E.9 Answers to Tutorial 9 tutorial

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E.9.1 Answers to Sect. 9.1

  1. The two groups are completely different; cannot determine how to pair. Indeed, the sample sizes are different.

    Each rat, of course, must be in just one group (they measure lifetimes!).

    It is paired: For each unit of analysis, there are two observations.

  2. The null hypothesis is that there is no difference in the mean lifetimes of the two groups of rats. In symbols (where \(\mu\) represents the mean lifetime in the population):

    \(H_0\): \(\mu_R = \mu_{FE}\); and \(H_1\): \(\mu_R > \mu_{FE}\); one-tailed, because of the RQ.

  3. If we took many samples of the same size from the population, the mean differences would vary from sample to sample. The ‘standard error of the difference’ tells us how much the sample mean will vary from sample to sample. It is is the standard deviation of this variation in sample means.

  4. The variation in the sample means is small: The population means are estimated quite precisely. The means look different.

  5. Either sampling variation, or the diets really are different.

  6. We always use the not-equal variance (Welch’s test) row: \(t=9.161\); \(\text{df}=154.94\); one-tailed \(P<0.0005\) (test is one-tailed). The evidence contradicts \(H_0\).

  7. The differences are defined as the Mean restricted-diet lifetime minus the Mean Free-eating diet lifetime; we see this as the (poorly-labelled) “Mean difference” column is positive, and the only way to get this value positive is to subtract them in this order.

  8. Very strong evidence exists in the sample (\(t=9.161\) for two independent samples; \(\text{df}=154.94\); one-tailed \(P<0.0005\)) that the population mean lifetime of rats on a restricted diet (mean lifetime: 968.75 days; std. dev.: 284.6) is greater than rats on a free-eating diet (684.01 days; 134.1) (95% CI for the difference from 223.3 to 346.1 days).

  9. The two samples are independent; the sample means have a normal distribution, so: the population has a normal distribution, and/or \(n>30\) or so.

  10. The sample sizes here are quite large so we should be OK, as the Figure suggests not very severe non-normality.

  11. Rats from the same litter are likely to be similar to each other. The litter would probably be the unit of analysis then, not the individual rat.

E.9.2 Answers to Sect. 9.2

We only need to use the 68–95–99.7 rule. The one-tailed \(P\)-values are half of the two-tailed \(P\)-values.
  1. Less than 0.05.
  2. Greater than 0.05.
  3. Less than 0.003.
  4. Greater than 0.05.
  5. Greater than 0.05.
  6. Very small.

E.9.3 Answers to Sect. 9.3

  1. Before and after measurement on same amputee.

  2. \(H_0\): \(\mu_d=0\), differences defined as ‘with’ minus ‘without’. \(H_1\): \(\mu_d>0\) the way I defined differences.

    We are explicitly seeking to test for an improvement, 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. The standard error of the mean is the standard deviation of the sample mean diff in plasma concentrations, a measurement of how precisely the sample mean difference estimates the population mean diff.

  4. Mean increase: \(\bar{x} = 24.9\)m; Standard deviation: \(s=813.034\)m; standard error: \(4.122\)m.

  5. \(t=6.041\) and since \(t\) is very large, expect \(P\) to be very small. (The one-tailed \(P\) less than \(0.0005\) from Table A.2.)

    The differences have just been defined in the opposite directions. (Notice the \(P\)-values are the same.)

  6. Very strong evidence exists in the sample (paired \(t=6.041\); one-tailed \(P\) less than \(0.0005\)) that the population mean 2MWT are higher after receiving the implant compared to without the implant (mean difference: \(24.9\)m higher after receiving the implant; standard deviation: 13.034m; 95% CI from \(15.58\) to \(34.224\)m).

  7. The population of differences has a normal distribution, and/or \(n>25\) or so.

  8. Since \(n<25\), must assume the population of differences has a normal distribution. The histogram suggests this is not unreasonable.

E.9.4 Answers to Sect. 9.4

Answers implied by H5P.

E.9.5 Answers to Sect. 9.5

Some answers embedded.

  1. See Table E.4.

  2. Using artificial limb: \(49/16 = 3.0625\). Not using artificial limb: \(21/19 = 1.105263\). The OR is \(3.0625/1.105263 = 2.771\); that is, the odds of being alive after five years is almost three times higher for those using an artificial limb compared to those who do not.

    See Table E.5.

  3. The CI for the OR is from \(1.198\) to \(6.411\).

  4. Chi-squared: \(5.836\); like \(z = \sqrt{5.836/1} = 2.42\): large; \(P\) about \(0.016\).

    The sample provides evidence to suggest that the odds of dying within five years is not the same between having a wearing an artificial limb and the five-year mortality rate in the population (\(\text{chi-square}=5.836\); \(\text{df}=1\); \(P=0.016\); OR: \(2.771\) and 95% CI from 1.198 to 6.411).

TABLE E.4: Five-year mortality for artifical limb users
Alive Dead Total
Used art. limb 49 16 65
Did not use art. limb 21 19 40
Total 70 35 105
TABLE E.5: Five-year mortality and use of an artificial limb: Numerical summary
Percentage alive after 5 years Odds alive after 5 years Sample size
Use artificial limb 75.4 3.06 65
Did not use artifical limb 52.5 1.11 40
Odds ratio 2.771 626

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