Chapter 8.5, Problem 8.26E

(a)

To determine

To find: the confidence interval for the population mean.

Answer to Problem 8.26E

The confidence interval is $(32.5530,35.4470)$ .

Explanation of Solution

Given:

The sample size is 38.

The level of significance ( $\alpha$ ) is (1-0.98) i.e. 0.01.

The sample mean is 0.34.

The sample variance is 12.

Formulae used:

The $100(1-\alpha )\%$ Confidence interval for the population mean is given by:-

  $\left(\overline{x}-\sqrt{\frac{s^2}{n}}{Z}_{\frac{\alpha }{2}},\overline{x}+\sqrt{\frac{s^2}{n}}{Z}_{\frac{\alpha }{2}}\right)$

Calculations:

The Z Critical value at 99% level of significance is 2.58 calculated from the standard normal table.

The $100(1-\alpha )\%$ Confidence interval can be computed as:- $\begin{array}{l}\left(\overline{x}-\sqrt{\frac{s^2}{n}}{Z}_{\frac{\alpha }{2}},\overline{x}+\sqrt{\frac{s^2}{n}}{Z}_{\frac{\alpha }{2}}\right)=\left(34-2.58\sqrt{\frac{12}{38}},34+2.58\sqrt{\frac{12}{38}}\right)\\ =(32.5530,35.4470)\end{array}$

Therefore, the required confidence interval is $(32.5530,35.4470)$ .

(b)

To determine

To find: the confidence interval for the population mean.

Answer to Problem 8.26E

The confidence interval is $(1047.5429,1050.4751)$ .

Explanation of Solution

Given:

The sample size is 65.

The level of significance ( $\alpha$ ) is (1-0.90) i.e. 0.1.

The sample mean is 1049.

The sample variance is 51.

Calculations:

The Z Critical value at 90% level of significance is 1.645 calculated from the standard normal table.

The $100(1-\alpha )\%$ Confidence interval can be computed as:-

  $\begin{array}{l}\left(\overline{x}-\sqrt{\frac{s^2}{n}}{Z}_{\frac{\alpha }{2}},\overline{x}+\sqrt{\frac{s^2}{n}}{Z}_{\frac{\alpha }{2}}\right)=\left(1049-1.645\sqrt{\frac{51}{65}},1049+1.645\sqrt{\frac{51}{65}}\right)\\ =(1047.5429,1050.4751)\end{array}$

Therefore, the required confidence interval is $(1047.5429,1050.4751)$ .

(c)

To determine

To find: the confidence interval for the population mean.

Answer to Problem 8.26E

The confidence interval is $(65.9728,66.6272)$ .

Explanation of Solution

Given:

The sample size is 89.

The level of significance ( $\alpha$ ) is (1-0.90) i.e. 0.1.

The sample mean is 66.3.

The sample variance is 2.48.

Calculations:

The Z Critical value at 95% level of significance is 1.96 calculated from the standard normal table.

The $100(1-\alpha )\%$ Confidence interval can be computed as:-

  $\begin{array}{l}\left(\overline{x}-\sqrt{\frac{s^2}{n}}{Z}_{\frac{\alpha }{2}},\overline{x}+\sqrt{\frac{s^2}{n}}{Z}_{\frac{\alpha }{2}}\right)=\left(66.3-1.96\sqrt{\frac{2.48}{89}},66.3-1.96\sqrt{\frac{2.48}{89}}\right)\\ =(65.9728,66.6272)\end{array}$

Therefore, the required confidence interval is $(65.9728,66.6272)$ .