It would help if you proofread your questions before you posted them.
I assume you are using "5" instead of "=".
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
Childhood participation in sports, cultural groups,
and youth groups appears to be related to improved
self-esteem for adolescents (McGee, Williams,
Howden-Chapman, Martin, & Kawachi, 2006). In
a representative study, a sample of n 5 100 adolescents
with a history of group participation is given
a standardized self-esteem questionnaire. For the
general population of adolescents, scores on this
questionnaire form a normal distribution with a mean
of m 5 50 and a standard deviation of s 5 15. The
sample of group-participation adolescents had an
average of M 5 53.8.
a. Does this sample provide enough evidence to conclude
that self-esteem scores for these adolescents
are significantly different from those of the general
population? Use a two-tailed test with a 5 .05.
b. Compute Cohen’s d to measure the size of the
difference.
c. Write a sentence describing the outcome of the
hypothesis test and the measure of effect size as it
would appear in a research report.
1 answer
1 answer