18. Numerous studies have found that males report higher self-esteem than females, especially for adolescents (Kling, Hyde, Showers, & Buswell, 1999). Typical results show a mean self-esteem score of M = 39.0 with SS = 60.2 for a sample of n =10 male adolescents and a mean of M =35.4 with SS= 69.4 for a sample of n =10 female adolescents.

a. Do the results indicate that self-esteem is significantly higher for males? Use a one-tailed test with = .01.
b. Use the data to make a 95% confidence interval estimate of the mean difference in self-esteem
between male and female adolescents.

a. To determine if the results indicate that self-esteem is significantly higher for males, we can perform a hypothesis test using a one-tailed test with a significance level of α = 0.01.

Step 1: State the null hypothesis (H0) and alternative hypothesis (Ha):
H0: There is no significant difference in self-esteem between males and females (μmale = μfemale)
Ha: Self-esteem is significantly higher for males (μmale > μfemale)

Step 2: Set the significance level (α) and locate the critical value(s):
Since we are performing a one-tailed test with α = 0.01, we need to find the critical value from the t-table or t-distribution with degrees of freedom (df) = n1 + n2 - 2. Since both groups have a sample size of 10, df = 10 + 10 - 2 = 18. The critical value for a one-tailed test with α = 0.01 and df = 18 is approximately 2.552.

Step 3: Calculate the test statistic:
First, calculate the pooled variance (Spooled) using the formula:
Spooled = [(n1 - 1)s1^2 + (n2 - 1)s2^2] / (n1 + n2 - 2)
where n1 = sample size of males, s1^2 = variance of male self-esteem, n2 = sample size of females, and s2^2 = variance of female self-esteem.
Plugging in the values, we get:
Spooled = [(10 - 1)(60.2) + (10 - 1)(69.4)] / (10 + 10 - 2) = 64.8

Next, calculate the test statistic (t):
t = (M1 - M2) / √((Spooled/n1) + (Spooled/n2))
where M1 = mean of male self-esteem, M2 = mean of female self-esteem, n1 = sample size of males, and n2 = sample size of females.
Plugging in the values, we get:
t = (39 - 35.4) / √((64.8/10) + (64.8/10))
= 3.6 / √(6.48 + 6.48)
= 3.6 / √12.96
≈ 3.6 / 3.603
≈ 1

Step 4: Make a decision:
Since the calculated test statistic (t = 1) is less than the critical value (2.552), we do not have enough evidence to reject the null hypothesis. Thus, the results do not indicate that self-esteem is significantly higher for males at the α = 0.01 level.

b. To make a 95% confidence interval estimate of the mean difference in self-esteem between male and female adolescents, you can use the formula:

CI = (M1 - M2) ± (t * SE)
where M1 - M2 = difference in means (39 - 35.4), t = critical value for a 95% confidence interval (from t-table or t-distribution), and SE = standard error of the difference in means.

First, calculate the standard error (SE) using the formula:
SE = √((s1^2 / n1) + (s2^2 / n2))
where s1^2 = variance of male self-esteem, s2^2 = variance of female self-esteem, n1 = sample size of males, and n2 = sample size of females.
Plugging in the values, we get:
SE = √((60.2 / 10) + (69.4 / 10))
= √(6.02 + 6.94)
= √13.96
≈ 3.734

Next, find the critical value for a 95% confidence interval with a two-tailed test. Since we have a one-tailed test, we need to divide α by 2, which gives us α = 0.05/2 = 0.025. Looking up the critical value for α = 0.025 and df = 18 in the t-table or t-distribution, we find it to be approximately 2.101.

Finally, calculate the confidence interval:
CI = (39 - 35.4) ± (2.101 * 3.734)
= 3.6 ± (2.101 * 3.734)
≈ 3.6 ± 7.83
≈ (-4.23, 11.43)

Thus, we can estimate with 95% confidence that the mean difference in self-esteem between male and female adolescents is between -4.23 and 11.43.