Chapter 6, Problem 124E

(a)

To determine

To find: The $95\%$ confidence intervals for each occupation by the use of the standard deviation of the sample proportion.

Answer to Problem 124E

Solution: The $95\%$ confidence intervals for professional is $\left(0.22966,0.23034\right)$, for managerial is $\left(0.21968,0.22032\right)$, for administrative is $\left(0.19698,0.17032\right)$, for sales is $\left(0.14964,0.15036\right)$, for mechanical is $\left(0.11966,0.12034\right)$, for service is $\left(0.12975,0.13025\right)$, for operator is $\left(0.11977,0.12023\right)$ and for farm is $\left(0.07906,0.08094\right)$.

Explanation of Solution

Given: The summary table is provided.

Calculation: The standard deviation of a sample proportion with estimated mean $\widehat{p}$ is given as:

$\sigma =\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}$.

The $95\%$ confidence interval can be given as:

$\widehat{p}\pm z^{*}\left(\frac{\sigma }{\sqrt{n}}\right)$.

The value of $z^{*}$ is 1.96 at 95% confidence interval and it is obtained from the standard normal probabilities table.

For the professional, $\widehat{p}=0.23$ and $n=2447$. Therefore, standard deviation can be calculated as:

$\begin{array}{c}\sigma =\sqrt{\frac{0.23\left(1-0.23\right)}{2447}}\\ =\sqrt{\frac{0.23\times 0.77}{2447}}\\ =\sqrt{\frac{0.1771}{2447}}\\ =0.008507\end{array}$

Thus, the $95\%$ confidence interval can be calculated as:

$\begin{array}{c}\widehat{p}\pm z^{*}\left(\frac{\sigma }{\sqrt{n}}\right)=0.23\pm 1.96\left(\frac{0.008507}{\sqrt{2447}}\right)\\ =0.23\pm 1.96\left(\frac{0.008507}{49.47}\right)\\ =0.23\pm 0.00034\end{array}$

Hence, the interval is $\left(0.23-0.00034,0.23+0.00034\right)$ that is $\left(0.22966,0.23034\right)$.For the managerial, $\widehat{p}=0.22$ and $n=2552$. Therefore, standard deviation can be calculated as:

$\begin{array}{c}\sigma =\sqrt{\frac{0.22\left(1-0.22\right)}{2552}}\\ =\sqrt{\frac{0.22\times 0.78}{2552}}\\ =\sqrt{\frac{0.1716}{2552}}\\ =0.0082\end{array}$

Thus, the $95\%$ confidence interval can be calculated as:

$\begin{array}{c}\widehat{p}\pm z^{*}\left(\frac{\sigma }{\sqrt{n}}\right)=0.22\pm 1.96\left(\frac{0.0082}{\sqrt{2552}}\right)\\ =0.22\pm 1.96\left(\frac{0.0082}{50.52}\right)\\ =0.22\pm 0.00032\end{array}$

Hence, the interval is $\left(0.22-0.00032,0.22+0.00032\right)$ that is $\left(0.21968,0.22032\right)$.For the administrative, $\widehat{p}=0.17$ and $n=2309$. Therefore, standard deviation can be calculated as:

$\begin{array}{c}\sigma =\sqrt{\frac{0.17\left(1-0.17\right)}{2309}}\\ =\sqrt{\frac{0.17\times 0.83}{2309}}\\ =\sqrt{\frac{0.1411}{2309}}\\ =0.00782\end{array}$

Thus, the $95\%$ confidence interval can be calculated as:

$\begin{array}{c}\widehat{p}\pm z^{*}\left(\frac{\sigma }{\sqrt{n}}\right)=0.17\pm 1.96\left(\frac{0.00782}{\sqrt{2309}}\right)\\ =0.17\pm 1.96\left(\frac{0.00782}{48.05}\right)\\ =0.17\pm 0.00032\end{array}$

Hence, the interval is $\left(0.17-0.00032,0.17+0.00032\right)$. That is, $\left(0.19698,0.17032\right)$.

For the sales, $\widehat{p}=0.15$ and $n=1811$. Therefore, standard deviation is,

$\begin{array}{c}\sigma =\sqrt{\frac{0.15\left(1-0.15\right)}{1811}}\\ =\sqrt{\frac{0.15\times 0.85}{1811}}\\ =\sqrt{\frac{0.1275}{1811}}\\ =0.00704\end{array}$

Thus, the $95\%$ confidence interval can be calculated as:

$\begin{array}{c}\widehat{p}\pm z^{*}\left(\frac{\sigma }{\sqrt{n}}\right)=0.15\pm 1.96\left(\frac{0.00704}{\sqrt{1811}}\right)\\ =0.15\pm 1.96\left(\frac{0.00782}{42.56}\right)\\ =0.15\pm 0.00036\end{array}$

Hence, the interval is $\left(0.15-0.00036,0.15+0.00036\right)$ that is $\left(0.14964,0.15036\right)$.

For the mechanical, $\widehat{p}=0.12$ and $n=1979$. Therefore, standard deviation can be calculated as:

$\begin{array}{c}\sigma =\sqrt{\frac{0.12\left(1-0.12\right)}{1979}}\\ =\sqrt{\frac{0.12\times 0.88}{1979}}\\ =\sqrt{\frac{0.1056}{1979}}\\ =0.0073\end{array}$

Thus, the $95\%$ confidence interval can be calculated as:

$\begin{array}{c}\widehat{p}\pm 1.96\left(\frac{\sigma }{\sqrt{n}}\right)=0.12\pm 1.96\left(\frac{0.0073}{\sqrt{1797}}\right)\\ =0.12\pm 1.96\left(\frac{0.0073}{42.39}\right)\\ =0.12\pm 0.00034\end{array}$

Hence, the interval is $\left(0.12-0.00034,0.12+0.00034\right)$ that is $\left(0.11966,0.12034\right)$.For the service, $\widehat{p}=0.13$ and $n=2592$. Therefore, standard deviation can be calculated as:

$\begin{array}{c}\sigma =\sqrt{\frac{0.13\left(1-0.13\right)}{2592}}\\ =\sqrt{\frac{0.13\times 0.87}{2592}}\\ =\sqrt{\frac{0.1131}{2592}}\\ =0.0066\end{array}$

Thus, the $95\%$ confidence interval can be calculated as:

$\begin{array}{c}\widehat{p}\pm z^{*}\left(\frac{\sigma }{\sqrt{n}}\right)=0.13\pm 1.96\left(\frac{0.0066}{\sqrt{2592}}\right)\\ =0.13\pm 1.96\left(\frac{0.0066}{50.91}\right)\\ =0.13\pm 0.00025\end{array}$

Hence, the interval is $\left(0.13-0.00025,0.13+0.00025\right)$ that is $\left(0.12975,0.13025\right)$.

For the operator, $\widehat{p}=0.12$ and $n=2782$. Therefore, standard deviation can be calculated as:

$\begin{array}{c}\sigma =\sqrt{\frac{0.12\left(1-0.12\right)}{2782}}\\ =\sqrt{\frac{0.12\times 0.88}{2782}}\\ =\sqrt{\frac{0.1056}{2782}}\\ =0.0062\end{array}$

Thus, the $95\%$ confidence interval can be calculated as:

$\begin{array}{c}\widehat{p}\pm z^{*}\left(\frac{\sigma }{\sqrt{n}}\right)=0.12\pm 1.96\left(\frac{0.0062}{\sqrt{2782}}\right)\\ =0.12\pm 1.96\left(\frac{0.0062}{52.74}\right)\\ =0.12\pm 0.00023\end{array}$

Hence, the interval is $\left(0.12-0.00023,0.12+0.00023\right)$ that is $\left(0.11977,0.12023\right)$.

For the farm, $\widehat{p}=0.08$ and $n=571$. Therefore, standard deviation is,

$\begin{array}{c}\sigma =\sqrt{\frac{0.08\left(1-0.08\right)}{571}}\\ =\sqrt{\frac{0.08\times 0.92}{571}}\\ =\sqrt{\frac{0.0736}{571}}\\ =0.0114\end{array}$

Thus, the $95\%$ confidence interval can be calculated as:

$\begin{array}{c}\widehat{p}\pm z^{*}\left(\frac{\sigma }{\sqrt{n}}\right)=0.08\pm 1.96\left(\frac{0.0114}{\sqrt{571}}\right)\\ =0.08\pm 1.96\left(\frac{0.0114}{23.896}\right)\\ =0.08\pm 0.00094\end{array}$

Hence, the interval is $\left(0.08-0.00094,0.08+0.00094\right)$ that is $\left(0.07906,0.08094\right)$.

Interpretation: From the above results, it is observed that the $95\%$ confidence interval of the farms is the largest of all.

(b)

To determine

Whether there are any groups of occupations with similar stress levels in the result of part (a).

Answer to Problem 124E

Solution: Yes, the stress level of the occupations professional and managerial are same.

Explanation of Solution

From part (a), the $95\%$ confidence intervals for different occupations are,

Occupations	$95\%$ confidence intervals
professional	$0.22\pm 0.00032$
managerial	$0.22\pm 0.00032$
administrative	$0.17\pm 0.00032$
sales	$0.15\pm 0.00036$
mechanical	$0.12\pm 0.00034$
service	$0.13\pm 0.00025$
operator	$0.12\pm 0.00023$
farm	$0.08\pm 0.00094$

From the above results, it is observed that the stress level for the occupations professional and managerial are same and for mechanical and operator the stress level is comparable and also the stress level of stress and administrative are almost comparable.

(c)

To determine

The reason for the concern about the standard deviation formula in part (a).

Answer to Problem 124E

Solution: The reason for the concern is that the result may not that accurate due to the consideration of less stressed people among the major stressed people.

Explanation of Solution

In the part (a), the result is not that accurate because there is no consideration for people with stress but not a lot of stress. People with a little stress is also counted among the people with major stress. Therefore, the use of binomial distribution in this problem is not so adequate.