Corporate sustainability of CPA firms. Refer to the Business and Society (March 2011) study on the sustainability behaviors of CPA corporations, Exercise 6.12 (p. 339). Recall that the level of support for corporate sustainability (measured on a quantitative scale ranging from 0 to 160 points) was obtained for each in a sample of 992 senior managers at CPA firms. The accompanying Minitab printout gives 90% confidence intervals for both the variance and standard deviation of level of support for all senior managers at CPA firms.

a. Locate the 90% confidence interval for ${\sigma }^2$on the printout. Interpret the result. b. Use the sample variance on the printout to calculate the 90% confidence interval for${\sigma }^2$. Does your result agree with the interval shown on the printout?

c. Locate the 90% confidence interval for on the printout.

d. Use the result, part a, to calculate the 90% confidence interval for$\sigma$. Does your result agree with the interval shown on the printout?

e. Give a practical interpretation of the 90% confidence interval for.$\sigma$

f. What assumption about the distribution of level of support is required for the inference, part e, to be valid? Is this assumption reasonably satisfied? (Use your answer to Exercise 4.125, p. 270.)

The level of support for corporate sustainability was obtained for each in a sample of 992 senior managers at CPA firms. So, the sample size is 992.

From the given MINITAB output, the 90% confidence interval is (672,779). Thus, the variance of level of support for all senior managers at CPA firms lies between 672 points squared and 779 points squared at 90% confidence.

b.

Since, the$\mathbf{100}\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{\alpha }\mathbf{)}\mathit{\%}$ confidence interval is given as,

$\mathbf{(}\frac{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}{\mathbf{s}}^{\mathbf{2}}}{{\mathbf{\chi }}_{\mathbf{0}\mathbf{.05}\frac{\mathbf{\alpha }}{\mathbf{2}}}^{\mathbf{2}}}\mathbf{\le }{\mathbf{\sigma }}^{\mathbf{2}}\mathbf{\le }\frac{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}{\mathbf{s}}^{\mathbf{2}}}{{\mathbf{\chi }}_{\mathbf{0}\mathbf{.95}}^{\mathbf{2}}}\mathbf{)}$

Let the confidence level be,

$\begin{array}{c}1-\alpha =0.90\\ \alpha =1-0.90\\ =0.10\end{array}$

Therefore,

$\frac{\alpha }{2}=0.05$

Also, the degrees of freedom is $(n-1)=991$

Now, from table critical-values are,

$\begin{array}{l}{\chi }_{0.05}^2\left(991\right)=1065.3\\ {\chi }_{0.95}^2\left(991\right)=918.90\end{array}$

Therefore,

role="math" localid="1658299573144" $\begin{array}{c}(\frac{(n-1)s^2}{{\chi }_{0.05\frac{\alpha }{2}}^2}\le {\sigma }^2\le \frac{(n-1)s^2}{{\chi }_{0.95}^2})=(\frac{(992-1)772}{1065.3}\le {\sigma }^2\le \frac{(992-1)722}{918.90})\\ =(\frac{\left(991\right)772}{1065.3}\le {\sigma }^2\le \frac{\left(991\right)722}{918.90})\\ =(671.6\le {\sigma }^2\le 778.6)\end{array}$

Hence, the 90% confidence interval for${\sigma }^2$ is (671.6, 778.6).

Yes, this result is near to the interval shown in the printout.

From the given MINITAB output, the 90% confidence interval forthe standard deviation of level of support for $\sigma$all senior managers at CPA firms lies between 25.9 points and 27.9 points.

From part a, the 90% confidence interval is (672,779). Take the square root on the lower and upper limit. That is,

$(\sqrt{672},\sqrt{779})=(25.9,27.9)$

This result is same as the interval shown in the printout.

The 90% confidence interval for the standard deviation of level of support for all senior managers at CPA firms lies between 25.9 points and 27.9 points.

f.

To ensure the appropriateness of the level of support for all senior managers at CPA firms of the population must possess the normal distribution. Thus, it can be concluded that the distribution of the data is approximately normal since the normal probability plot do not shows deviation from the line.