91影视

For Sample 1

$$\begin{gathered} {\stackrel{炉}{x}}_1=43.6 \\ {n}_1=15 \\ {蟽}_1=5.47 \end{gathered}$$

For Sample 2

$$\begin{gathered} {\stackrel{炉}{x}}_2=53.625 \\ {n}_2=16 \\ {蟽}_2=5.41 \end{gathered}$$

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

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

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

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 $({\mathrm{渭}}_1-{\mathrm{渭}}_2)>10$.

The 98% confidence interval for the difference in means

$$\begin{gathered} =\left({\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2\right)卤t_{伪/2}脳s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}} \\ =\left(43.6-53.625\right)卤2.46脳5.44\sqrt{\frac{1}{15}+\frac{1}{16}} \\ =\left(10.025\right)卤\left(2.46脳5.44脳0.36\right) \\ =10.025卤4.82 \end{gathered}$$

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