A.9 Answer: TW 9 tutorial

Answers for Sect. 9.2

Answers implied by H5P.

Null hypotheses:

Is the mean length of a \(12\)-inch sub really \(12\) inches? F: \(\mu = 12\).
Is the mean length of a \(12\)-inch sub different for white and wholemeal subs? B: \(\mu_{\text{white}} - \mu_{\text{wholemeal}} = 0\).
Is the proportion of \(12\)-inch subs shorter than \(12\) inches different for white, wholemeal subs? A: \(p_{\text{white}} - p_{\text{wholemeal}} = 0\).
Is the mean length of a \(12\)-inch sub longer for white (compared to wholemeal) subs? B: \(\mu_{\text{white}} - \mu_{\text{wholemeal}} = 0\).

Alternative hypotheses:

Is the mean length of a \(12\)-inch sub really \(12\) inches? I: \(\mu \ne 12\).
Is the mean length of a \(12\)-inch sub different for white and wholemeal subs? G: \(\mu_{\text{white}} - \mu_{\text{wholemeal}} \ne 0\).
Is the proportion of \(12\)-inch subs shorter than \(12\) inches different for white, wholemeal subs? K: \(p_{\text{white}} - p_{\text{wholemeal}}\ne 0\).
Is the mean length of a \(12\)-inch sub longer for white (compared to wholemeal) subs? L: \(\mu_{\text{white}} - \mu_{\text{wholemeal}} > 0\).

Answers for Sect. 9.3

How much further patients walk with the implant.
Repeated-measures: Every subject has two TWMT recorded.
\(\mu_d\) is the mean difference in the target population; \(\bar{d}\) is the mean difference in this sample.
Each measurement is measured after and before on the same subject.
\(32\); \(12\); \(24\); \(30\); \(8\); \(14\); \(14\); \(28\); \(38\); \(49\). (Differences in the other direction are also acceptable; it just changes the signs of these differences and so on. Importantly, the direction should be stated somewhere.)
It makes more sense to define directions this way, so that the difference is the increase in 2MWT.
\(\bar{d} = 24.9\); \(s_d=13.03372\,\text{m}\).
\(\text{s.e.}(\bar{d}) = s_d/\sqrt{n} = 13.03372/\sqrt{10} = 4.121623\,\text{m}\).
This is the standard deviation of the sample mean difference, a measurement of how precisely the sample mean difference measures the population mean difference.
Almost impossible. Sample means vary every time we take a sample around the true mean difference, with a normal distribution with standard error \(4.12\). Since we don't know \(\mu\), the best we can say is that the sample mean will vary about our best guess of the population mean; in other words, the sample means vary around \(24.9\) with a standard deviation of about \(4.12\).
Normal; mean \(\mu_d\), std deviation is the standard error of \(4.121\).
\(24.9\pm (2\times 4.121623)\), or \(24.9\pm 8.243246\), or from \(16.65675\) to \(33.143245\,\text{m}\).
\(H_0\): \(\mu_d = 0\), differences defined as 'with' minus 'without'. \(H_1\): \(\mu_d > 0\) the way I defined differences.
\(t = 6.041\) and since \(t\) is very large, expect \(P\) to be very small.
The differences have just been defined in the opposite directions. (Notice the \(P\)-values are the same.)
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: \(24.9\,\text{m}\) higher after receiving the implant; standard deviation: \(13.034\,\text{m}\); \(95\)% CI from \(16.66\) to \(33.14\,\text{m}\)).
The population of differences has a normal distribution, and/or \(n > 25\) or so.
Since \(n < 25\), must assume the population of differences has a normal distribution. The histogram suggests this is not unreasonable.

Answers for Sect. 9.5

The two groups are different rats.
The parameter of interest is the difference between the population mean lifetimes, say \(\mu_R - \mu_F\). Measure how much longer the lifetimes are for rats on a restricted diet.
Table not shown.
The \(95\)% CI is the bottom one: from \(223.34\) to \(346.13\) days.
The best of these is Option (e)... but in practice, we usually think about CIs in terms of Option (d) so Option (d) is fine.
The CI explanation can be improved by (i) indicating which diet leads to larger average lifetimes; and (ii) providing sample summary info. Here is a better answer:

"The \(95\)% confidence interval for the difference between the populations mean lifetimes of rats on the restricted diet (sample mean: \(968.8\) days; std dev: \(284.6\) days) and on the free-eating diet (\(684.0\) days; std dev: \(134.1\) days) is that rats on a restricted diet live between \(223.34\) and \(346.13\) days longer."
Boxplots: shows the variation in the lifetimes of individual rats. Error bar chart: displays the variation that the sample means would be expected to show from sample to sample.
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.
Either sampling variation explain the difference, or the diets really are different.
Use the not-equal variance (Welch's test) row: \(t = 9.161\); one-tailed \(P < 0.0005\) (test is one-tailed).
\(t = (284.73 - 0)/31.08 = 9.16\), as per output.
The evidence contradicts \(H_0\). 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.: \(284.6\) days) 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).
The the sample means have a normal distribution. Since both sample sizes are greater than \(25\), this is true. The figure suggests not very severe non-normality.
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.