Question

Gravetter/Wallnau/Forzano, Essentials - Chapter 8 - End-of-chapter question 12 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 - 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 us - 50 and a standard deviation of 0 – 15. The sample of group participation adolescents had an average of M - 53.8. 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-talled test with a - .05. The null hypothesis is He: even group participation The standard error is (to two decimal places). Using the Distributions tool, the critical value of - Standard Normal Durban

Answer 1

After conducting a two-tailed z-test, it's concluded that there is sufficient evidence to reject the null hypothesis, indicating a significant difference in average self-esteem scores between group-participating adolescents and the general adolescent population.

Step-by-step explanation:

The question asks whether the mean self-esteem scores for adolescents who have participated in group activities such as sports are statistically different from the general population. Given the information, we will conduct a two-tailed hypothesis test using a z-test for a sample mean. The null hypothesis (μ) is that there is no difference in self-esteem scores between the sample group and the general population of adolescents.

Step-by-Step Explanation

1. First, identify the parameters given; the population mean (μ) is 50, and the population standard deviation (σ) is 15. The sample mean (M) is 53.8 with a sample size (n) of 100.
2. Second, calculate the standard error (SE) of the sample mean. SE = σ/√n = 15/10 = 1.5.
3. Next, calculate the z-score for the sample mean, which is (M - μ)/SE = (53.8 - 50)/1.5 = 2.53.
4. Compare the calculated z-score to the critical values for a two-tailed test with α = .05. The critical z-values are approximately ±1.96.
5. Since the calculated z-score (2.53) is greater than the critical z-value (1.96), we reject the null hypothesis.

Therefore, there is sufficient evidence to conclude that the average self-esteem score for adolescents with a history of group participation is significantly different from the general population of adolescents.