A realtor wants to compare the mean sales-to-appraisal ratios of residential pro

Question

A realtor wants to compare the mean sales-to-appraisal ratios of residential properties sold in four neighborhoods (A, B, C, and D). Four properties are randomly selected from each neighborhood and the ratios recorded for each, as shown below.

Neighborhood A

Neighborhood B

Neighborhood C

Neighborhood D

1.2

2.5

1.0

0.8

1.1

2.1

1.5

1.3

0.9

1.9

1.1

1.1

0.4

1.6

1.3

0.7

The realtor decided to compare the means of the sales-to-appraisal ratios of the properties sold in the four neighborhoods (A, B, C, and D) by using the Tukey-Kramer procedure with an overall level of significance of 0.05.

a)How many pairwise comparisons that can be made?   ______________________

b)Using an overall level of significance of 0.05, the critical value of the Studentized range Q used in calculating the critical range for the Tukey-Kramer procedure is ________.

c)Based on the Tukey-Kramer procedure with an overall level of significance of 0.05, the realtor would decide that there is a significant difference between which pairs of neighborhood properties.

_______________________________________________________________________________

d)Based on the Tukey-Kramer procedure with an overall level of significance of 0.05, the realtor would decide that there is no significant difference between which pairs of neighborhood properties.

_______________________________________________________________________________

Neighborhood A

Neighborhood B

Neighborhood C

Neighborhood D

1.2

2.5

1.0

0.8

1.1

2.1

1.5

1.3

0.9

1.9

1.1

1.1

0.4

1.6

1.3

0.7

Explanation / Answer

(a) How many pairwise comparisons that can be made? = 4C2= 6

(b) Critical value of  Studentized range Q test for alpha = 0.05 and DF = (4-1) *4 = 12 and number of treatments k = 4

so Qcritical= 4.199

(c) HEre table for means for All FOur Neighbourhood

Here is the table of Mean difference for pairs

Q = (ymax- ymin)/ sqrt [MSW/n]

Here ymax = maximum mean among 2 given mean

ymin= minimum mean among 2 given mean

MSW = mean square within treatements =1/dF * [ (nA-1) Var(A) + (nB-1) Var(B) + (nC -1) Var(C) + (nD -1) Var(D) ] = 1.1825/ 12 = 0.098542

n= 4

(i) Q value for Mean values for NEighbourhood A and Neighbourhood B

Q = (ymax - ymin)/ sqrt [MSW/n] = (1.125)/ sqrt (0.098542 /4) = 1.125/ 0.157 = 7.167

(ii) Q value for Mean values for NEighbourhood A and Neighbourhood C

Q = (ymax - ymin)/ sqrt [MSW/n] = (0.325)/ sqrt (0.098542 /4) = 0.325/ 0.157 = 2.07

(iii) Q value for Mean values for NEighbourhood A and Neighbourhood D

Q = (ymax - ymin)/ sqrt [MSW/n] = (0.075)/ sqrt (0.098542 /4) = 0.075/ 0.157 = 0.48

(iv) Q value for Mean values for NEighbourhood B and Neighbourhood C

Q = (ymax - ymin)/ sqrt [MSW/n] = (0.8)/ sqrt (0.098542 /4) = 0.8/ 0.157 = 5.10

(v) Q value for Mean values for NEighbourhood B and Neighbourhood D

Q = (ymax - ymin)/ sqrt [MSW/n] = (1.05)/ sqrt (0.098542 /4) = 1.05/ 0.157 = 6.69

(vi) Q value for Mean values for NEighbourhood C and Neighbourhood D

Q = (ymax - ymin)/ sqrt [MSW/n] = (0.25)/ sqrt (0.098542 /4) = 0.25/ 0.157 = 1.59

so for Neighbour hood pair A & B , Pair B & C and Pair B & D , there is significant difference.

(d) And there is no significant difference between pair A & C , A& D and C & D.

so we can also conclude that group B is most statistically significant group.

Groups Count Sum Average Variance Neighborhood A 4 3.6 0.9 0.126667 Neighborhood B 4 8.1 2.025 0.1425 Neighborhood C 4 4.9 1.225 0.049167 Neighborhood D 4 3.9 0.975 0.075833

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