Question

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 0.05 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.

Answer 1

Answer:

Given,

sample n = 16

standard deviation s = 2

xbar = 1.8

Null hypothesis Ho : u1 = u2

Alternative hypothesis Ha : u1 != u2

significance level = 0.05

Degree of freedom = n - 1

= 16 - 1

= 15

Critical value corresponding to t(alpha/2 , df) is +/- 2.13

test statistic = (xbar - u)/(s/sqrt(n))

substitute values

= 1.8/(2/sqrt(16))

= 3.6

Corresponding p value = 0.0026

Here we observe that, test statistic > critical value, so we reject Ho

So there is sufficient evidence.