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A researcher sampled 16 couples and measured the mean difference in their marital satisfaction. each couple was paired and the differences in their ratings (on a 7-point scale) were taken. if the mean difference in satisfaction ratings for this sample was 1.8±2.0 (md /- sd), then what is the decision at a .05 level of significance?

User Ganpaan
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Final answer:

A hypothesis test is conducted to determine if the mean difference in marital satisfaction between husbands and wives is negative.

Step-by-step explanation:

In this question, a hypothesis test is being conducted to determine if the mean difference in marital satisfaction between husbands and wives is negative. The researcher sampled 16 couples and measured the mean difference in their satisfaction ratings. The mean difference in satisfaction ratings for this sample was found to be 1.8 with a standard deviation of 2.0.

To make a decision at a 0.05 level of significance, we need to compare the calculated test statistic, which is equal to the mean difference divided by the standard deviation, to the critical value from the t-distribution with n-1 degrees of freedom.

If the calculated test statistic falls in the critical region (the area beyond the critical value), we reject the null hypothesis and conclude that there is evidence to support that the mean difference in marital satisfaction is negative. Otherwise, if the calculated test statistic falls in the non-critical region, we fail to reject the null hypothesis and do not have enough evidence to conclude that the mean difference is negative.

User Npinti
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