In a respresentative study,a sample of n = 100 adolescents with a history of group participation is given a standardized self-esteem questionnaire. For the general po;ulation of adolescents, scores on this questionnaire form a normal distribution with a mean of u = 40 and a standard deviation of 12. The sample of group-participation adolescents had an average of M = 43.84.

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 standard error = .01
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.

I thought college students knew the difference between the name of their school and the name of their school subject. (Sigh!)

For part a: Use the appropriate hypothesis test for your data.

For part b: Cohen's d is the difference between two means divided by a standard deviation.

Calculate Cohen's d (d) and the effect-size correlation (r) using the following formulas:

d = (M1 - M2) / s

r = d / √(d^2 + 4)

With your data:

d = (43.84 - 40) / 12 = 0.32

r = 0.32 / √(0.32^2 + 4) = 0.078

Check these formulas and calculations.

I'll let you take it from here.

A. To determine whether the self-esteem scores of the group-participation adolescents are significantly different from those of the general population, we can perform a hypothesis test using a two-tailed test with a significance level of α = 0.01.

First, we need to set up the null and alternative hypotheses:
- Null hypothesis (H0): The self-esteem scores of the group-participation adolescents are not significantly different from those of the general population (μ = 40).
- Alternative hypothesis (Ha): The self-esteem scores of the group-participation adolescents are significantly different from those of the general population (μ ≠ 40).

Next, we can compute the standard error of the mean (SE) using the formula SE = σ/√n, where σ is the population standard deviation (12) and n is the sample size (100). Thus, SE = 12/√100 = 1.2.

We can then compute the test statistic (z-score) using the formula z = (X - μ)/SE, where X is the sample mean (43.84) and μ is the population mean (40). Hence, z = (43.84 - 40)/1.2 = 3.03.

We can compare this z-score to the critical value(s) for a two-tailed test with a significance level of α = 0.01. The critical value(s) can be found using a standard normal distribution table or a statistical software. For a two-tailed test with α = 0.01, the critical values are approximately ±2.576.

Since the calculated z-score (3.03) is greater than the critical value (2.576), we can reject the null hypothesis. This means that there is enough evidence to conclude that the self-esteem scores for the group-participation adolescents are significantly different from those of the general population.

B. Cohen's d is a measure of effect size that quantifies the difference between two means by standardizing it with the population standard deviation. Cohen's d is calculated using the formula d = (X1 - X2)/σ, where X1 is the mean of the group-participation adolescents (43.84), X2 is the mean of the general population (40), and σ is the population standard deviation (12). Hence, d = (43.84 - 40)/12 = 0.3033.

C. In a research report, the outcome of the hypothesis test and the measure of effect size could be described as follows: "A representative study of n = 100 adolescents with a history of group participation provided sufficient evidence to conclude that self-esteem scores for these adolescents are significantly different from those of the general population (t(99) = 3.03, p < 0.01). Cohen's d was calculated to measure the size of the difference, revealing a small effect size (d = 0.3033)."