Solution to Exercise 7.42.

  Using the formulae in the box on p. 305, the lower limit of the 95% CI for $\sigma^2$ is

$$(n-1) s^2 / \chi^2_{\alpha/2,n-1} \; = \; (9-1) (2.81)^2 / \chi^2_{.025,9-1} \; = \; 8 * (2.81)^2 / 17.534$$

$$\; = \; 63.1688 / 17.534 \; = \; 3.602646 .$$

The upper limit is

$$(n-1) s^2 / \chi^2_{1-\alpha/2,n-1} \; = \; (9-1) (2.81)^2 / \chi^2_{.975,9-1} \; = \; 8 * (2.81)^2 / 2.180$$

$$\; = \; 63.1688 / 2.180 \; = \; 28.97651 .$$

In summary,

$$3.60 \; \le \; \sigma^2 \; \le \; 28.98 , \quad \mbox{with 95\% confidence}.$$

For the standard deviation, we can take square roots all the way through:

$$\sqrt{3.602646} \; \le \; \sigma \; \le \; \sqrt{28.97651} ,$$

or

$$1.898064 \; \le \; \sigma \; \le \; 5.382983 ,$$

or, rounding off to the appropriate number of significant digits,

$$1.90 \; \le \; \sigma \; \le \; 5.38 , \quad \mbox{with 95\% confidence}.$$

Not that the sample standard deviation s = 2.81 is in the interval, of course, but is not the midpoint. It is further from the upper limit, indicating there is more uncertainty on the ``high side,'' which will always be the case with this kind of confidence interval.