91影视

$$\begin{gathered} {蟽}_{\left({\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2\right)}=\sqrt{\frac{{{蟽}_1}^2}{n_1}+\frac{{{蟽}_2}^2}{n_2}} \\ =\sqrt{\frac{9}{100}+\frac{16}{100}} \\ =\sqrt{\frac{25}{100}} \\ =\frac{5}{10} \\ =0.5 \end{gathered}$$

Therefore, the value of${蟽}_{\left({\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2\right)}$ is 0.5.

The mean of the sampling distribution $\left({\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2\right)$is the population mean ${渭}_1-{渭}_2$.

That is ${渭}_{\left({\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2\right)}={渭}_1-{渭}_2$, which is independent of sample size.

So, the diagrammatic presentation of ${渭}_1-{渭}_2=10$will be as below 鈥�

The value$\left({\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2\right)$ will be

$$\begin{gathered} =26.6-15.5 \\ =11.1 \end{gathered}$$

Now, we will locate it on the graph below 鈥�

The observed value$\left({\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2\right)$ contradicts the null hypothesis $H_0:{渭}_1-{渭}_2=10$.

Null Hypothesis:$H_0:{渭}_1-{渭}_2=10$

Alternate Hypothesis:$H_0:{渭}_1-{渭}_2鈮�10$

Confidence level = 0.95, so$伪=1-0.95=0.05$

Level of significance =5%

The critical value Z at 5% the level of significance is 1.96.

So, the rejection region will be as below 鈥�

If , $z>z_{\frac{伪}{2}}\left(=1.96\right)$then reject the null hypothesis.

If ,$z>-z_{\frac{伪}{2}}\left(=-1.96\right)$ then reject the null hypothesis.

Null Hypothesis:$H_0:{渭}_1-{渭}_2=10$

Alternate Hypothesis:$H_0:{渭}_1-{渭}_2鈮�10$

Confidence level = 0.95, so$伪=1-0.95=0.05$

Level of significance = 5%

The critical value of Z the level of significance is 1.96.

$$\begin{gathered} Z=\frac{\left({\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2\right)-\left({渭}_1-{渭}_2\right)}{\sqrt{\frac{{{蟽}_1}^2}{n}+\frac{{{蟽}_2}^2}{n}}} \\ =\frac{\left(26.6-15.5\right)-10}{\sqrt{\frac{9}{100}+\frac{16}{100}}} \\ =\frac{1.5}{0.5} \\ =3 \end{gathered}$$

So, the value p of is.0027 .

As the value is less than the significance level, so the null hypothesis is rejected.

The 90% confidence interval for the difference in means

$$\begin{gathered} =\left({\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2\right)卤z_{伪/2}脳\sqrt{\frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2}} \\ =\left(26.6-15.5\right)卤1.96脳\sqrt{\frac{9}{100}+\frac{16}{100}} \\ =\left(11.1\right)卤\left(1.96脳0.5\right) \\ =11.1卤0.98 \end{gathered}$$

Therefore, the confidence interval for the difference of means is 10.12 to 12.08.

The confidence interval tells us the specific limit within which the difference between the population means is expected to lie with 95% confidence, whereas the hypothesis testing presents the situation where we can tell that ${渭}_1-{渭}_2鈮�10$without specifying any value of the difference between the population means.

Therefore, the confidence interval provides more information ${渭}_1-{渭}_2$.