Question

11) A company that employs a large number of salespeople is interested in learning which of the salespeople sell the most: those strictly on commission, those with a fixed salary, or those with a reduced fixed salary plus a commission. The previous month's records for a sample of salespeople are inspected and the amount of sales (in dollars) is recorded for each, as shown in the table. Commissioned $492 $520 $462 $483 $466 $475 Fixed Salary $450 $418 $437 $432 $444 Commission plus Salary $425 $492 $470 $506 Test to determine if a difference exists in the mean sales (in dollars) among the three compensation systems. Test using a=0.10. Include the following steps in your test: a. State the null and alternative hypothesis. b. State the rejection region. (draw a picture) Calculate the test statistic (you may use your calculator test). d. State the p-value (you may use your calculator test). e. Make a decision whether to reject the null hypothesis. f. Interpret your decision. c.

Answer 1

Commissioned Fixed Salary Commission plus Salary 492 450 425 520 418 492 462 437 470 483 432 506 466 444 475

We are given the data for three variables.

We determine if a difference exist in amean sales amoung the tree compensation systems.

We are given the level of significance ( alpha) = 0.10

So here we need to use the one way ANOVA.

a )

Null hypotheis ( H0) : Mean for all three compensation system are same.

Alternative hypothesis ( H1) : At least one mean is different.

b)

We can construct one way ANOVA using formula or technology.

Let me give you the answer using excel.

First please enter given data in excel.

Please go through Data >> Data Analysis >> one way ANOVA >> then give input and output range >> Click on OK

Then you will get the given output.

Anova: Single Factor Count Sum SUMMARY Groups Commissioned Fixed Salary Commission plus Salary 6 2898 Average 483 436.2 473.25 Variance 448.8 150.2 1254.25 5 2181 4 1893 SS df MS ANOVA Source of Variation Between Groups Within Groups F P-value Fcrit 5.777799638 0.017479239 3.885294 6362.85 2 3181.425 6607.55 12 550.6291667 Total 12970.4 14

c )

Test statistic = 5.7778

d )

P - value = 0.0175

e)

Decision rule :

If p value is < or equal to alpha ( 0.10), we can reject the null hypothesis.

If p value > 0.10, we can fail to reject null hypothesis.

So here p value is greater than the 0.10 so we can fail to reject the null hypothesis.

f )

Conclusion :

So here all the compensation system has same mean.