Question

In: Statistics and Probability

Need; Hypotheses: Test statistic and value: P-value: and Conclusion: A) Past sales records indicate that sometimes...

Need; Hypotheses: Test statistic and value: P-value: and Conclusion:

A) Past sales records indicate that sometimes there is a large difference between the selling prices that different sales reps are able to negotiate for robots. As sales manager, you decide to randomly select four sales over the past year for each of your three sales reps and observe the actual selling price of the robots. The following table shows the amounts at which the robots sold in thousands of dollars. Based on these data, can you conclude that there is a difference between the mean selling prices for these three salespeople at a level of significance of 0.05? Assume that the three population distributions are approximately normal with equal population variances.

Selling Prices of Robots (in thousands of dollars)

Salesperson 1

Salesperson 2

Salesperson 3

10

11

11

14

16

13

13

14

12

12

15

15

Solutions

Expert Solution

Here Hypothesis are

H0: There is no signfiicant diference in mean selling prices for three salespeople.

Ha : There is at least one population mean selling price for a salesperson is different from the other two.

Here is the summary data regarding sample mean and sample standard deviation for selling price for all three salesperson.

Groups Sum Average Variance
Salesperson 1 49 12.25 2.916667
Salesperson 2 56 14 4.666667
Salesperson 3 51 12.75 2.916667

So, here we will perform one way ANOVA .

Here Degree of freedom Between groups = 3 -1 = 2

Degree of freedom within groups = (4-1) + (4-1) + (4-1) = 9

First we will calculate Grand mean xGM = [4 * 12.25 + 4 * 14 + 4 * 12.75] /12 = 13

SS(Between Groups) =

= 4 * (12.25 - 13)2 + 4 * (14 - 13)2 + 4 * (12.75 - 13)2

= 6.5

SS(Within Groups) =

SS(W) = (4-1) * 2.916667 + (4-1) * 4.66667 + (4-1) * 2.916667 = 31.5

Now MSB = SSB/dF =6.5/2 = 3.25

MSW = SSW/dF =31.5/9 = 3.5

Here F = 3.25/3.5 = 0.9286

Pr(F >0.9286) = FDIST(0.9286; 2; 9) = 0.43

Fcritical= F0.05,2,9= 4.2565

ANOVA
Source of Variation SS df MS F P-value F crit
Between Groups 6.5 2 3.25 0.928571 0.429904 4.256495
Within Groups 31.5 9 3.5
Total 38 11

Here F <  Fcritical then we fail to  reject the null hypothesis so we shall say that There is no signfiicant diference in mean selling prices for three salespeople.


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