Chapter 8.3, Problem 22E

To determine

a.

To find:

Effect on margin of error for confidence interval for a population mean if sample size decreases from 500 to 387 people.

Answer to Problem 22E

Solution:

When the sample size decreases, there is an increase in the margin of error for a population mean.

Explanation of Solution

Given:

The sample size decreases from $n_1=500$ to $n_2=387$.

Description:

Given a population with standard deviation $\sigma$, while estimating a population parameter, it is better to consider a range of values using samples of the same size (say $n$), also known as an interval estimate.

The true population parameters probability lies in this interval range is known as the confidence level $\left(c=1-\alpha \right)$.

At a certain level of confidence, the interval in which the maximum error can be observed is known as the confidence interval.

The maximum distance from point estimate that the confidence interval covers is margin of error and is given by:

$E=\frac{z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times \sigma }{\sqrt{n}}$,

where, $z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ is the critical value for the level of confidence $c=1-\alpha$.

Calculation:

Thus, the ratio of margins of error is:

$\begin{array}{rll}\frac{E_1}{E_2}& =& \frac{\left(\frac{z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times \sigma }{\sqrt{n_1}}\right)}{\left(\frac{z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times \sigma }{\sqrt{n_2}}\right)}\\ & =& \frac{\sqrt{n_2}}{\sqrt{n_1}}\end{array}$

where $E_1$ is the margin of error without decrease in sample size and $E_2$ is the margin of error after the sample size decreases.

The ratio is calculated as follows:

$\begin{array}{rll}\frac{E_1}{E_2}& =& \frac{\sqrt{n_2}}{\sqrt{n_1}}\\ \text{or},\frac{E_1}{E_2}& =& \frac{\sqrt{387}}{\sqrt{500}}\\ \text{or},\frac{E_1}{E_2}\approx \frac{19}{22}\end{array}$

which gives:

$\begin{array}{r}\frac{E_1}{E_2}<1\\ \text{or},E_1<E_2\end{array}$

Conclusion:

When the sample size decreases, there is an increase in the margin of error for a population mean.

To determine

b.

To find:

Effect on width of confidence interval for a population mean if sample size decreases from 500 to 387 people.

Answer to Problem 22E

Solution:

When the sample size decreases, there is an increase in the width of confidence interval for a population mean.

Explanation of Solution

Given:

The sample size decreases from $n_1=500$ to $n_2=387$.

Description:

Given a population with standard deviation $\sigma$, while estimating a population parameter, it is better to consider a range of values using samples of the same size (say $n$), also known as an interval estimate.

The true population parameters probability lies in this interval range is known as the confidence level $\left(c=1-\alpha \right)$.

At a certain level of confidence, the interval in which the maximum error can be observed is known as the confidence interval.

The maximum distance from point estimate that the confidence interval covers is margin of error and is given by:

$E=\frac{z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times \sigma }{\sqrt{n}}$,

where, $z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ is the critical value for the level of confidence $c=1-\alpha$.

Given the sample mean $\left(\overline{x}\right)$, the population mean confidence interval is calculated as:

$\left(\overline{x}-E,\overline{x}+E\right)$,

and width of confidence interval is:

$w=2\times E$.

Calculation:

Thus, the ratio of widths of respective confidence intervals is:

$\begin{array}{rll}\frac{w_1}{w_2}& =& \frac{2\times E_1}{2\times E_2}\\ & =& \frac{2\times \left(\frac{z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times \sigma }{\sqrt{n_1}}\right)}{2\times \left(\frac{z_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times \sigma }{\sqrt{n_2}}\right)}\\ & =& \frac{\sqrt{n_2}}{\sqrt{n_1}}\end{array}$

where $w_1$ is the width of confidence interval without decrease in sample size and $w_2$ is the width of confidence interval after the sample size decreases.

The ratio is calculated as follows:

$\begin{array}{rll}\frac{w_1}{w_2}& =& \frac{\sqrt{n_2}}{\sqrt{n_1}}\\ \text{or},\frac{w_1}{w_2}& =& \frac{\sqrt{387}}{\sqrt{500}}\\ \text{or},\frac{w_1}{w_2}& \approx & \frac{19}{22}\end{array}$

which gives:

$\begin{array}{rll}\frac{w_1}{w_2}& <& 1\\ \text{or},w_1& <& w_2\end{array}$

Conclusion:

When the sample size decreases, there is an increase in the width of confidence interval for a population mean.