Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Similar to problem 11.7 in the text, based on section 11.2 of the text. An indiv

ID: 3179576 • Letter: S

Question

Similar to problem 11.7 in the text, based on section 11.2 of the text. An individual owns 3 automobile dealerships and is interested in determining whether average daily sales for the 3 dealerships are the same A random selection of daily sales at the 3 dealerships is presented in the table below. State the appropriate null and alternative hypotheses for this analysis. What is the between-column (explained) variance? Report the Mean Square Error (MSE) for this ANOVA? What is the value of the test-statistic for the test specified above? What are the degrees of freedom for the hypothesis that average sales for each dealership are equal? a. 2 for numerator. 9 for denominator b. 9 for numerator. 2 for denominator c. 2 for numerator. 11 for denominator d. 11 for numerator. 9 for denominator In testing the hypothesis that average sales for each dealership arc equal, the appropriate conclusion to be drawn would be: a. reject H_0 at the 10% significance level. b. the test is significant at the 88% level. c. the test is not significant at the 5% level. d. cannot reject the null hypothesis at the 10% level. c. accept the null hypothesis at the 88% level. What is the critical value for this test if the decision maker is willing to take a 10% chance of error? In one sentence, make a statement to interpret the results of this analysis.

Explanation / Answer

Part-1

Null hypothesis H0: Mu1=Mu2=Mu3

Alternative hypothesis H1: at least one pair of mean is different

Here Mu1,Mu2,Mu3 are the true population mean daily sales for the dealerships 1,2,3 respectively.

Part-3

The between column explained variance=MS between groups=1.1667

Part-4

MSE=8.9630

Part-5

Test statistic F=0.1302

Part-6

Degree of freedom of numerator=2

Degree of freedom of denominator=9

Part-8

p-value=0.8796 =0.88

as p-value=0.88>0.10, we can not reject the null hypothesis at 10% level

Part-9

Critical value =F(0.10,2,9)=3.0065 using excel function =FINV(0.1,2,9)

Part-10

We conclude that the true population mean daily sales for the dealerships 1,2,3 are not different and sales is almost same in all 3 dealerships

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote