## 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 \(s=651.1\):

*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.