Question

The following is an incomplete F-table summarizing the results of a study of the variance of life satisfaction scores among unemployed, retired, part-time, and full-time employees.

Source of Variation	SS	df	MS	F
Between groups			18
Within groups (error)		36
Total	130

(a) Complete the F-table. (Round your values for mean squares and F to two decimal places.)

Source of Variation	SS	df	MS	F
Between groups			18
Within groups (error)		36
Total	130

(b) Compute omega-squared

(ω²).

(Round your answer to two decimal places.)
ω² =  

(c) Is the decision to retain or reject the null hypothesis? (Assume alpha equal to 0.05.)

Retain the null hypothesis.Reject the null hypothesis.     

Answer 1

a)

There are 4 groups:unemployed, retired, part-time, and full-time employees. So, k=4

Here' how the ANOVA F table can be calculated by hand:

and

	SS	df	MS	F
Between	SS(B)	k-1	SS(B)/(k-1)	MS(B)/MS(W)
Within	SS(W)	N-k	SS(W)/(N-k)
Total	SS(W)+SS(B)	N-1

Using the basic mathematics i have filled table: So the table looks like:

df_b = 4-1 =3. SS_b =df*MS = 3*18 =54

SS_t = 130, so SS_e = 130-54=76, MS_e = SS_e/df_e = 2.1111

F = MS_b/MS_e   = 18/2.111 = 8.526

	SS	df	MS	F
Between groups	54	3	18	8.526316
Within groups (error)	76	36	2.111111
Total	130	39

b)

Where SS_b = 54, k=4, MS_e = 2.111, SS_t = 130.

We get ω² = 0.36

c)

We reject the null hypothesis as probability against F(3,36,8.526) is 0.9998.

because this lies well within the rejection region of alpha=.05( 1-0.05 = 0.95).