If two random samples of sizes 30 and 32 are selected independently from two populations with means 121 and 109,then the mean of the sampling distribution of the sample mean difference, $\bar { X } _ { 1 } - \bar { X } _ { 2 }$ ,equals 12.

Question

Quizplus · Accepted Answer

The mean of the sampling distribution of the sample mean difference is given by:

$$\mu_{\bar{X}_1 - \bar{X}_2} = \mu_1 - \mu_2$$

where $\mu_1$ and $\mu_2$ are the population means and assuming that the two samples are independent and the sample sizes are sufficiently large, the central limit theorem states that:

$$\bar{X}_1 \sim N(\mu_1, \frac{\sigma_1^2}{n_1})$$ 
and
$$\bar{X}_2 \sim N(\mu_2, \frac{\sigma_2^2}{n_2})$$

Then, the difference of two normally distributed variables is also normally distributed, that is:

$$\bar{X}_1 - \bar{X}_2 \sim N(\mu_{\bar{X}_1 -\bar{X}_2}, \frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2})$$

Substituting the given values, we have:

$$\mu_{\bar{X}_1 - \bar{X}_2} = \mu_1 - \mu_2 = 121 - 109 = 12$$

Therefore, the mean of the sampling distribution of $\bar{X}_1 - \bar{X}_2$ is 12, which makes the statement true.

If Two Random Samples of Sizes 30 and 32 Are Xˉ1−Xˉ2\bar { X } _ { 1 } - \bar { X } _ { 2 }Xˉ1​−Xˉ2​

If Two Random Samples of Sizes 30 and 32 Are $\bar { X } _ { 1 } - \bar { X } _ { 2 }$