Question

Childhood participation in sports, cultural groups and youth groups appears to be related to improved self-esteem for adolescents (McGee et al., 2006). A sample of n =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 μ = 40 and a standard deviation of σ = 12. The sample of group participation adolescents had an average of M = 43.84.

Is there 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 α=.01.

(A) The null hypothesis in words is

Self-esteem scores of children who are engaged in group participation are not significantly greater than those of the general population.

Self-esteem scores of children who are engaged in group participation are significantly greater than those of the general population.

Self-esteem scores of children who are engaged in group participation are significantly different from those in the general population

Self-esteem scores of children who are engaged in group participation are not significantly different from those of the general population.

(B) The critical z values are

(D) Cohen's D?

(C) Your decision is

Group of answer choices

Reject the null hypothesis because your z statistic is not beyond the critical z values

Reject the null hypothesis because your z statistic is beyond the critical z values

Fail to reject the null hypothesis because your z statistic is beyond the critical z values

Fail to reject the null hypothesis because your z statistic is not beyond the critical z values

Answer 1

A) The null hypothesis in words is Self-esteem scores of children who are engaged in group participation are not significantly different from those of the general population.

B) At alpha = 0.01, the critical values are +/- z_0.005 = +/- 2.58

C) Cohen's D = (M - )/

= (43.84 - 40)/12

= 0.32

D) The test statistic z = (M - )/()

= (43.84 - 40)/(12/)

= 3.2

Reject the null hypothesis because the z statistic is beyond the critical z values.