A study9 compared personality characteristics between adult children of alcoholics and a control group matched on age and gender. For the 29 pairs of women, the authors reported a mean of 24.8 on the well-being measure for the children of alcoholics, and a mean of 29.0 for the control group. They reported t = 2.67 for the test comparing the means. Assuming that this is the result of a dependent-samples analysis, identify the df for the t test statistic, report the P-value, and interpret.

df.= 28

P - value = 0.0124 < 0.05

There is a difference between the mean of  the well-being measure for the children of alcoholics, and a mean of the control group.

Step-by-step explanation:

As the given test is dependent or paired test the degrees of freedom is n-1 = 20-1 = 28

The reported t = 2.67

The P value for ∝= 0.05 at  calculated t= 2.67 and 28 d.f is 0.0124 which is less than 0.05 .

The null hypothesis is rejected.

The null hypothesis is that there is no difference between the means of the two samples.

H0: u1 = u2

The alternate hypotheis is accepted that there is  difference between the means of the two samples.

Ha: u1≠ u2

There is a difference between the mean of  the well-being measure for the children of alcoholics, and a mean of the control group.

ou would have 12 of each

its 60

4 x 12 = 48

3 x 4 = 12

48 + 12 = 60

yeeeet