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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, s shown below. A: 12, 11, 09, 0.4 C 1.0, 1.5, 1.1, 1.3 B: 2.3, 2.1,1.9, 1.6 D0.8,1.3, 1.1, 0.7 The ressults of the analysis are Sumarized in the following ANOVA table F P-value 1818 1.0606 Neighbarhood Error Total 0001 12 4.3644 (a) (Spt) Fill in the blanks in the ANOVA table. (Show your steps for partial credits (b) (2pt) Bsed on the P-value in the ANOVA, you should rejesct) Hp then conclude thnt (reject or not (c) (2pt) The realtor decided to compare the 4 population mes by using the a multiple comparison procedure. In this case, you prefer to uTukey or Bonferroni) because (d) (1pt) Given the multiple comparison procesdr comparisons that can be made in (c), there are

Sagot :

The solutions are,

(a) 1.1825

(b) 0.0985

(c) Reject the null hypothesis since the test's p-value (0.001) is less than 0.05. One might draw the conclusion that there is evidence of a difference in the mean ratios for the 4 neighborhoods based on the sample data.

Given data,

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 ;

(a) The within-group sum of squares for this analysis will be

4.3644 - 3.1819 = 1.1825

Hence, within the group sum of squares for this analysis is 1.1825.

(b) There are four groups so df for within groups is 12 so the within-group mean squares for this analysis will be,

[tex]\frac{1.1825}{12}=0.0985[/tex]

(c) Since the p-value of the test (0.001) is less than 0.05, reject the null hypothesis. That is based on sample evidence one can conclude that there is evidence of a difference in the mean ratios for the 4 neighborhoods.

To learn more about mean click here:

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