Question: Independent random samples from approximately normal populations produced the results shown below.

Sample 1	Sample 2
$$\begin{gathered} \mathbf{\text{52 33 42 44}} \\ \mathbf{\text{41 50 44 51}} \\ \mathbf{\text{45 38 37 40}} \\ \mathbf{\text{44 50 43}} \end{gathered}$$	$$\begin{gathered} \mathbf{\text{52 43 47 56}} \\ \mathbf{\text{62 53 61 50}} \\ \mathbf{\text{56 52 53 60}} \\ \mathbf{\text{50 48 60 55}} \end{gathered}$$

a. Do the data provide sufficient evidence to conclude that $\left({\mu }_1-{\mu }_2\right)\mathbf{>}\mathbf{}\mathbf{10}$? Test using$\mathit{\alpha }\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{1}$.

b. Construct a confidence interval for $\mathbf{(}{\mathbf{\mu }}_{\mathbf{1}}\mathbf{-}{\mathbf{\mu }}_{\mathbf{2}}\mathbf{)}$. Interpret your result.

For Sample 1

$$\begin{gathered} {\overline{x}}_1=43.6 \\ {n}_1=15 \\ {\sigma }_1=5.47 \end{gathered}$$

For Sample 2

$$\begin{gathered} {\overline{x}}_2=53.625 \\ {n}_2=16 \\ {\sigma }_2=5.41 \end{gathered}$$

The null hypothesis is $H_0:({\mathrm{\mu }}_1-{\mathrm{\mu }}_2)=10$, the alternate hypothesis is $H_0:({\mathrm{\mu }}_1-{\mathrm{\mu }}_2)>10$and the level of significance is 0.01 .

The pooled standard deviation is$s_p=\sqrt{\frac{\left(n_1-1\right){{\sigma }_1}^2+\left(n_2-1\right){{\sigma }_2}^2}{n_1+n_2-2}}$

$$\begin{gathered} =\sqrt{\frac{\left(15-1\right){\left(5.47\right)}^2+\left(16-1\right){\left(5.41\right)}^2}{15+16-2}} \\ =5.44 \\ t=\frac{\left(\overline{x_1}-{\overline{x}}_2\right)-\left({\mu }_1-{\mu }_2\right)}{s_p\sqrt{\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}} \\ =\frac{\left(43.6-53.625\right)-10}{5.44\sqrt{\left(\frac{1}{15}+\frac{1}{16}\right)}} \\ =\frac{-0.025}{1.955} \\ =\frac{-1.24}{0.52} \\ =-0.01 \end{gathered}$$

The degree of freedom of test is $15+16-2=29$.

From the t-distribution table, the critical value at 0.10 the level of significance for 29 degrees of freedom is 2.46.

Since, , so the null hypothesis will not be rejected.

Therefore, we cannot conclude that $({\mathrm{\mu }}_1-{\mathrm{\mu }}_2)>10$.

The 98% confidence interval for the difference in means

$$\begin{gathered} =\left({\overline{x}}_1-{\overline{x}}_2\right)\pm t_{\alpha /2}\times s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}} \\ =\left(43.6-53.625\right)\pm 2.46\times 5.44\sqrt{\frac{1}{15}+\frac{1}{16}} \\ =\left(10.025\right)\pm \left(2.46\times 5.44\times 0.36\right) \\ =10.025\pm 4.82 \end{gathered}$$

Thus, the confidence interval for the disparity of means is $5.205\text{to}14.845$.