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 (M_D ± S_D),then what is the decision at a .05 level of significance? A) Satisfaction ratings do not significantly differ, t(14) = 0.90. B) Satisfaction ratings do not significantly differ, t(15) = 0.90. C) Satisfaction ratings significantly differ, t(14) = 3.60. D) Satisfaction ratings significantly differ, t(15) = 3.60.

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Quizplus · Accepted Answer

The t-value is calculated as (1.8 - 0) / (2.0 / sqrt(16)) = 3.60. With 15 degrees of freedom (n-1 for paired samples), this t-value is significant at the .05 level, indicating a significant difference in satisfaction ratings.

A Researcher Sampled 16 Couples and Measured the Mean Difference