A.9 Answers to Teaching Week 9 tutorial

Answers to Sect. 9.2

  1. The two groups are completely different.

  2. The parameter of interest is the difference between the population mean lifetimes, say \(\mu_R - \mu_F\).

  3. The 95% CI is the bottom one: from \(223.34\) to \(346.13\) days.

  4. The best of these is Option (e)... but in practice, we usually think about CIs in terms of Option (d).

  5. The CI explanation can be improved by:

    1. indicating which diet leads to larger average lifetimes; and
    2. 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."

  6. Since the sample is large, we must have that the two samples are independent (which is reasonable). (The figure is not needed.)

  7. The boxplots show the variation in the lifetimes of individual rats. The error bar chart displays the variation that the sample means would be expected to show from sample to sample.

Answers to Sect. 9.3

Some answers embedded.

  1. See Table A.2.
  2. Use a side-by-side barchart, for example, if necessary.
  3. Odds of boys maturing late: \(352\div(2\,864-352) = 0.1401\). Thus boys are 0.1401 times more likely to mature late than not.
  4. Odds of girls maturing late: \(336\div(2\,328) = 0.1443\). Thus girls are 0.1443 times more likely to mature late than not.
  5. Hence, to compare boys to girls: \(0.1401\div 0.1443 = 0.971\).
  6. The parameter of interest is the population odds ratio of late maturing, comparing boys to girls.
  7. From software: OR is 0.971, and 95% CI is from 0.828 to 1.139.
  8. See Table A.3. 1, The difference could be explained by sampling variation, or because there is a real difference...
TABLE A.2: Maturation and gender
Matured late Did not mature late Total
Males 352 2512 2864
Females 336 2328 2664
Total 688 4840 5528
TABLE A.3: Maturation and gender: Numerical summary (Enter percentages to one decimal place. Enter odds and odds ratio to three decimal places)
Percentage maturing late Odds maturing late Sample size
Males 12.3 0.1401 2864
Females 12.6 0.1443 2664
Odds ratio 0.97088

Answers to Sect. 9.4

  1. The parameter of interest is the population mean pizza diameter, say \(\mu\).

  2. From the output: \(\bar{x} = 11.486\) and \(s=0.247\).

  3. Use \(\text{s.e.}(\bar{x}) = s/\sqrt{n} = 0.247/\sqrt{125} = 0.02205479\).

  4. \(s\) tells us the variation in diameter from pizza to pizza. \(\text{s.e.}(\bar{x})\) tells us how much the sample mean is likely to vary from sample to sample, in sample of size 125.

  5. The hypotheses are:

    \(H_0\): The mean diameter is 12 inches; or \(\mu=12\).
    \(H_1\): The mean diameter is not 12 inches; or \(\mu\ne 12\)

  6. Two-tailed, because the RQ asks if the diameter is 12 inches, or not.

  7. The normal distribution has a mean of 12, and a standard deviation of \(\text{s.e.}(\bar{x}) = 0.02205\).

  8. \(t = -23.3\).

  9. \(t=(11.486-12)/0.02205 = -23.3\).

  10. The \(t\)-value is huge (and negative), so the \(P\)-value is very small. (Two-tailed \(P<0.001\) from the table or from software.)

  11. Very strong evidence exists in the sample (\(t=-23.3\); \(\text{df}=124\); two-tailed \(P\) less than 0.001) that the population mean pizza diameter of pizzas from Eagle Boys' is less than 12 inches (sample mean diameter: \(11.49\) inches; std. dev.: \(0.246\); 95% CI from \(11.44\) to \(11.53\) inches; approximate 95% CI is \(11.486\pm 0.0441\)) (Note: The CI was computed in the Week~8 tutorial; or use \(t=1.98\).)

  12. Since \(n\) is much larger than \(30\), we do not require that the population has a normal distribution, just that the population is not grossly non-normal; the sample means will still have an approximate normal distribution.

  13. Very unlikely!

A.9.1 Answers to Sect. 9.5

1. \(s=7.145\)m; \(\text{s.e.}(\bar{x}) = s/\sqrt{n} = 7.145/\sqrt{44} = 1.077\)m. The first measures variation in the original data; the second measures precision of the sample mean when estimating the population mean. 2. The CI is from 13.85 to 18.19m. 3. Strong to moderate evidence exists in the sample (one sample \(t=2.714\); \(\text{df}=43\); two-tailed \(P=0.010\)) that the population mean guess (mean: 16.02m; standard deviation: 7.145m) of the students is not 13.1m (95% CI for population mean guess: 13.85 to 18.19m). 4. The population of differences has a normal distribution, and/or \(n>30\) or so. 5. Since \(n>30\), all OK if the histogram isn't severely skewed. Probably OK. 6. Not really! But was this just due to using metric units? Perhaps students are just very poor at estimating widths... In fact, the Professor also had the students estimate the width of the hall in imperial units also, as a comparison.

A.9.2 Answers to Sect. 9.6.

1. Observational. 2. Relational. 3. Two completely separate samples are compared. 4. \(-0.774\) to \(-0.560\) inches (differences are Dominos less Eagle Boys). 5. The 95% CI for the difference in population means pizza diameters between EB and DOM pizzas from \(0.774\) to \(0.560\) inches, larger for EB. 6. Since the sample sizes are large (both 125), we do not require that the populations have normal distributions. 7. The sample sizes are large (\(n=125\) in each), so we don't need the populations to be normally distributed; we don't need the histogram. 8. Probably yes. Amount of topping on the pizza? Which tastes better? Whether the samples were randonly selected or not?

A.9.3 Answers to Sect. 9.7

A.9.4 Answers to Sect. 9.8

1. \(H_0\): No association between income level and opinion of GM foods in the population; \(H_1\): An association between income level and opinion of GM foods in the population. 2. Odds HIE: \(263/151 = 1.742\). HIE is 1.74 times more likely to be for GM foods than against. 3. Odds LIE: \(258\div222 = 1.162\). LIE is 1.16 times more likely to be for GM foods than against. 4. \(\text{OR}(\text{HIE in favour})\div\text{OR}(\text{LIE in favour}) = 1.742/ 1.162 = 1.5\) (\(1.499\) in table). The odds of HIE being for GM food 1.5 times the odds that a LIE for GM foods. 5. From the sample, we estimate the OR in the population to be between \(1.145\) to \(1.961\). (Loosely, though technically incorrect: the true OR is likely to be between 1.145 and 1.961.) Importantly, this interval does not include 1.