Certainly! Let's dive into the solution for each part of the given question.
### Part (a): Calculate the sample statistics.
#### Female Students:
- Sample Mean ([tex]\(\mu_f\)[/tex]):
[tex]\[
\mu_f = 73.0
\][/tex]
- Sample Standard Deviation ([tex]\(s_f\)[/tex]):
[tex]\[
s_f = 8.17124413839354
\][/tex]
- Sample Size ([tex]\(n_f\)[/tex]):
[tex]\[
n_f = 14
\][/tex]
#### Male Students:
- Sample Mean ([tex]\(\mu_m\)[/tex]):
[tex]\[
\mu_m = 82.1
\][/tex]
- Sample Standard Deviation ([tex]\(s_m\)[/tex]):
[tex]\[
s_m = 9.085397319019375
\][/tex]
- Sample Size ([tex]\(n_m\)[/tex]):
[tex]\[
n_m = 10
\][/tex]
### Part (b): Assume that the population variances are equal and calculate the standard error of the difference of the means.
Given the assumption that the population variances are equal, we calculate the Pooled Variance ([tex]\(s_p^2\)[/tex]) as follows:
[tex]\[
s_p^2 = \frac{(n_f - 1) s_f^2 + (n_m - 1) s_m^2}{n_f + n_m - 2}
\][/tex]
Then, the Standard Error of the Difference of the Means using pooled variance is:
[tex]\[
\text{Standard Error}_{\text{equal variances}} = \sqrt{s_p^2 \left( \frac{1}{n_f} + \frac{1}{n_m} \right)}
\][/tex]
Thus,
[tex]\[
\text{Standard Error}_{\text{equal variances}} = 3.5429461656180345
\][/tex]
### Part (c): Assume that the population variances are not equal and calculate the standard error of the sampling distributions of the mean.
When population variances are not equal, the Standard Error is calculated as:
[tex]\[
\text{Standard Error}_{\text{unequal variances}} = \sqrt{\frac{s_f^2}{n_f} + \frac{s_m^2}{n_m}}
\][/tex]
Thus,
[tex]\[
\text{Standard Error}_{\text{unequal variances}} = 3.608832943442411
\][/tex]
### Part (d): Assuming population variances are not equal, calculate the 90% confidence interval for the difference of the means.
To calculate the 90% confidence interval for the difference in means, we need the critical t-value for a 90% confidence level, and the smallest sample size's degrees of freedom ([tex]\(df\)[/tex]):
[tex]\[
t_{\alpha/2, (n_{\text{smaller}} - 1)} \quad \text{where} \quad \alpha = 0.10 \quad \text{and} \quad n_{\text{smaller}} = 10
\][/tex]
With a 90% confidence level, the t-value is approximately [tex]\(t_{0.05, 9} = 1.833\)[/tex]. Using this, we calculate the Margin of Error (ME) as:
[tex]\[
\text{ME} = t_{\alpha/2} \times \text{Standard Error}_{\text{unequal variances}}
\][/tex]
[tex]\[
\text{ME} = 1.833 \times 3.608832943442411 = 6.612398340410758
\][/tex]
The confidence interval for the difference in means is then:
[tex]\[
CI = (\mu_f - \mu_m) \pm \text{ME}
\][/tex]
[tex]\[
CI = (73.0 - 82.1) \pm 6.612398340410758
\][/tex]
[tex]\[
CI = -9.1 \pm 6.612398340410758
\][/tex]
[tex]\[
CI = [-15.715398340410758, -2.4846016595892317]
\][/tex]
Thus, the 90% confidence interval for the difference in means between female and male students is between [tex]\(-15.715\)[/tex] and [tex]\(-2.485\)[/tex].
