E.8 Answers to Lecture 8 tutorial

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E.8.1 Answers to Sect. 8.1

  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.

  1. The normal distribution has a mean of 12, and a standard deviation of \(\text{s.e.}(\bar{x}) = 0.02205\).
  2. \(t=(11.486-12)/0.02205 = -23.3\).
  3. 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.)
  4. 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 \(t=1.98\).)
  5. 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.
  6. Very unlikely!

E.8.2 Answers to Sect. 8.2

  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.

E.8.3 Answers to Sect. 8.3

Some answers embedded.

  1. See Table E.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 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 E.3.
TABLE E.2: Maturation and gender
Matured late Did not mature late Total
Males 352 2512 2864
Females 336 2328 2664
Total 688 4840 5528
TABLE E.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

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