Question

A variable of two populations has a mean of $40$and a standard deviation of $12$for one of the populations and a mean of $40$and a standard deviation of $6$for the other population. Moreover, the variable is normally distributed on each of the two populations.

a. For independent samples of sizes $9$and $4$, respectively, determine the mean and standard deviation of $\overline{x_1}-\overline{x_2}$.

b. Can you conclude that the variable $\overline{x_1}-\overline{x_2}$is normally distributed? Explain your answer.

c. Determine the percentage of all pairs of independent samples of sizes $9$and $4$, respectively, from the two populations with the property that the difference $\overline{x_1}-\overline{x_2}$between the sample means is between $-10$and $10$.

Short answer

Part a. For the variable $\overline{x_1}-\overline{x_2}$, the mean is $0$and the standard deviation is $5$.

Part b. It can be concluded that the variable role="math" localid="1652702776288" $\overline{x_1}-\overline{x_2}$is normally distributed.

Part c. The confidence interval will be $95.44\%$.

Step-by-step solution

Part (a) Step 1. Given Information

We are given data of two populations:

For the first population, the sample size is $n_1=9$, mean is ${\mu }_1=40$, and the standard deviation is ${\sigma }_1=12$.

For the second population, the sample size is $n_2=4$, mean is ${\mu }_2=40$, and the standard deviation is ${\sigma }_2=6$.

Part (a) Step 2. Find the mean and the standard deviation

The mean for the variable $\overline{x_1}-\overline{x_2}$is given as

$$\begin{gathered} {\mu }_{\overline{x_1}-\overline{x_2}}={\mu }_1-{\mu }_2 \\ {\mu }_{\overline{x_1}-\overline{x_2}}=40-40 \\ {\mu }_{\overline{x_1}-\overline{x_2}}=0 \end{gathered}$$

And the standard deviation is given as

$$\begin{gathered} {\sigma }_{\overline{x_1}-\overline{x_2}}=\sqrt{\frac{{{\sigma }_1}^2}{n_1}+\frac{{{\sigma }_2}^2}{n_2}} \\ {\sigma }_{\overline{x_1}-\overline{x_2}}=\sqrt{\frac{{12}^2}{9}+\frac{6^2}{4}} \\ {\sigma }_{\overline{x_1}-\overline{x_2}}=\sqrt{16+9} \\ {\sigma }_{\overline{x_1}-\overline{x_2}}=\sqrt{25} \\ {\sigma }_{\overline{x_1}-\overline{x_2}}=5 \end{gathered}$$

Part (b) Step 1. Whether the variable x1¯-x2¯ is normally distributed

As it is given that the variable is normally distributed on each of the two populations. So we can conclude using it that the variable $\overline{x_1}-\overline{x_2}$is normally distributed.

Part (c) Step 1. Find the confidence interval

We have that the variable $\overline{x_1}-\overline{x_2}$is normally distributed and ${\mu }_{\overline{x_1}-\overline{x_2}}=0,{\sigma }_{\overline{x_1}-\overline{x_2}}=5$.

Using the $68.26-95.44-99.74$rule, we determine the percentage of all pairs of independent samples of sizes $9$and $4$respectively, from the two populations with the property that the difference role="math" localid="1652703305588" $\overline{x_1}-\overline{x_2}$between the sample means is between $-10$and $10$as:

$$\begin{gathered} {\mu }_{\overline{x_1}-\overline{x_2}}-2{\sigma }_{\overline{x_1}-\overline{x_2}}=0-2\times 5 \\ =-10 \\ {\mu }_{\overline{x_1}-\overline{x_2}}+2{\sigma }_{\overline{x_1}-\overline{x_2}}=0+2\times 5 \\ =10 \end{gathered}$$

Therefore, we can conclude that $95.44\%$of all pairs of independent samples of sizes $9$and $4$respectively, from the two populations with the property that the difference $\overline{x_1}-\overline{x_2}$between the sample means is between $-10$and $10$.