Question

SPSS Homework 1 Instructions: Review and One-Way ANOVAs (40 points)

Please note that for all problems in this course, the standard cut-off (alpha) for a test of significance will be .05, and you always report the exact power unless SPSS output states p=.000 (you’d report p<.001). Also, remember that we divide the p value in half when reporting one-tailed tests with 1 – 2 groups.

Problem Set 2: Pearson’s correlation (7 pts) Research Scenario: Is there a positive relationship between grit and dieting success? A researcher examined this issue by having people complete a grit inventory using a Likert-based scale (range 1 – 7), where higher numbers indicate more “grit”. Dieting success was measured using a likert-based inventory as well, with higher numbers indicating more success (range 1-7). Enter the data shown here into SPSS to run an analysis to test whether increased grit is associated with higher dieting success. Grit Success 3.5 2.3 2.0 1.1 1.7 0.3 1.8 0.1 2.7 0 3.5 0.7 3.9 3.7 4.1 5.5 3.5 3.7 4.7 6.7 5.1 5.9 3.2 5.3 5.6 4.4 4.1 2.8

1.     Paste relevant SPSS output. (2 pts)

2.     Create an appropriate graph to display this data. (2 pts)

3.     Write an APA-style Results section based on your analysis. All homework “Results sections” should follow the examples provided in the presentations and textbooks. They should include the statistical statement within a complete sentence that mentions the type of test conducted, whether the test was significant, and if relevant, coefficient of determination and its interpretation. Don’t forget to include a decision about the null hypothesis. (3 pts)

Problem Set 3: One-Way Randomized ANOVA (7 pts) Research Scenario: A developmental psychologist is looking into the effects of play group size on empathy in four year old children. She divides the children into three separate groups for the semester during structured free time: play alone (n = 8), dyads (n = 8), or groups of 4 (n = 8). At the end of the semester, empathy is measured using the Young Children’s Empathy Measure during the parent-teacher conference (Poresky, 1990), which uses an accuracy scale of 0 – 4, with higher scores indicating more cognitive and affective empathy. The scores for each group are shown in the table below. Input the data into SPSS (remember you will enter the data only using 2 columns – one for group (make sure you label the values) and one for empathy. Conduct a one-way ANOVA to determine the effect of these different play group assignments on empathy. ALONE 2.7 2.3 3.1 2.5 2.6 2.0 2.9 2.4 DYAD 2.3 2.5 3.3 2.4 3.0 2.7 2.1 2.4 FOUR 2.4 2.6 2.6 3.1 3.2 3.0 2.7 2.6

4.     Paste relevant SPSS output. (2 pt)

5.    Create an appropriate graph to display this data. (2 pts)

6.    Write an APA-style Results section based on your analysis. All homework “Results sections” should follow the examples provided in the presentations and textbooks. They should include the statistical statement within a complete sentence that mentions the type of test conducted, whether the test was significant, and if relevant, effect size and/or post hoc analyses. Don’t forget to include a decision about the null hypothesis. (3 pts)

Answer 1

Correlations: Grit, Success Pearson correlation of Grit and Success = 0.777 P-Value 0.001

x2 Y2 Y X X*Y 2.3 8.05 12.25 5.29 3.5 2.0 2.2 4 1.1 1.21 1.7 0.3 0.51 2.89 0.09 0.18 1.8 0.1 3.24 0.01 2.7 7.29 0.0 0 0 0.7 2.45 0.49 3.5 12.25 3.9 15.21 3.7 14.43 13.69 4.1 22.55 5.5 16.81 30.25 3.5 3.7 12.95 12.25 13.69 4.7 6.7 22.09 31.49 44.89 5.1 5.9 30.09 26.01 34.81 5.3 3.2 16.96 10.24 28.09 5.6 4.4 31.36 19.36 24.64 7.84 4.1 2.8 11.48 16.81 199.71 Sum 49.4 42.5 177.98 192.7 LC LO LC LC t st LO

The correlation coefficient r is computed using the following expression: SSXY r = where x(£x) (&») 77 17 1 sSxY XY X3 n =1 i=1 77 1 n i-1 ( 2 n n i=1 In this case, based on the data provided, we get that (49.4 x 42.5) 14 SSXY 177.98 28.016 (49.4)2 14 SSXX 18.389 192.7 1 (42.5) SSyy 199.71 70.692

Therefore, based on this information, the sample correlation coefficient is computed as follows SSXY 28.016 0.777 V18.389 x 70.692

2 12 -2 14 The sample size is n 14, so then the number of degrees of freedom is df The corresponding critical correlation value re for a significance level of a 0.05, for a two-tailed test is: Te = 0.532 Observe that in this case, the null hypothesis is rejected if r>Te = 0.532

We reject the null hypothesis that p 0 p-value is less than 0.05 as given by output.

2)

One-way ANOVA: ALONE, DYAD, FOUR Source DF SS MS Factor 2 0.216 0.108 0.91 0.420 Error 21 2.503 0.119 Fotal 23 2.718 R-Sq (adj) S 0.3452 R-Sg 7.94% 0.00% - Individual 95% CIs For Mean Based on Pooled StDev Level Mean StDev --+- ALONE 8 2.5625 0.3462 (- 0.3944 2.5875 DYAD FOUR 8 2.7750 0.2866 -+- --+---- 2.40 2.60 2,80 3.00 0 0 0X

Group 2 Group 1 Group 3 2.7 2.3 2.4 2.3 2.5 2.6 3.3 3.1 2.6 2.5 2.4 3.1 2.6 3.2 2 2.7 2.9 2.1 2.7 2.4 2.4 2.6 Sum 20.5 20.7 22.2 2.775 Average = 2.563 2.588 Σ Χ 53.37 54.65 62.18 0.346 0.394 St. Dev. 0.287 SS = 0.83875000000002 1.08875 0.575 8 8 n =

24. Therefore, the total degrees of freedom are: The total sample size is N dftotal 24 1 = 23 Also, the between-groups degrees of freedom are dfbetween = 3 - 1 2, and the within-groups degrees of freedom are: 21 dfwithin = dftotal - dfbetween = 23 2 First, we need to compute the total sum of values and the grand mean. The following is obtained Xij = 20.5 +20.7 +22.2 = 63.4 ij Also, the sum of squared values is 53.37 +54.65 62.18 =170.2 X i.j Based on the above calculations, the total sum of squares is computed as follows 2 63.42 S Stotal = 170.2 = 2.718 24 Xj

The within sum of squares is computed as shown in the calculation below: Σs SSapithin Swithingroups = 0.83875000000002 1.08875 +0.575 = 2.503 The between sum of squares is computed directly as shown in the calculation below

S Sbetween dfbetween 0.216 MSbetween 0.108 2 S Supithin dfwithin 2.503 = 0.119 with 21 Finally, with having already calculated the mean sum of squares, the F-statistic is computed as follows: M Sbetween F MSyithin 0.108 _ 0.119 0.906 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: H1H2 P3 Ha: Not all means are equal The above hypotheses will be tested using an F-ratio for a One-Way ANOVA. (2) Rejection Region Based on the information provided, the significance level is a 0.05, and the degrees of freedom are dfi 2 and df2 = 2, therefore, the rejection region for this F-test is R = {F:F > Fe = 3.467}

(3) Test Statistics M Sbetween F 0.108 0.906 0.119 within (4) Decision about the null hypothesis Since it is observed that F 0.906 < Fe 3.467, it is then concluded that the null hypothesis is not rejected Using the P-value approach: The p-value is p 0.4195, and since p 0.4195 > 0.05, it is concluded that the null hypothesis is not rejected (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that not all 3 population means are equal, at the a 0.05 significance level.