11. A researcher sampled 14 couples and measured the mean difference in their marital satisfaction. Each...
11. A researcher sampled 14 couples and measured the mean difference in their marital satisfaction. Each couple was paired and the differences in their rating (on a 7-point scale) were taken. If the mean difference in satisfaction rating for this sample was 1.5±2.0 (MD±SD), then what is the decision at a 0.01 level of significance?
A) Satisfaction ratings do not significantly differ, t(14)=2.83
B) Satisfaction ratings do not significantly differ, t(12)=3.6
C) Satisfaction ratings do not significantly differ, t(13)=2.83
D) Satisfaction ratings significantly differ, t(14)=3.6
E) Satisfaction ratings significantly differ, t(13)=2.83
Homework Answers
Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (1.5 - 0)/(2/sqrt(14))
t = 2.83
P-value Approach
P-value = 0.0142
As P-value >= 0.01, fail to reject null hypothesis.
C) Satisfaction ratings do not significantly differ, t(13)=2.83
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