A researcher tests how well participants know their spouse while under hypnosis,and again without hypnosis.It was hypothesized that participant responses would be more accurate (greater number of correct responses)when under hypnosis.The mean number of correct responses was M = 5.30 under hypnosis and M = 3.40 without hypnosis in a sample of 30 participants.If the estimated standard error for difference scores was 2.52,then what is the decision at a .05 level of significance (two-tailed test)?
A) Retain the null hypothesis; responses were more accurate under hypnosis.
B) Retain the null hypothesis; responses were not more accurate under hypnosis.
C) Reject the null hypothesis; responses were more accurate under hypnosis.
D) Reject the null hypothesis; responses were not more accurate under hypnosis.

Question

Quizplus · Accepted Answer

The mean difference is 5.30 - 3.40 = 1.90. The t-value is 1.90 / 2.52 = 0.754. The critical t-value for a two-tailed test with 29 degrees of freedom and a .05 level of significance is 2.045. Since the t-value is less than the critical t-value, we fail to reject the null hypothesis.

A Researcher Tests How Well Participants Know Their Spouse While