91Ó°ÊÓ

Given in the question that

Mean $=40$

Standard deviation role="math" localid="1651243475125" $12$in sample size $9$

Standard deviation $6$in sample size$4$

Here we need to find the mean for ${\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2$

So,

$$\begin{gathered} {\mathrm{μ}}_{{\stackrel{¯}{x}}_1−{\stackrel{¯}{x}}_2}={\mathrm{μ}}_1−{\mathrm{μ}}_2 \\ =40−40 \\ =0 \end{gathered}$$

Now the standard deviation of ${\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2$is

$$\begin{gathered} {\mathrm{σ}}_{x_1−{\stackrel{¯}{x}}_2}=\sqrt{\frac{{\mathrm{σ}}_1^2}{n_1}+\frac{{\mathrm{σ}}_2^2}{n_2}} \\ =\sqrt{\frac{{12}^2}{9}+\frac{6^2}{4}} \\ =\sqrt{\frac{144}{9}+\frac{36}{4}} \\ =\sqrt{16+9} \\ =5 \end{gathered}$$

Given in the question that

Mean $=40$

Standard deviation $12$in sample size $9$

Standard deviation $6$in sample size $4$

Here we need to conclude that the ${\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2$is normally distributed

Regardless of the distribution of variables on an individual population, the distribution of mean and standard deviation of ${\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2$remains the same.

The standard deviation formula for ${\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2$is also based on the assumption that the samples are independent. Because each individual population is normally distributed, ${\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2$is also normally distributed, and because of the central limit theorem, this holds true for samples of huge sizes.

Given in the question that

Mean $=40$

Standard deviation $12$in sample size $9$

Standard deviation $6$in sample size $4$

Now we need to find the pooled standard deviation

$$\begin{gathered} s_p=\sqrt{\frac{\left(n_1−1\right)s_1^2+\left(n_2−1\right)s_2^2}{n_1+n_2−2}} \\ =\sqrt{\frac{(9−1)(12{)}^2+(4−1\left)\right(6{)}^2}{9+4−2}} \\ =\sqrt{\frac{888(122.9{)}^2+1118(108.1{)}^2}{2006}} \\ =10.7 \end{gathered}$$

Here we need to find the confidence interval

role="math" localid="1651246914263" $$\begin{gathered} \text{Confidence interval}=\left({\stackrel{¯}{x}}_1−{\stackrel{¯}{x}}_2\right)Â\pm t_{\mathrm{Î\pm }/2}×s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}} \\ −10\text{to}10=(40−40)Â\pm t_{\mathrm{Î\pm }/2}×10.7\sqrt{\frac{1}{9}+\frac{1}{4}} \\ Â\pm 10=Â\pm t_{\mathrm{Î\pm }/2}×6.43 \\ {t}_{\mathrm{Î\pm }/2}=1.555 \end{gathered}$$

Critical value role="math" localid="1651246880570" $t_{\mathrm{Î\pm }/2}=1.555$

$$\begin{gathered} t_{\mathrm{Î\pm }/2}=t_{0.05} \\ \mathrm{Î\pm }=0.10 \\ 1−\mathrm{Î\pm }=0.90 \end{gathered}$$