Chapter 6.1, Problem 13E

Section 1

To determine

To graph: The diagram which shows the effect of sample size on the width of a 95% interval.

Explanation of Solution

Calculation: The confidence interval for the population mean $\mu$ at a significance level of $"\alpha "$ can be calculated by using the following formula:

$\begin{array}{c}\text{Lower}\text{bound}\text{=}\overline{x}-m\\ =\overline{x}-Z_{\alpha /2}\times \frac{\sigma }{\sqrt{n}}\end{array}$

And,

$\begin{array}{c}\text{Upper}\text{bound}\text{=}\overline{x}+m\\ =\overline{x}+Z_{\alpha /2}\times \frac{\sigma }{\sqrt{n}}\end{array}$

Where,

$\begin{array}{c}m=\text{Margin of error}\\ \overline{x}=\text{Sample mean}\\ \sigma \text{=Standard deviation}\\ n=\text{Sample size}\end{array}$

$$

Here, $Z_{\alpha /2}$ is the critical value which is obtained from standard normal table. The critical value is 1.96 at 5% level of significance. The confidence interval for population mean when the sample size is $n_1=10$ can be calculated as:

$\begin{array}{c}\text{Lower bound = }\overline{x}-Z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n_1}}\\ =73-1.96\times \frac{28}{\sqrt{10}}\\ =73-10.37\\ =62.63\end{array}$

And,

$\begin{array}{c}\text{Lower bound = }\overline{x}+Z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n_1}}\\ =73+1.96\times \frac{28}{\sqrt{10}}\\ =73+10.37\\ =83.37\end{array}$

Therefore, the required confidence interval is $\left(62.63,83.37\right)$.The confidence interval when $n_2=20:$

$\begin{array}{c}\text{Lower bound = }\overline{x}-Z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n_2}}\\ =73-1.96\times \frac{28}{\sqrt{20}}\\ =73-8.77\\ =64.23\end{array}$

And,

$\begin{array}{c}\text{Lower bound = }\overline{x}+Z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n_2}}\\ =73+1.96\times \frac{28}{\sqrt{20}}\\ =73+8.77\\ =81.77\end{array}$

Therefore, the required confidence interval is $\left(64.23,\text{81}\text{.77}\right)$.The confidence interval when $n_3=40:$

$\begin{array}{c}\text{Lower bound = }\overline{x}-Z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n_3}}\\ =73-1.96\times \frac{28}{\sqrt{40}}\\ =73-12.4\\ =60.6\end{array}$

And,

$\begin{array}{c}\text{Lower bound = }\overline{x}+Z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n_3}}\\ =73+1.96\times \frac{28}{\sqrt{40}}\\ =73+12.4\\ =85.4\end{array}$

Therefore, the required confidence interval is $\left(60.6,85.4\right)$.The confidence interval when $n_4=80:$

$\begin{array}{c}\text{Lower bound = }\overline{x}-Z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n_4}}\\ =73-1.96\times \frac{28}{\sqrt{80}}\\ =73-17.53\\ =55.47\end{array}$

And,

$\begin{array}{c}\text{Lower bound = }\overline{x}+Z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n_4}}\\ =73+1.96\times \frac{28}{\sqrt{80}}\\ =73+17.53\\ =90.54\end{array}$

Therefore, the required confidence interval is $\left(55.46,90.54\right)$.

Graph:

Section 2

To determine

To explain: The results.

Answer to Problem 13E

Solution: The width of the confidence interval reduces due to hike in the sample size.

Explanation of Solution

The growth of the sample size leads to a fall in the confidence interval because the margin of error is directly proportional to the sample size which means that if the sample size increases the margin of error decreases. This leads to reduction in the width of the confidence interval.