18.05% of 47 is 8.48. See Table A.2. 5. \(25\div 88 = 0.284\). 6. \(22\div 491 = 0.0448\). 7. The OR is \(0.284\div 0.0448 = 6.3\). This OR means that the odds of having Hep. C is 6.3 times greater for students who have a tattoo, compared to those who do not have a tattoo.
TABLE A.2: Five-year mortality for artifical limb users
Had Hep. C Did not have Hep. C Total
Had tattoo 25 88 113
Did not have tattoo 22 491 513
Total 47 579 626

A.9.3 Answers to Sect. 9.8

Answers implied by H5P.

A.9.4 Answer for Sect. 9.9

No evidence of a difference in the mean number fatalities between female- and male-named hurricanes.

A.9.5 Answers to Sect. 9.10

Answers implied by H5P.

A.9.6 Answers for Sect. 9.11

1. \(H_0\): \(\mu_E = \mu_U\) (that is, the means are the same in the two populations) and \(H_1\): \(\mu_E\ne\mu_U\), which can also be expressed in words. Two-tailed. 2. The precision with which the sample mean battery life estimate the population mean battery life. 3. Use the second row (though it matters little here): \(t=-0.486\); \(\text{df}=13.1\) and \(P=0.635\). 4. The sample presents no evidence (\(t=-0.486\); \(\text{df}=13.0\) and \(P=0.635\)) of a difference in the mean lifetimes (mean difference: \(-0.0544\); s.e.: \(0.112\)) of the batteries in the population. (95% CI for the difference from \(-0.30\) to \(0.19\)mins in favour of Ultracell.) 5. Both population have a normal distribution, and/or \(n>30\) or so. 6. Since \(n<30\) for both samples, we must assume the populations both have a normal distribution. Stem-and-leaf plots aren’t convincing (possible outliers) but the samples are too small to know anything for sure.

A.9.7 Answers to Sect. 9.12

1. The third one. The first is about samples. The second is about means. \(H_0\): \(\text{Pop. odds having byss.}_{\text{Smokers}} = \text{Pop. odds having byss.}_{\text{Non-Smokers}}\) or \(\text{OR}=1\); \(H_1\): \(\text{Pop. odds having byss.}_{\text{Smokers}} > \text{Pop. odds having byss.}_{\text{Non-Smokers}}\) or \(\text{OR}>1\) for the OR suitably defined. (One-tailed, from RQ). 3. Smokers: \(\hat{p} = 125/3,189 = 0.0392\); Non-smokers: \(\hat{p} = 40/2,230 = 0.0179\). 4. For smokers: \(\text{Odds having byssinosis} = 125/3,064 = 0.0408\).
This means: smokers are 0.041 times as likely to have byssinosis than not (or, inverting the ratio, 24.5 times as likely not to have byssinosis than have it). For non-smokers: \(\text{odds} = 40/2,190 = 0.01826\). Non-smokers are 0.018 times as likely to have byssinosis than not (or, inverting the ratio, 54.8 times as likely not to have byssinosis than have it). 5. The OR is \(0.04080/0.01826 = 2.2\), so the odds of a smoker having byssinosis are 2.2 times the odds of a non-smoker having byssinosis. 6. OR not one could be due to chance or because of a real difference in the population, due to smoking and/or other reasons. 7. The sample provides strong evidence (one-tailed \(P<0.001)\); \(\text{df}=1\); \(\text{chi-square}=20.092\); OR: \(2.234\) (\(95\)% CI: \(1.6\) to \(3.2\))) that the population proportion of smokers with byssinosis (\(\hat{p} = 0.0392\)) is greater than the population proportion of non-smokers with byssinosis (\(\hat{p} = 0.0179\)). 8. Valid, since expected counts all exceed five (which they do: no SPSS warnings given for example).

A.9.8 Answers to Sect. 9.13

1. The null hypothesis is that the proportion (or the odds) of successful attempts is the same using EGTA and LM, in the two populations. 2. Using symbols, based on using proportions: \(H_0\): \(p_{\text{EGTA}} = p_{\text{LM}}\); \(H_1\): \(p_{\text{EGTA}} \ne p_{\text{LM}}\) (two-tailed). 3. The sample presents insufficient evidence (\(\text{chi-square}=2.63\); \(\text{df}=1\); \(P=0.104\)) that the success rates between EGTA and LM are different in the population. 4. Yes: All expected counts are greater than 5 (no SPSS warning messages). 5. No evidence of a difference, so perhaps consider other issues such as cost, longevity, pain associated with insertion etc.

A.9.9 Answers to Sect. 9.14

1. Before and after measurement on same runner. 2. \(H_0\): \(\mu_d=0\), differences defined as ‘after’ minus ‘before’. \(H_1\): \(\mu_d>0\) the way I defined the differences.
We are explicitly seeking to test for an increase, so the test is one-tailed. (Diffs in the other direction are also OK; the signs of the differences and some other subsequent things change.) 3. Mean increase: \(\bar{x} = 18.73\) pmol/litre; Standard deviation: \(s=8.3297\) pmol/litre; standard error: \(2.511\) pmol/litre. 4. \(t=7.46\) and \(\text{df}=11-1=10\). Since \(t\) is very large, expect \(P\) to be very small. (The one-tailed \(P\) less than \(0.0005\) from Table A.2.) 5. Very strong evidence exists in the sample (paired \(t=7.46\); \(\text{df}=10\); one-tailed \(P\) less than \(0.0005\)) that the population mean \(\beta\) plasma concentrations are higher after the race compared to before the race (mean difference: \(18.73\) pmol/litre higher after the race; std. dev.: 8.33 pmol/litre; 95% CI from \(13.14\) to \(24.33\) pmol/litre). 6. The population of differences has a normal distribution, and/or \(n>30\) or so. 7. Since \(n<30\), must assume the population of differences has a normal distribution. The histogram suggests this is not unreasonable.