In: Statistics and Probability
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 | 24 | |||
Within groups (error) | 36 | |||
Total | 152 |
(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 | 24 | |||
Within groups (error) | 36 | |||
Total | 152 |
(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.
You may need to use the appropriate table in Appendix C to answer this question.
webassign.net/priviterastats3/priviterastats3_appendix_c.pdf
Given:
a)
The following is an incomplete F-table summarizing the results of a study of the variance of life satisfaction scores among 1) unemployed, 2) retired, 3) part-time, and 4) full-time employees.
Thus k = Number of groups = 4
Source of Variation | SS | df | MS | F |
---|---|---|---|---|
Between groups | 24 | |||
Within groups (error) | 36 | |||
Total | 152 |
We have to complete the F-table.
dfbetween = k - 1
dfbetween = 4 - 1
dfbetween = 3
MSbetween = 24
thus
MSbetween = SSbetween / dfbetween
24 = SSbetween / 3
SSbetween = 24 X 3
SSbetween = 72
We have:
SStotal = 152
thus
SSwithin = SStotal - SSbetween
SSwithin = 152 - 72
SSwithin = 80
and
dfwithin = 36
thus
MSwithin =SSwithin / dfwithin
MSwithin = 80 / 36
MSwithin = 2.22
and
dftotal = dfbetween + dfwithin
dftotal = 3 + 36
dftotal = 39
Thus
F = MSbetween / MSwithin
F = 24 / 2.22
F = 10.8
thus
Source of Variation | SS | df | MS | F |
---|---|---|---|---|
Between Groups | 72 | 3 | 24 | 10.8 |
Within Groups | 80 | 36 | 2.22 | |
Total | 152 | 39 |
b)
Compute omega-squared (ω2).
= 72 - (3 *2.22) / (152+2.22)
= (72 - 6.66) / 154.22
= 0.42
c) p value 0.000 < 0.05 ()
so, reject the null hypothesis.