Solution to Exercise 7.32.

  Assuming a normal distribution means we can use the t-distribution for confidence intervals. The basic formula for such a CI (confidence interval) is

$$\bar{x} \, \pm \, t_{\alpha/2 , \nu } \star s / \sqrt{n} ,$$

where

$$\bar{x} \; = \; \mbox{sample mean} ,$$

$$t_{\alpha/2 , \nu } \; = \; \mbox{quantile from $t$-distribution}$$

$$1-\alpha \; = \; \mbox{desired level of confidence}$$

$$\nu \; = \; n-1 \quad \mbox{ where $n$\space is sample size}$$

$$s \; = \; \mbox{sample standard deviation}$$

We want 95% confidence so $\alpha$ = .05 and n=8, so we use

$$t_{.025,7} \; = \; 2.365 .$$

Thus, given the data, the 95% CI is

$$30.2 \, \pm \, 2.365 * 3.1 / \sqrt{8} \; = \; 30.2 \pm 2.591 \; = \; 30.2 \pm 2.6 \; = \; (27.6,32.8) .$$

Thus, with 95% confidence,

$$27.6 \; \le \; \mu \; \le \; 32.8 .$$