Question

A social psychologist is interested in how optimism is related to life satisfaction. A sample of individuals categorized as optimistic were asked about past, present, and projected future satisfaction with their lives. Higher scores on the life satisfaction measure indicate more satisfaction. Below are the data. What can the psychologist conclude with α = 0.01?

past	present	future
22 24 27 26 28	27 28 29 30 30	24 27 25 28 29

a) What is the appropriate test statistic?
---Select--- na one-way ANOVA within-subjects ANOVA two-way ANOVA

b) Compute the appropriate test statistic(s) to make a decision about H₀.
critical value =  ; test statistic =  
Decision:  ---Select--- Reject H0 Fail to reject H0

c) Compute the corresponding effect size(s) and indicate magnitude(s).
η² =  ;  ---Select--- na trivial effect small effect medium effect large effect

d) Make an interpretation based on the results.

At least one of the temporal means is different on life satisfaction.None of the temporal means is different on life satisfaction.    

e) Conduct Tukey's Post Hoc Test for the following comparisons:
1 vs. 3: difference =  ; significant:  ---Select--- Yes No
2 vs. 3: difference =  ; significant:  ---Select--- Yes No

f) Conduct Scheffe's Post Hoc Test for the following comparisons:
2 vs. 3: test statistic =  ; significant:  ---Select--- Yes No
1 vs. 3: test statistic =  ; significant:  ---Select--- Yes No

Answer 1

Calculations for F, Tukeys and Scheffes are at the end.

(a) The Appropriate Test Statistic is A one Way ANOVA

(b) Critical value = 6.927, The

Test statitic, F = 3.78

Since F test is < F critical, we Fail to reject H0

(c) = SS Between / SS Total = 29.73 / 76.93 = 0.3865

The effect is medium

(d) Conclusion: At least one of the temporal means is different on life satisfaction.

(e) For Tukey

For 1 vs 3 : Difference = 1.35, Significant: No

For 2 vs 3 : Difference = 2.48, Significant: No

(f) For Sceffe

2 vs 3: Test Statistic = 3.45: Significant : No

1 vs 3:Test Statistic = 3.45: Significant : No

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Calculation of F

The Summary statistics are as below

	A	B	C
Total	127	144	133
n	5	5	5
Mean	25.400	28.800	26.6
Sum Of Squares	23.2	6.8	17.2

The Overall Mean = (127 + 144 + 133) / 15 = 26.93

SS treatment = SUM n* ( - overall mean)² = 5 * (25.4 - 7.072)² + 5 * (28.8 - 26.93)² + 5 * (26.6 - 26.93)² = 29.73

df1 = k - 1 = 3 - 1 = 2

MSTR = SS treatment/df1 = 29.73 / 2 = 14.87

SSerror = SUM [Sum of Squares] = 23.2 + 6.8 + 17.2 = 47.2

df2 = N - k = 15 - 3 = 12

Therefore MS error = SSerror/df2 = 47.2 / 12 = 3.933

F = MSTR / MS Error = 14.87 / 9.333 = 3.78

Source	SS	DF	Mean Square	F	Fcv	p
Between	29.73	2	14.87	3.780	6.9266	0.0533
Within/Error	47.20	12	3.9333
Total	76.93	14

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Tukeys Post Hoc

= Sqrt(3.933 / 5) = 0.887

Tukeys critical for = 0.01, df = 12, k = 3 is 5.05

For 1 vs 3: HSD = Abs(25.4 - 26.6) / 0.887 = 1.35. Since this is < 5.05, difference is not significant

For 2 vs 3: HSD = Abs(28.8 - 26.6) / 0.887 = 2.48. Since this is < 5.05, difference is not significant

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Scheffes Test

2 vs 3: The Difference in means = 2.2.

The Test Statistic = = Sqrt( 2 * 3.78 * 3.933 * 0.4) = 3.45
Since The difference is less than 3.45, the difference is not significant.

1 vs 3: The Difference in means = 1.2

The Test statistic is the same as n does not change.

Since The difference is less than 3.45, the difference is not significant.

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