I am not sure what i am missing here.

Use your SPSS output to help you to make a decision about the null hypotheses, as it actually tells whether or not this value from the analysis is significant. Be sure to provide the APA formatted results for with your response.

Identify the IV and DV in the study.

Independent variable (IV) - clients who are exposed to media during therapy

Dependent variable (DV) - depression score / level or depression after therapy

1. State the null hypothesis and the directional (one-tailed) alternative hypothesis.

Null hypothesis: no significant difference in depression scores between group 1 who were told not to watch or read news for two weeks and group 2 who continued watch or read news.

Alternative hypothesis: a significant difference in depression scores between group 1 who were told not to watch or read news for two weeks and group 2 who continued watch or read news.

1. Can you reject the null hypothesis at = .05? Explain why or why not.

Based on the t-value of -2.58 and a critical t-value of +/- 2.201 - the null hypothesis can be rejected.

(DataSet1) Group Statistics Sod. Error Groups McEn Scores No NEWS 7 34.4571 4.31072 Treatment Only 54.1429 20.45069 7 71190 Independent Samples Test Levene's Test for Equality of 1-best for Equality of Means 95% Confidence Interval of 5y. 12- the Difference F 3g. Oneranch Citternca Upper Scores Equal variances .212 -2.580 12 024 -24.25571 11.35123 -54.01791 -4.55352 Equal variance: no. -2.580 1 1 038 .024 -24.26571 11.35123 -54.03217 -4.53926 Independent Samples Effect Sizes Standardizera Point 95% Confidence Interval ESTIMATE Lower Upper Scores Cohen's d 21.21620 1.373 -2.534 -.175 Hedges' correction 22.05906 -1.291 -2.375 -.164 Glass's delia 20.45669 -1. 432 -2.709 -.083 a, The denominator used in est mating the effect woes, Cohen's d uses the pooled standard deviation. Hodges' correction uses the pooled standard deviadon, plus a correction factor. Glas53 delia woes the sample standard deviation of the comrol group