Chapter 13, Problem 2CP

Compensation for Sales Professionals

Suppose that a local chapter of sales professionals in the greater San Francisco area conducted a survey of its membership to study the relationship, if any, between the years of experience and salary for individuals employed in inside and outside sales positions. On the survey, respondents were asked to specify one of three levels of years of experience: low (1–10 years), medium (11–20 years), and high (21 or more years). A portion of the data obtained follow. The complete data set, consisting of 120 observations, is contained in the file named SalesSalary.

Managerial Report

1. Use descriptive statistics to summarize the data.

2. Develop a 95% confidence interval estimate of the mean annual salary for all salespersons, regardless of years of experience and type of position.

3. Develop a 95% confidence interval estimate of the mean salary for inside salespersons.

4. Develop a 95% confidence interval estimate of the mean salary for outside salespersons.

5. Use analysis of variance to test for any significant differences due to position. Use a .05 level of significance, and for now, ignore the effect of years of experience.

6. Use analysis of variance to test for any significant differences due to years of experience. Use a .05 level of significance, and for now, ignore the effect of position.

7. At the .05 level of significance test for any significant differences due to position, years of experience, and interaction.

To determine

Use descriptive statistics to summarize the data.

Explanation of Solution

Calculation:

The data represents the survey results obtained to study the relationship, if any, between the years of experience and salary for individuals employed in inside and outside sales positions. The respondents were asked to specify one of the three levels of years of experience: low, medium and high.

Software procedure:

Step by step procedure to obtain descriptive statistics using the MINITAB software:

• Choose Stat> Tables >Descriptive statistics.
• In Categorical variables, for rows enter Position.
• In Categorical variables,for columns enter Experience.
• In Categorical variables click on Counts.
• In Associated variables enter Salary.
• Under Display click on Means, Standard deviations.
• Click OK.

Output using the MINITAB software is given below:

Thus, the descriptive statistics for the years of experience and salary for individuals employed in inside and outside sales positions is obtained.

The mean annual salary for sales persons regardless of years of experience and type of position is $64,925.48 and the standard deviation is $10,838.67. The mean salary for ‘Inside’ sales persons is $56,020.52 and the standard deviation is $3589.83. The mean salary for ‘Outside’ sales persons is $73,830.43 and the standard deviation is $7,922.96. Themean salary and standard deviation for ‘Outside’ sales persons is higher comparing with themean salary for ‘Inside’ sales persons.

The mean salary for sales persons who have ‘Low’ years of experience is $59,819.63 and the standard deviation is $6,005.06.

The mean salary for sales persons who have ‘Medium’ years of experience is $68,618.13 and the standard deviation is $13,621.38.

The mean salary for sales persons who have ‘High’ years of experience is $66,338.68 and the standard deviation is $9,699.51.

The mean salary and standard deviation for sales persons who have ‘Medium’ years of experience is higher compared with the mean salary for sales persons who have ‘Low’ years of experience and ‘High’ years of experience.

To determine

Develop a 95% confidence interval estimate of the mean annual salary for all sales persons regardless of years of experience and type of position.

Answer to Problem 2CP

The 95% confidence interval estimate of the mean annual salary for all sales persons regardless of years of experience and type of position is (62,966.41, 66,884.55).

Explanation of Solution

Calculation:

Here, 120 observations is considered as the sample and the population standard deviation is not known. Hence, t-test can be used for finding confidence intervals for testing population means.

The level of significance is 0.05.

Hence,

$\begin{array}{c}{t}_{\frac{\alpha }{2}}=t_{\frac{0.05}{2}}\\ =t_{\frac{0.025}{2}}\end{array}$

The 95% confidence interval for the mean annual salary for all sales persons regardless of years of experience and type of position is, $\overline{x}-t_{0.025,119}\left(\frac{s}{\sqrt{n}}\right)<\mu <\overline{x}+t_{0.025,119}\left(\frac{s}{\sqrt{n}}\right)$.

From part (a), substitute, $\overline{x}=\$64,925.48,s=10,838.67,n=120$ in the above formula.

$64,925.48-t_{0.025,119}\left(\frac{10,838.67}{\sqrt{120}}\right)<\mu <64,925.48+t_{0.025,119}\left(\frac{10,838.67}{\sqrt{120}}\right)$

Software procedure:

Step by step procedure to obtain $t_{0.025,119}$ using the MINITAB software:

• Choose Graph > Probability Distribution Plot >View Single, and then clickOK.
• From Distribution, choose ‘t’ distribution.
• Under Degrees of freedom enter 119.
• Click the Shaded Area tab.
• Choose Probability and Both tail for the region of the curve to shade.
• Enter the value as 0.05.
• Click OK.

Output using MINITAB software is given below:

The value $t_{0.025,119}=1.980$.

The 95% confidence interval for the mean is,

$\begin{array}{c}64,925.48-t_{0.025,119}\left(\frac{10,838.67}{\sqrt{120}}\right)<\mu <64,925.48+t_{0.025,119}\left(\frac{10,838.67}{\sqrt{120}}\right)\\ 64,925.48-1.980\left(\frac{10,838.67}{\sqrt{120}}\right)<\mu <64,925.48+1.980\left(\frac{10,838.67}{\sqrt{120}}\right)\\ 64,925.48-1,959.07<\mu <64,925.48+1,959.07\\ 62,966.41<\mu <66,884.55\end{array}$

Thus, the 95% confidence interval estimate of the mean annual salary for all sales persons regardless of years of experience and type of position is (62,966.41, 66,884.55).

To determine

Develop a 95% confidence interval estimate of the mean salary for inside sales persons.

Answer to Problem 2CP

The 95% confidence interval estimate of the mean salary for inside sales persons is (56,947.87, 55,093.17).

Explanation of Solution

Calculation:

From part (a), substitute $\overline{x}=\$56020.52,$$s=3,589.83,n=60$ in the above formula.

$56,020.52-t_{0.025,59}\left(\frac{3,589.83}{\sqrt{60}}\right)<\mu <56,020.52+t_{0.025,59}\left(\frac{3,589.83}{\sqrt{60}}\right)$

Step by step procedure to obtain $t_{0.025,59}$ using the MINITAB software:

• Choose Graph > Probability Distribution Plot >View Single, and then clickOK.
• From Distribution, choose ‘t’ distribution.
• Under Degrees of freedom enter 59.
• Click the Shaded Area tab.
• Choose Probability and Both tail for the region of the curve to shade.
• Enter the value as 0.05.
• Click OK.

Output using MINITAB software is given below:

The value $t_{0.025,59}=2.001$.

The 95% confidence interval for the mean is,

$\begin{array}{c}56,020.52-t_{0.025,59}\left(\frac{3,589.83}{\sqrt{60}}\right)<\mu <56,020.52+t_{0.025,59}\left(\frac{3,589.83}{\sqrt{60}}\right)\\ 56,020.52-2.001\left(\frac{3,589.83}{\sqrt{60}}\right)<\mu <56,020.52+2.001\left(\frac{3,589.83}{\sqrt{60}}\right)\\ 56,020.52-927.35<\mu <56,020.52-927.35\\ 56947.87<\mu <55093.17\end{array}$.

Thus, the 95% confidence interval estimate of the mean salary for inside sales persons is (56,947.87, 55,093.17).

To determine

Develop a 95% confidence interval estimate of the mean salary for outside sales persons.

Answer to Problem 2CP

The 95% confidence interval estimate of the mean salary for outside sales persons is (75,877.15, 71,783.71).

Explanation of Solution

Calculation:

The 95% confidence interval for the mean salary for outside sales persons is, $\overline{x}-t_{0.025,59}\left(\frac{s}{\sqrt{n}}\right)<\mu <\overline{x}+t_{0.025,59}\left(\frac{s}{\sqrt{n}}\right)$.

From part (a), substitute $\overline{x}=\$73,830.43,$$s=7,922.96,n=60$ in the above formula.

$73,830.43-t_{0.025,59}\left(\frac{3,589.83}{\sqrt{60}}\right)<\mu <56,020.52+t_{0.025,59}\left(\frac{3,589.83}{\sqrt{60}}\right)$

The 95% confidence interval for the mean is,

$\begin{array}{c}73,830.43-t_{0.025,59}\left(\frac{3,589.83}{\sqrt{60}}\right)<\mu <56,020.52+t_{0.025,59}\left(\frac{3,589.83}{\sqrt{60}}\right)\\ 73,830.43-2.001\left(\frac{3,589.83}{\sqrt{60}}\right)<\mu <56,020.52+2.001\left(\frac{3,589.83}{\sqrt{60}}\right)\\ 73,830.43-2,046.72<\mu <73,830.43-2,046.72\\ 75,877.15<\mu <71,783.71\end{array}$.

Thus, the 95% confidence interval estimate of the mean salary for inside sales persons is (56947.87, 55093.17).

To determine

Check whether there are any significant differences due to position at $\alpha =0.05$ level of significance by ignoring the effect of years of experience.

Answer to Problem 2CP

There is sufficient evidence to conclude that there is significant difference in the mean of positions at $\alpha =0.05$ level of significance.

Explanation of Solution

Calculation:

State the hypotheses:

Null hypothesis:

$H_0:{\mu }_1={\mu }_2$

Alternative hypothesis:

$H_a:$ Not all the population means are equal.

The level of significance is 0.05.

Software procedure:

Step by step procedure to obtain One-Way ANOVA using the MINITAB software:

• Choose Stat > ANOVA > One-Way.
• In Response, enter the column of values.
• In Factor, enter the column of Position.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the F-ratio is 251.54 and the p-value is 0.000.

Decision:

If $p\text{-value}\le \alpha ,$ reject the null hypothesis $H_0$.

If $p\text{-value}>\alpha ,$ fail to reject the null hypothesis $H_0$.

Conclusion:

Here, the p-value is less than the level of significance.

That is, $p\text{-value}\left(=0.000\right)<\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is rejected.

There is sufficient evidence to conclude that there is significant difference in the mean of positions at $\alpha =0.05$ level of significance.

To determine

Check whether there are any significant differencesdue to years of experience at $\alpha =0.05$ level of significance by ignoring the effect of position.

Answer to Problem 2CP

There is sufficient evidence to conclude that there is significant difference in the mean of years of experience at $\alpha =0.05$ level of significance.

Explanation of Solution

Calculation:

State the hypotheses:

Null hypothesis:

$H_0:{\mu }_1={\mu }_2={\mu }_3$

Alternative hypothesis:

$H_a:$ Not all the population means are equal.

The level of significance is 0.05.

Software procedure:

Step by step procedure to obtain One-Way ANOVA using the MINITAB software:

• Choose Stat > ANOVA > One-Way.
• In Response, enter the column of values.
• In Factor, enter the column of Experience.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the F-ratio is 7.93 and the p-value is 0.001.

Conclusion:

Here, the p-value is less than the level of significance.

That is, $p\text{-value}\left(=0.001\right)<\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is rejected.

There is sufficient evidence to conclude that there is significant difference in the mean of years of experience at $\alpha =0.05$ level of significance.

To determine

Test for any significant differences due to position, years of experience and interaction at $\alpha =0.05$ level of significance.

Answer to Problem 2CP

The main effect of factor A (Position) is significant.

The main effect of factor B (Experience) is significant.

The interaction is significant.

Explanation of Solution

Calculation:

Factor A is Position (Inside, Outside). Factor B is Experience (Low, Medium, High).

The testing of hypotheses is as follows:

State the hypotheses:

Main effect of factor A:

Null hypothesis:

$H_0:$ There is no significant difference in the means of “Position”.

Alternative hypothesis:

$H_a:$ There is significant difference in the means of “Position”.

Main effect of factor B:

Null hypothesis:

$H_0:$ There is no significant difference in the means of “Experience”.

Alternative hypothesis:

$H_a:$ There is significant difference in the means of “Experience”.

Interaction:

Null hypothesis:

$H_0:$ There is no interaction between the Factor A (Position) and Factor B (Experience).

Alternative hypothesis:

$H_a:$ There is an interaction between the Factor A (Position) and Factor B (Experience).

Software procedure:

Step-by-step procedure to obtain two-way ANOVA using MINITAB software is given below:

• Choose Stat > ANOVA > Two-Way.
• In Response, enter the column of Salary.
• In Row Factor, enter the column of Position. Click on display means.
• In Column Factor, enter the column of Experience. Click on display means.
• Click on Store residuals and Store fits.
• Click OK.

Output obtained by MINITAB procedure is as follows:

For Factor A (Position), the F-test statistic is 751.36 and the p-value is 0.000.

For Factor B (Experience), the F-test statistic is 65.86 and the p-value is 0.000.

For interaction, the F-test statistic is 53.38 and the p-value is 0.000.

Decision:

If $p\text{-value}\le \alpha ,$ reject the null hypothesis $H_0$.

If $p\text{-value}>\alpha ,$ fail to reject the null hypothesis $H_0$.

Conclusion:

Factor A:

Here, the p-value is less than the level of significance.

That is, $p\text{-value}\left(=0.000\right)<\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is rejected.

That is, the main effect of factor A (Position) is significant.

Factor B:

Here, the p-value is less than the level of significance.

That is, $p\text{-value}\left(=0.000\right)<\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is rejected.

That is, the main effect of factor B (Experience) is significant.

Interaction:

Here, the p-value is less than the level of significance.

That is, $p\text{-value}\left(=0.000\right)<\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is rejected.

Thus, the interaction is significant.