Question

Anova 7. A large marketing firm uses many photocopying machines, several of each of four different models. During the last six months, the office manager has tabulated for each machine the average number of minutes per week that it is out of service due to repairs, resulting in the following data: Model G: 56 61 68 42 82 Model H: 74 77 92 63 54 Model K: 25 36 29 56 44 Model M: 78 105 89 112 61 Perform an analysis of variance to decide whether the difference among the means of the four samples can be attributed to chance.

State the null and alternative hypothesis

Give the p-value

Give a conclusion for the hypothesis test.

Answer 1

Null Hypothesis H0: The average minutes per week that machine is out of service due to repairs is equal for all the models.

Alternative Hypothesis Ha: Not all average minutes per week that machine is out of service due to repairs is equal for all the models.

Let T_i be the total time for model i, n_i be number of observations of model i.

Let G be the total time of all observations and N be total number of observations.

ΣX^² is sum of squares for all observations.

T1 = 309, T2 = 360 , T3 = 190 , T4 = 445

G = 309 + 360 + 190 + 445 = 1304

ΣX² = 19969 + 26754 + 7834 + 41295 = 95852

SST = ΣX² - G²/N = 95852 - 1304^²/20 = 10831.2

SSTR = ΣT²/n - G²/N = (309^² /5 + 360^² /5 + 190^² /5 + 445^2 / 5) - 1304^²/20 = 6820.4

SSE = 10831.2 - 6820.4 = 4010.8

Degree of freedom of group, dfTR = Number of level - 1 = 4 - 1 = 3

Degree of freedom of error, dfE = Number of observations - Number of level = 20 - 4 = 16

MSTR = SSTR / dfTR = 6820.4 / 3 = 2273.467

MSE = SSE / dfE = 4010.8 / 16 = 250.675

F = MSTR / MSE = 2273.467 / 250.675 = 9.069

P-value = P(F > 9.069 , df = 3, 16) = 0.00097

Conclusion for the hypothesis test -

As, p-value is less than the significance level 0.05, we reject the null hypothesis and conclude that there is significant evidence that not all average minutes per week that the machine is out of service due to repairs is equal for all the models.