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There are four auto body shops in Bangor, Maine, and all claim to promptly repair cars....

There are four auto body shops in Bangor, Maine, and all claim to promptly repair cars. To check if there is any difference iThe critical value is F =State the decision regarding the null hypothesis. Decision: HO. Conclude that at least one mean waiting time is the differenc

There are four auto body shops in Bangor, Maine, and all claim to promptly repair cars. To check if there is any difference in repair times, customers are randomly selected from each repair shop and their repair times in days are recorded. The output from a statistical E software package is: ptly repaircare. To check if there is any difference in te pa Count Groups Body Shop A Body Shop B Body Shop C Body Shop D Average 5.133333, Summary Sum 15.4 32 25.2 25. 9 6 AUW Variance 0.323333 1.433333 0.748 0.595833 5.04 .475 ANOVA df F Source of Variation Between Groups Within Groups Total SS 23.37321 9.726167 33.09938 MS 7.791069 0.810514 p-value 0.001632 9.612506 oÑ Is there evidence to suggest a difference in the mean waiting times at the four body shops? Use the 0.05 significance level. Compute the critical value. (Round your answer to 2 decimal places.)
The critical value is F =
State the decision regarding the null hypothesis. Decision: HO. Conclude that at least one mean waiting time is the difference.
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Answer #1

Solution-

at 0.05 and degree of freedom d1 = 3 and d2 = 12 critical value is

F = 3.49

F Distribution: Pr(F > 3.49) = 0.05 0.20 0.10 0.00 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

by the output provided.

p - value is less than significance level.

Also, F value is greater than F critical value.

So, Null hypothesis is rejected.

Decision:

Reject Ho, conclude that atleast one mean waiting time is the difference.

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