Two independent random samples have been selected, 100 observations from population 1 and 100 from population 2 . Sample means $\bar{x}_{1}=74$ and $\bar{x}_{2}=52$ were obtained. From previous experience with these populations, it is known that the variances are $\sigma_{1}^{2}=100$ and $\sigma_{2}^{2}=49$. Complete parts a through $g$ below. Click the icon to view the table of normal curve areas. D. Sketch the approximate sampling distribution $\left(x_{1}-x_{2}\right)$, assuming that $\left(\mu_{1}-\mu_{2}\right)=5$. A. c. B. D.
Solution
Step 1 ：The problem is asking to sketch the approximate sampling distribution of the difference in sample means, assuming that the difference in population means is 5.
Step 2 ：The mean of the sampling distribution of the difference in sample means is equal to the difference in population means, which is given as 5.
Step 3 ：The standard deviation of the sampling distribution of the difference in sample means can be calculated using the formula \(\sqrt{\frac{\sigma_{1}^{2}}{n_{1}} + \frac{\sigma_{2}^{2}}{n_{2}}}\), where \(\sigma_{1}^{2}\) and \(\sigma_{2}^{2}\) are the variances of the two populations and \(n_{1}\) and \(n_{2}\) are the sizes of the two samples.
Step 4 ：In this case, \(\sigma_{1}^{2}=100\), \(\sigma_{2}^{2}=49\), \(n_{1}=100\) and \(n_{2}=100\).
Step 5 ：Calculating the standard deviation gives approximately 1.22.
Step 6 ：The approximate sampling distribution of the difference in sample means is a normal distribution with a mean of 5 and a standard deviation of approximately 1.22.
Step 7 ：\(\boxed{\text{The approximate sampling distribution of the difference in sample means is a normal distribution with a mean of 5 and a standard deviation of approximately 1.22.}}\)
