## A.8 Answers to Lecture 8 tutorial

### A.8.1 Answers to Sect. 8.4

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.8.2 Answers to Sect. 8.5.

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.8.3 Answers to Sect. 8.6

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.