Below is listed the partial results of an ANOVA table A loca

Below is listed the partial results of an ANOVA table. A local auto shop is testing the difference between the average life expectancy of the car batteries they sell. What is the null hypothesis What is the alternative hypothesis Fill out the ANOVA table above. Assume a significance level of .05. What is the critical F-value for this test What is the statistical decision What is the answer to the actual question

Solution

A. Null hypothesis : There is no difference between the average life expectancy of the car bateries they sell.

B : Not all the average life of expectancy of the car bateries they sell are equal.

C.

Source SS DF MS F

Treatment 8400 7 1200 4

Error 30000 100 300

Total 38400 107

MS (Treatment) = MS(Error) *F = 300*4 = 1200

SS(Treatment) = DF(treatment) * MS(treatment) = 1200*7 = 8400

SS(Total) = SS(Treatment) + SS(Error) = 8400 +30000 = 38400

DF(Error) = DF(Total) - DF(Treatment) = 107 - 7 = 100

D. At a significance level of 0.05, F critical = f(df1=7, df2=100) = 2.10

E. We reject null hypothesis as F =4 > F critical 2.10

F. At the = 0.05 level of significance, there exists enough evidence to conclude that the average life expectancy of the car bateries they sell differs.