Question

PSY 2030 One Sample t-Test SPSS Assignment 5

For Question 1, include all eight Steps of Hypothesis Testing

Assume that the mean number of violent acts in television shows is 10.2 per episode (this is historical average and may be treated as a population parameter). A random sample of shows from a new network has the following number of acts violence. The question motivating this research is Does the number of acts of violence from these new shows differ on average from the overall mean of 10.2?

13, 11, 14,10, 9, 13,14, 11,13,17,16, 12

Keep alpha=.05, have SPSS do the one -sample t test for you. The output of this command will be Step 5 and 6. Use the p-value (sig) to decide whether to reject or not reject the null hypothesis . For a nondirectional (two-tailed) test, which is the kind we're doing here, the rules is simple: If p < a, we reject H0. If p>a, we fail to reject H0.

Annotatetobs and the p-value on the printout. Turn your written steps and the annotated SPSS analysis.

Answer 1

Null hypothesis: Ho:u=10.2 That is, the mean number of violent acts in television shows is 10.2 per episode. Alternative hypothesis: H2:4 10.2 That is the mean number of violent acts in television differs on average from the overall mean of 10.2. Rejection rule: If p-value sal=0.05), then reject the null hypothesis H. SPSS output:

One-Sample Statistics Mean 12.75 Std. Error Mean .676 Std. Deviation 2.340 12 Sample One-Sample Test Test Value = 10.2 Mean Difference 2.550 df 11 Sample 95% Confidence Interval of the Difference Lower Upper 1. 06 4 .04 Sig. (2-tailed) | .003 3.774 The value of the test statistic is t = 3.774. p-value = 0.003. Conclusion: Use a significance level, a = 0.05. Here, p-value is 0.003, which is less than the value of a = 0.05. That is, p-value (=0.003) sa(=0.05). Therefore, by the rejection rule, reject the null hypothesis H.

Interpretation: There is sufficient evidence at level of significance a = 0.05. That is, the mean number of violent acts in television differs on average from the overall mean of 10.2.