## D.21 Answers: CIs for one mean

Answers to exercises in Sect. 22.9.

Answer to Exercise 22.1: Standard error: $$\text{s.e.} = s/\sqrt{n} = 0.43/\sqrt{45} = 0.06410062$$ (keeping lots of decimal places in the working). Approximate 95% CI: $$2.85 \pm(2\times 0.06410062)$$, or $$2.85\pm 0.1282012$$, or from 2.72 litres to 2.98 litres.
Answer to Exercise 22.2: Standard error: $$\text{s.e.} = s/\sqrt{n} = 7571.74/\sqrt{58} = 994.2182$$ (keeping lots of decimal places in the working). Approximate 95% CI is: 4967.984 micrograms to 8944.86 micrograms.
Answer to Exercise 22.3: Approximate 95% CI for the mean brushing time: 29.9 seconds to 36.1 seconds.

Answer to Exercise 22.4: 1. Standard error: $$\text{s.e.}(\bar{x}) = 651.1/\sqrt{199} = 46.15526$$; approximate 95% CI: 754.1ml to 938.7ml. 2. They don’t seem very good at estimating (the article reports that the guesses ranged from 50ml to 3000ml). 3. The sample size is much larger than 25; the CI should be statistically valid. 4. Using the margin-of-error as 50, and $$s=651.1$$:

$\left( \frac{2\times 651.1}{50}\right)^2 = 678.2899.$ We would need about 679 participants (remembering to round up).

5. Using margin-of-error as 25, and $$s=651.1$$:

$\left( \frac{2\times 651.1}{25}\right)^2 = 2713.16.$ Need about 2714 participants (remembering to round up). 6. To halve the width of the margin of error, four times as many subjects are needed.
Answer to Exercise 22.5: None of these interpretations are acceptable. 1. CIs are not about how individual observations vary; they are about how a statistic varies (in this case, the sample mean). In addition, CIs are about populations and not samples. 2. CIs are not about how individual observations vary; they are about how a statistic varies (in this case, the sample mean. 3. This doesn’t make sense: samples can’t vary between two values. Sample statistics vary. In addition, CIs are about populations, not samples. 4. This doesn’t make sense: populations can’t vary between two values. Even population parameters don’t vary. 5. The population parameter does not vary. It is a fixed (but unknown) value to be estimated. (If the value of the population mean was constantly changing, it would be very hard to estimate…) 6. We know exactly what the sample mean is ($$\bar{x}=1.3649$$mmol/L: We don’t need a interval for the sample mean. 7. We know exactly what the sample mean is ($$\bar{x}=1.3649$$mmol/L: We don’t need a interval for the sample mean.
Answer to Exercise 22.6: Neither is correct. To learn about the variation in individuals trees, use the standard deviation rather than the standard error. The standard error tells us about the sample mean diameter, not about individual trees.