Question: Two independent random samples have been selected—100 observations from population 1 and 100 from population 2. Sample means ${\overline{\mathbf{\text{x}}}}_{\mathbf{1}}\mathbf{\text{=26.6,}}{\overline{\mathbf{\text{x}}}}_{\mathbf{\text{2}}}\mathbf{\text{= 15.5}}$ were obtained. From previous experience with these populations, it is known that the variances are${{\mathit{\sigma }}_{\mathbf{1}}}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{9}\mathbf{}\mathit{a}\mathit{n}\mathit{d}{{\mathbf{\sigma }}_{\mathbf{2}}}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{16}$ .

a. Find ${\mathit{\sigma }}_{\left({\overline{\text{x}}}_{\text{1}}\text{-}{\overline{\text{x}}}_{\text{2}}\right)}$.

b. Sketch the approximate sampling distribution for $\left({\overline{\text{x}}}_{\text{1}}-{\overline{\text{x}}}_{\text{2}}\right)$, assuming $\mathbf{(}{\mathbf{\mu }}_{\mathbf{1}}\mathbf{-}{\mathbf{\mu }}_{\mathbf{2}}\mathbf{)}\mathbf{=}\mathbf{10}$.

c. Locate the observed value of $\mathbf{(}{\overline{\mathbf{\text{x}}}}_{\mathbf{\text{1}}}\mathbf{-}{\overline{\mathbf{\text{x}}}}_{\mathbf{\text{2}}}\mathbf{)}$the graph you drew in part

b. Does it appear that this value contradicts the null hypothesis ${\mathit{H}}_{\mathbf{0}}\mathbf{:}\left({\mu }_1-{\mu }_2\right)\mathbf{=}\mathbf{10}$?

d. Use the z-table to determine the rejection region for the test against${\mathit{H}}_{\mathbf{0}}\mathbf{:}\mathbf{(}{\mathbf{\mu }}_{\mathbf{1}}\mathbf{-}{\mathbf{\mu }}_{\mathbf{2}}\mathbf{)}\mathbf{\ne }\mathbf{10}$. Use$\mathit{\alpha }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}$.

e. Conduct the hypothesis test of part d and interpret your result.

f. Construct a confidence interval for $\left({\mathbf{\mu }}_{\mathbf{1}}\mathbf{-}{\mathbf{\mu }}_{\mathbf{2}}\right)$. Interpret the interval.

g. Which inference provides more information about the value of $\left({\mathbf{\mu }}_{\mathbf{1}}\mathbf{-}{\mathbf{\mu }}_{\mathbf{2}}\right)$— the test of hypothesis in part e or the confidence interval in part f?

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

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

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

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

So, the diagrammatic presentation of ${\mu }_1-{\mu }_2=10$will be as below –

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

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

Now, we will locate it on the graph below –

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

Null Hypothesis:$H_0:{\mu }_1-{\mu }_2=10$

Alternate Hypothesis:$H_0:{\mu }_1-{\mu }_2\ne 10$

Confidence level = 0.95, so$\alpha =1-0.95=0.05$

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 –

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

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

Null Hypothesis:$H_0:{\mu }_1-{\mu }_2=10$

Alternate Hypothesis:$H_0:{\mu }_1-{\mu }_2\ne 10$

Confidence level = 0.95, so$\alpha =1-0.95=0.05$

Level of significance = 5%

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

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

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

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

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 ${\mu }_1-{\mu }_2\ne 10$without specifying any value of the difference between the population means.

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