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CSS Statistics QUESTION #1978
Question 1
Given \(n_1 = 2\) samples with replacement from population \(\{4, 6, 8\}\) and \(n_2 = 2\) from \(\{1, 2, 3\}\), the variance of the sampling distribution of \(\bar{X}_1 - \bar{X}_2\) equals:
  • \(\dfrac{\sigma_1^2}{n_1} - \dfrac{\sigma_2^2}{n_2}\)
  • \(\dfrac{\sigma_1^2}{n_1} + \dfrac{\sigma_2^2}{n_2}\)✔️
  • \(\sigma_1^2 + \sigma_2^2\)
  • \(\dfrac{\sigma_1^2 + \sigma_2^2}{n_1 + n_2}\)
Correct Answer Explanation
For independent samples, \(\sigma^2_{\bar{X}_1 - \bar{X}_2} = \dfrac{\sigma_1^2}{n_1} + \dfrac{\sigma_2^2}{n_2}\). For population \(\{4,6,8\}\): \(\mu_1=6\), \(\sigma_1^2 = \frac{8}{3}\). For \(\{1,2,3\}\): \(\mu_2=2\), \(\sigma_2^2=\frac{2}{3}\). So variance \(= \frac{8/3}{2}+\frac{2/3}{2} = \frac{4}{3}+\frac{1}{3}=\frac{5}{3}\).