## 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.

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.