Question

This has been previously answered but I don't understand, no one did the Ho and Ha, could someone please explain step by step with formulas?

A realtor wishes to compare the square footage of houses of similar prices in 4 different cities. He took a sample of 5 houses from City 1, 4 houses from City 2, 6 houses from City 3, and 7 houses from City 4. Sum of Squares Total is 71.06 and Mean Squares Between is 8.75. Can the realtor conclude that the mean square footage is the same in all four cities? Assume that the level of significance is 0.01. [6 marks]

(Hint: Set up an ANOVA table with the given information and then complete the ANOVA table for the missing information. Also, include the key steps of a test of hypothesis question). [4 marks]

Answer 1

Null Hypothesis H0: Mean square footage is the same in all four cities.

Alternative Hypothesis Ha: At least one of the city has different mean square footage.

First we will find degree of freedom.

df Total = Number of observations - 1 = (5 + 4 + 6 + 7) - 1 = 21

df Between = Number of groups (cities) - 1 = 4 -1 = 3

df Within = df Total - df Between = 21 - 3 = 18

SS Between = MS Between * df Between = 8.75 * 3 = 26.25

SS Within = SS Total - SS Between = 71.06 - 26.25 = 44.81

MS Within = SS Within / df Within = 44.81 / 18 = 2.49

F = MS Between / MS Within

= 8.75 / 2.49

= 3.51

Anova Table is,

Source Of Variation	SS	df	MS	F
Between Sum of Squares	26.25	3	8.75
Within sum of Squares (Error)	44.81	18	2.49
Total	71.06	21

Numerator df = df Between = 3

Denominator df = df Within = 18

Critical value of F at level of significance is 0.01 and df = 3, 18 is 5.09

Since the observed F (3.51) is less than the critical value, we fail to reject the null hypothesis H0 and conclude that there is no strong evidence to reject the claim that mean square footage is the same in all four cities.