Question

Independent random samples of newly completed apartments in four regions of a country yielded the data...

Independent random samples of newly completed apartments in four regions of a country yielded the data on monthly​ rents, in​ dollars and the summary statistics is given below:

Region n Mean Std.deviation
MW 6 752.33 89.37
NE 5 1051 146.10
S 4 862.25 159.37
W 5 1055.60 122.62

Do the data provide sufficient evidence to conclude that a difference exists in mean monthly rents among newly completed apartments in the four​ regions?

A partially filled in ANOVA table is presented below

Source SS DF MS F
Between 359293.92
Within XX
Total 620951.20 XX XX

What is the value of the F critical value at the α=0.05α=0.05 level? Report 4 decimal places.

Homework Answers

Answer #1
Source SS DF MS F
Between 359293.92 4 - 1 = 3 359293.92 / 3 = 119764.64 119764.64 / 16353.58 = 7.3234
Within 620951.20 - 359293.92 = 261657.28 19 - 3 = 16 261657.28 / 16 = 16353.58 XX
Total 620951.20 20 - 1 = 19 XX XX

For critical value of F at α=0.05

F( 3, 16 ) at α=0.05 = 3.239

Do the data provide sufficient evidence to conclude that a difference exists in mean monthly rents among newly completed apartments in the four​ regions?

Since, F = 7.3234 > F( 3, 16 ) at α=0.05 = 3.239 therefore, we will reject the null hypothesis of equality

Yes, a difference exists in mean monthly rents among newly completed apartments in the four​ regions

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