Question

Exercise 13-39 Static Given a recent outbreak of illness caused by E. coli bacteria, the mayor in a large city is concerned that some of his restaurant inspectors are not consistent with their evaluations of a restaurant's cleanliness. In order to investigate this possibility, the mayor has five restaurant inspectors grade (scale of 0 to 100) the cleanliness of three restaurants. The results are shown in the accompanying table. (You may find it useful to reference the g table.) Restaurant Inspector 72 68 73 69 75 54 84 85 80 82 84 60 56 a-1. Construct an ANOVA table. (Round intermediate calculations to at least 4 decimal places. Round "SS", "MS", "p-value" to 4 decimal places and "F to 3 decimal places.) ANOVA Source of Variation Rows Columns Error Total df MS p-value 1,723.6000 1,723.6000 76.0000 3,523.2000 861.8000 861.8000 6.3333 3.885 5.0068 3.885 5.0068 a-2. At the 5% significance level, can you conclude that the average grades differ by restaurant? Yes, since the p-value associated with restaurant is less than significance level. Yes, since the p-value associated with restaurant is greater than significance level. No, since the p-value associated with restaurant is less than significance level. No, since the p-value associated with restaurant is greater than significance level.

Answer 1

We will use R software to answer the question,

the ANOVA table is

a-1

Source of Variation	Df	Sum Sq	Mean Sq	F-value	Pr(>F)
Column	2	1723.6	861.8	102.7992	1.97E-06
Row	4	8.93	2.23	0.2664	0.8915
Residuals	8	67.07	8.38
Total	14	1799.60

a-2

here we get column (restaurant) p-value = 0.000 < 0.05 so we can reject the claim of all restaurant has same mean. Hence option A is correct.

b)

Mean difference	diff	lower	upper	p-value	Does the average differ
mu1-mu2	14.6	9.367417	19.832583	0.0001164	YES
mu1-mu3	-11.6	-16.832583	-6.367417	0.0005777	YES
mu2-mu3	-26.2	-31.432583	-20.967417	0.0000015	YES

c-1

no. since p-value=0.8915>0.05 . hence option D is correct

c-2

No

Please up-vote the solution.