Certainly! To determine the standard deviation of the sampling distribution for the difference in sample means ([tex]\(\bar{x}_A - \bar{x}_C\)[/tex]), we can follow these steps:
1. Identify the given parameters:
- Mean and standard deviation for Alex (Group A):
[tex]\[\mu_A = 5.28\][/tex]
[tex]\[\sigma_A = 0.38\][/tex]
- Sample size for Alex:
[tex]\[n_A = 10\][/tex]
- Mean and standard deviation for Chris (Group C):
[tex]\[\mu_C = 5.45\][/tex]
[tex]\[\sigma_C = 0.2\][/tex]
- Sample size for Chris:
[tex]\[n_C = 15\][/tex]
2. Recall the formula for the standard deviation of the difference in sample means ([tex]\(\sigma_{\bar{x}_A - \bar{x}_C}\)[/tex]):
[tex]\[
\sigma_{\bar{x}_A - \bar{x}_C} = \sqrt{\left(\frac{\sigma_A^2}{n_A}\right) + \left(\frac{\sigma_C^2}{n_C}\right)}
\][/tex]
3. Substitute the given values into the formula:
- Variance for Alex:
[tex]\[
\sigma_A^2 = (0.38)^2 = 0.1444
\][/tex]
- Variance for Chris:
[tex]\[
\sigma_C^2 = (0.2)^2 = 0.04
\][/tex]
- Calculation for Alex’s contribution to the standard deviation:
[tex]\[
\frac{\sigma_A^2}{n_A} = \frac{0.1444}{10} = 0.01444
\][/tex]
- Calculation for Chris’s contribution to the standard deviation:
[tex]\[
\frac{\sigma_C^2}{n_C} = \frac{0.04}{15} = 0.0026667
\][/tex]
4. Sum the contributions:
[tex]\[
\left(\frac{\sigma_A^2}{n_A}\right) + \left(\frac{\sigma_C^2}{n_C}\right) = 0.01444 + 0.0026667 = 0.0171067
\][/tex]
5. Take the square root to find the standard deviation:
[tex]\[
\sigma_{\bar{x}_A - \bar{x}_C} = \sqrt{0.0171067} \approx 0.13
\][/tex]
So, the standard deviation of the sampling distribution for [tex]\(\bar{x}_A - \bar{x}_C\)[/tex] is approximately [tex]\(0.13\)[/tex].
Hence, the correct answer is [tex]\(0.13\)[/tex].
