A.7 Answers to Lecture 7 tutorial

A.7.1 Answers to Sect. 7.4

1. \(\bar{x}=16.02\)m. 2. \(s=7.145\)m; \(\text{s.e.}(\bar{x}) = s/\sqrt{n} = 7.145/\sqrt{44} = 1.077\)m. The first is a measure of the variation in the original data; the second is a measure of the precision of the sample mean when estimating the population mean. 3. The CI is from 13.85 to 18.19m. 4. 95% CI for population mean guess: 13.85 to 18.19m. 5. The population of differences has a normal distribution, and/or \(n>30\) or so. 6. Since \(n>30\), all OK if the histogram isn’t severely skewed. Probably OK. 7. Not really; the CI doesn’t contain the true width. But was this just due to the metric units… or perhaps students are just very poor at estimating widths in general! In fact, the Professor also had the students estimate the width of the hall in imperial units also, as a comparison.

A.7.2 Answers to Sect. 7.5

\(n=(2\times 7.145\div 0.5)^2 = 816.8\), so use guesses from 817 students.

A.7.3 Answers to Sect. 7.6

Answers implies by H5P.

A.7.4 Answers to Sect. 7.7

1. Because each method is used in each sea state. 3. \(\bar{d}=0.06167\); \(s_d = 0.2901\). The mean difference is positive: Method 1 measurements slightly higher (on average) than Method 2. 4. \(\text{s.e.}({\bar{d}})= s_d/\sqrt{n} = 0.2901/\sqrt{18} = 0.0684\). \(\text{s.e.}({\bar{d}})\) measures the precision with which the sample mean difference estimates the population mean difference. 5. \(0.06167\pm(2\times 0.0684)\), which is \(0.06167 \pm 0.137\), or from \(-0.075\) to \(0.199\) Newton–metres. 6. Since \(n<30\), we require that the differences in the population have a normal distribution. 7. The stem-and-leaf plot of the sample doesn’t suggest the population is non-normal. 8. Since the CI includes zero, possibly the population mean difference could be zero.

A.7.5 Answers to Sect. 7.8

1. \(\mu_d\) is the mean difference in the target population; \(\bar{d}\) is the mean difference in this sample. 2. Each plasma \(\beta\) measurement is measured after and before on the same runner. 3. The direction of the difference should be clearly stated. 4. \(\bar{d} = 18.736\); \(s_d=8.3297\) 5. \(\text{s.e.}(\bar{d}) = s_d/\sqrt{n} = 8.3297/\sqrt{11} = 2.5115\). This is the standard deviation of the sample mean difference, a measurement of how precisely the sample mean difference measures the population mean difference. 6. Almost impossible. The sample means would vary every time we took a sample, around the true mean difference with a normal distribution having a standard error of about \(2.51\). Since we don’t know the population mean, 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 will vary around \(8.33\) with a standard deviation of about \(2.51\). 7. \(18.736\pm (2\times 2.5115)\), or \(18.736\pm5.023\), or from \(13.7\) to \(23.8\) pmol/litre. 8. We are 95% confident that the population \(\beta\) plasma concentration increases by a mean amount between \(13.7\) and \(23.8\) pmol/litre, during the fun run. 9. Either (or both) of these must be true: the population has a normal distribution, and/or the sample size is large enough so that the sample means have a normal distribution, so about larger than 25. 10. Since \(n<25\), we need to assume the population of differences has a normal distribution. The stem-and-leaf plot suggests this is not unreasonable, so the sample means quite possibly have an approx. normal distribution: 11. Looks likely that the plasma \(\beta\) concentrations are higher after the race. 12. \(n=(2\times 8.3297\div 2.5)^2 = 43.45\), so need data from \(44\) runners.