Question

Application Exercise: 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 an a of 0.05? past present future 22 24 27 26 28 27 28 29 30 30 24 27 26 28 23 a) What is the appropriate test statistic? -Select- b) Obtain/compute the appropriate values to make a decision about Hg. critical value = 1; test statistic = Decision: ---Select- c) Compute the corresponding effect size(s) and indicate magnitude(s). on? --Select- d) Make an interpretation based on the results. O At least one of the temporal means is different on life satisfaction. O None of the temporal means is different on life satisfaction. e) Regardless of the Ho decision in b), conduct Tukey's post hoc test for the following comparisons: 2 vs. 3: difference ; significant: ---Select- 1 vs. 2: difference = i significant: ---Select... f) Regardless of the Ho decision in b), conduct Scheffe's post hoc test for the following comparisons: 1 vs. 2: test statistics ; significant: -Select- 1 vs. 3: test statistic = significant: ---Select

a) Drop-down options: na, one-way ANOVA, within subjects ANOVA, or two-way ANOVA

b) Drop-down options: Yes or No

c) Drop-down options: na, Trivial Effect, Small Effect, Medium Effect, Large Effect

e & f) Drop-down options: Yes or No

Answer 1

(a). Approriate test for hypothesis is one-way anova because we want to compare the average past, present and future satisfaction scores of individuals and we have only one factor which is of temporal (i.e., past, present and future). So, our hypothesis are:

• Null Hypothesis (H_o) : μ₁ = μ₂ = μ₃
• Alternative Hypothesis (H₁) : means are not equal.

(b) for computation of anova in spss follow the given steps:

1. Firstly enter the data into spss in the one column named, SatisfactionScore..
2. Now in the second column named, factor, type the values 1,2,3 corresponding to past, present and , future data.
3. ,Now go to Analyze > Compare means > One-way Anova. Then a dialoque box will appear.
4. Then in Dependent list move data column and in Factor move factor column.
5. Then click on "Post-hoc" and then in the dialog box appear, select Turkey's and Scheffe's and set the significance level = 0.05
6. Then finally click OK

х *Untitled1 [DataSeto] - IBM SPSS Statistics Data Editor File Edit View Data Transform Analyze Direct Marketing Graphs Utilities Add-ons Window Help FLUR VE MU A AAC HERE HT REF: 12: Visible: 2 of 2 Variables Factor var var var var var var var var var var var var var var var 1 One-Way ANOVA X 2 3 Dependent List: SatisfactionScore 4 5 Contrasts.. Post Hoc. Options... Bootstrap 6 7 Satisfactions core 22.00 24.00 27.00 26.00 28.00 27.00 28.00 29.00 30.00 30.00 24.00 27.00 26.00 28.00 23.00 One-Way ANOVA: Post Hoc Multiple Comparisons X 8 1.00 1.00 1.00 1.00 1.00 2.00 2.00 2.00 2.00 2.00 3.00 3.00 3.00 3.00 3.00 9 Factor: Factor 10 ............ 11 OK Paste Reset Waller-Duncan Type I/Type ll Error Ratio: 100 Dunnett Control Category: Last Cancel Help Equal Variances Assumed LSD S-N-K Bonferroni ✓ Tukey Sidak Tukey's-b Scheffe Duncan R-E-G-WF Hochberg's GT2 R-E-G-WQ Gabriel 12 13 14 Test 15 2-sided <Control > Control 16 17 18 Equal Variances Not Assumed Tamhane's T2 Dunnett's T3 Games-Howell Dunnett's C 19 Significance level: 0.05 20 21 Continue Cancel Help 22 Data View Variable View IBM SPSS Statistics Processor is ready

OUTPUT:

ANOVA SatisfactionScore df Sum of Squares 36.400 47.200 83.600 F 4.627 Mean Square 18.200 3.933 Sig. .032 Between Groups Within Groups Total 2 12 14 Post Hoc Tests Multiple Comparisons Dependent Variable: SatisfactionScore 95% Confidence Interval (1) Factor 1.00 Tukey HSD Mean Difference (1- J) -3.40000 - 20000 3.40000 3.20000 20000 Sig. .046 .986 (J) Factor 2.00 3.00 1.00 3.00 1.00 Lower Bound -6.7464 -3.5464 Upper Bound -.0536 3.1 464 6.7464 6.5464 3.5464 2.00 Std. Error 1.25433 1.25433 1.25433 1.25433 1.25433 1.25433 1.25433 1.25433 .0536 .046 .061 .986 - 1464 3.00 -3.1464 2.00 -3.20000 .061 -6.5464 .1464 Scheffe 1.00 2.00 .057 -6.8965 .0965 -3.40000 - 20000 .987 -3.6965 3.2965 3.00 1.00 2.00 3.40000 .057 6.8965 1.25433 1.25433 -.0965 - 2965 3.00 3.20000 6.6965 .074 .987 3.00 1.00 .20000 1.25433 1.25433 -3.2965 -6.6965 3.6965 .2965 2.00 -3.20000 .074 *. The mean difference is significant at the 0.05 level.

• Now the critical value for f-distribution corresponding to 2,12 degree of freedom at 0.05 level of significance is 3.885 (By using F-distribution table).
• From the ouput we can see that our test statistics is 4.627.

(c). Formula of effect size is: SS_between / SS_total = 36.400/83.600 = 0.435. Therefore magnitude of effect size is large.

(d). Since our significance (p-value) for anova test is less than 0.05. So, we may reject our null hypothesis at 0.05 level of significance and conclude that "Atleast one of the temporal means is different on life satisfaction"

(e).By using SPSS output:

1. 2 vs 3 difference = 3.20000 ; significant : NO (because sig. (0.061) is greater than 0.05)
2. 1 vs 2 difference = -3.40000 ; significant : YES (because sig. (0.046) is less than 0.05)