91Ó°ÊÓ

We are given data of two populations:

For the first population, the sample size is $n_1=3$, mean is ${\mathrm{μ}}_1=7.9$, and the standard deviation is ${\mathrm{σ}}_1=5.4$.

For the second population, the sample size is$n_2=6$, mean is${\mathrm{μ}}_2=7.1$, and the standard deviation is ${\mathrm{σ}}_2=4.6$.

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

$$\begin{gathered} {\mathrm{μ}}_{\stackrel{¯}{x_1}-\stackrel{¯}{x_2}}={\mathrm{μ}}_1-{\mathrm{μ}}_2 \\ {\mathrm{μ}}_{\stackrel{¯}{x_1}-\stackrel{¯}{x_2}}=7.9-7.1 \\ {\mathrm{μ}}_{\stackrel{¯}{x_1}-\stackrel{¯}{x_2}}=0.8 \end{gathered}$$

And the standard deviation is given as

$$\begin{gathered} {\mathrm{σ}}_{\stackrel{¯}{x_1}-\stackrel{¯}{x_2}}=\sqrt{\frac{{{\mathrm{σ}}_1}^2}{n_1}+\frac{{{\mathrm{σ}}_2}^2}{n_2}} \\ {\mathrm{σ}}_{\stackrel{¯}{x_1}-\stackrel{¯}{x_2}}=\sqrt{\frac{5.4^2}{3}+\frac{4.6^2}{6}} \\ {\mathrm{σ}}_{\stackrel{¯}{x_1}-\stackrel{¯}{x_2}}=\sqrt{9.72+3.53} \\ {\mathrm{σ}}_{\stackrel{¯}{x_1}-\stackrel{¯}{x_2}}≈3.64 \end{gathered}$$

The formulas in part (a) can be used regardless of the distributions of the variables of the two populations.

So the variables under consideration may or may not be normally distributed on each of the two populations.

If the variable is normally distributed on each of the two populations then only it can be concluded that the variable $\stackrel{¯}{x_1}-\stackrel{¯}{x_2}$is normally distributed.

Also, the sample size is small so the distribution type matters.

Thus it cannot be concluded for sure that the variable $\stackrel{¯}{x_1}-\stackrel{¯}{x_2}$is normally distributed.