4. The president of an all-female school stated in an interview that she was sure that
the students at her school studied more on average, than the students at a neighboring all-male school. The president of the all-male school responded that he thought the mean study time for each student body was undoubtedly about the same and suggested that a study be undertaken to clear up the controversy. Accordingly independent samples were taken at the two schools with the following results.
SCHOOL SAMPLE SIZE MEAN STUDY TIME ST DEV
ALL FEM 65 18.56 4.35
ALL MALE 75 17.95 4.87
Determine at the 2% level of significance if there is significant difference between the mean studying times of the students in the two schools based on these samples.
Stat – test – 2- SampTtest - stats: ( x1, s1, n, x2, s2, n2) select pooled, select hypothesis. In our case we will input: 18.56, 4.65, 65, 17.95, 4.87, 75. Now select the hypothesis. The claim is that the mean study time is the same. Our hypothesis has not direction so the null will be that there is no difference between samples. It is pooled because we have the standard deviation, then select calculate.
t= .755
p=.452
df = 138
In conclusion our alpha is .02 and our p - value is .452 and thus p > alpha so we fail to reject the null meaning that the mean study times are similar.
