Chapter 6.1, Problem 14E

Section 1

To determine

To graph: The diagram that illustrates the effect of the confidence level on the width of the interval.

Explanation of Solution

Calculation: The confidence intervals based on different sample sizes need to be determined in order to draw the required diagram. The lower and upper bounds of the confidence interval for the population mean $\mu$, at a significance level of $\alpha$ is calculated using the following formula:

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

And,

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

Where $\overline{x}$ and m represent the sample mean and margin of error respectively and $\sigma$ denotes the population standard deviation and n is the size of the sample. $Z_{\frac{\alpha }{2}}$ is the critical value of the standard normal distribution at a significance level of $\alpha$, whose value is obtained from Table A provided in the book. For a confidence level of $80\%$, the level of significance $\alpha =0.20$. Therefore,

$\begin{array}{c}\text{Lower bound = }\overline{x}-Z_{\alpha /2}\times \frac{\sigma }{\sqrt{n}}\\ =78-Z_{{}_{0.20/2}}\times \frac{20}{\sqrt{64}}\\ =78-1.282\times \frac{20}{\sqrt{64}}\\ =74.80\end{array}$

And,

$\begin{array}{c}\text{Upper bound = }\overline{x}+Z_{{}_{\alpha /2}}\times \frac{\sigma }{\sqrt{n}}\\ =78+Z_{{}_{0.20/2}}\times \frac{20}{\sqrt{64}}\\ =78+1.282\times \frac{20}{\sqrt{64}}\\ =81.21\end{array}$

Therefore, the required confidence interval is $\left(74.80,\text{81}\text{.21}\right)$.

For a confidence level of $90\%$, the level of significance $\alpha =0.10$. Therefore,

$\begin{array}{c}\text{Lower bound = }\overline{x}-Z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n}}\\ =78-Z_{\frac{0.10}{2}}\times \frac{20}{\sqrt{64}}\\ =78-1.645\times \frac{20}{\sqrt{64}}\\ =73.89\end{array}$

And,

$\begin{array}{c}\text{Upper bound = }\overline{x}+Z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n}}\\ =78+Z_{\frac{0.10}{2}}\times \frac{20}{\sqrt{64}}\\ =78+1.645\times \frac{20}{\sqrt{64}}\\ =82.11\end{array}$

Therefore, the required confidence interval is $\left(73.89,\text{82}\text{.11}\right)$. For a confidence level of $95\%$, the level of significance $\alpha =0.05$. Therefore,

$\begin{array}{c}\text{Lower bound = }\overline{x}-Z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n}}\\ =78-Z_{\frac{0.05}{2}}\times \frac{20}{\sqrt{64}}\\ =78-1.96\times \frac{20}{\sqrt{64}}\\ =73.1\end{array}$

And,

$\begin{array}{c}\text{Upper bound = }\overline{x}+Z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n}}\\ =78+Z_{\frac{0.05}{2}}\times \frac{20}{\sqrt{64}}\\ =78+1.96\times \frac{20}{\sqrt{64}}\\ =82.9\end{array}$

Therefore, the required confidence interval is $\left(73.1,\text{82}\text{.9}\right)$. For a confidence level of $99\%$, the level of significance $\alpha =0.01$. Therefore,

$\begin{array}{c}\text{Lower bound = }\overline{x}-Z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n}}\\ =78-Z_{\frac{0.01}{2}}\times \frac{20}{\sqrt{64}}\\ =78-2.58\times \frac{20}{\sqrt{64}}\\ =71.55\end{array}$

And,

$\begin{array}{c}\text{Upper bound = }\overline{x}+Z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n}}\\ =78+Z_{\frac{0.01}{2}}\times \frac{20}{\sqrt{64}}\\ =78+2.58\times \frac{20}{\sqrt{64}}\\ =84.45\end{array}$

Therefore, the required confidence interval is $\left(71.55,\text{84}\text{.45}\right)$. Hence, the diagram which shows the effect of the confidence level on the width of the interval is shown below:

Graph:

Section 2

To determine

To explain: The results.

Answer to Problem 14E

Solution: The width of the confidence interval expands due to hike in the level of confidence.

Explanation of Solution

The margin of error rises with a rise in the confidence level because there is a direct relationship between the level of confidence and margin of error. This leads to an expansion in the width of the confidence interval. Hence, it can be concluded that as level of confidence increases, the width of the confidence interval increases.