In each of Exercises 11.100-11.105, we have provided the numbers of successes and the sample sizes for independent simple random samples from two populations. In each case, do the following tasks.

a. Determine the sample proportions.

b. Decide whether using the fwe-proportions $z$-procedures is appropriate. If so, also do parts (c) and (d).

c. Use the two-pmoportions $z$-test to conduct the required hypothesis test.

d. Use the fwo-proportions $z$-interval procedure to find the specified confidence interval.

$x_1=30,n_1=80,x_2=15,n_2=20;$two tailed test$\alpha =0.05;95\mathrm{\%}$confidence interval

Given in the question that, $x_1=30,n_1=80,x_2=15,n_2=20;$

We have to determine the sample proportions

We have to calculate the value of ${\tilde{p}}_1$as follow:

$$\begin{gathered} {\tilde{p}}_1=\frac{x_1}{n_1} \\ =\frac{30}{80} \\ =0.375 \end{gathered}$$

Then compute the value of ${\tilde{\mathit{p}}}_2$as follow:

$$\begin{gathered} {\tilde{p}}_2=\frac{x_2}{n_2} \\ =\frac{15}{20} \\ =0.75 \end{gathered}$$

Given in the question that,

$x_1=30,n_1=80,x_2=15,n_2=20,\alpha =0.05$

We have to determine whether the $z$-interval technique for the two proportions is adequate or not.

To begin, calculate $n_1-x_1\text{and}{n}_2-x_2$. After that, compare the outcome to $5$. The two-proportion $z$-test technique is appropriate if it is more than or equal to $5.$

We can compute the value of $n_1-x_1$as follow:

$$\begin{gathered} n_1-x_1=80-30 \\ =50 \end{gathered}$$

Let's compute the value of $n_2-x_2$as follow:

$$\begin{gathered} n_2-x_2=20-15 \\ =5 \end{gathered}$$

The two-proportion $z$-test technique is appropriate because the values are higher than or equal to$5.$

Given in the question that, $x_1=30,n_1=80,x_2=15,n_2=20,\alpha =0.05,\text{and}95\mathrm{\%}$confidence interval.

Using the $z$-test approach, find the needed hypothesis test.

We have to find the value of ${\tilde{p}}_p$as follow:

$$\begin{gathered} {\tilde{p}}_p=\frac{x_1+x_2}{n_1+n_2} \\ =\frac{30+15}{80+20} \\ =\frac{45}{100} \\ =0.45 \end{gathered}$$

Let's find the value of $z$by using the given formula:

$$\begin{gathered} z=\frac{{\tilde{p}}_1-{\tilde{p}}_2}{\sqrt{{\tilde{p}}_p\left(1-{\tilde{p}}_p\right)}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \\ =\frac{0.375-0.75}{\sqrt{0.45(1-0.45)}\left(\sqrt{\frac{1}{80}+\frac{1}{20}}\right)} \\ =\frac{-0.375}{0.125} \\ =-3 \end{gathered}$$

Perform the test at a significance level of $5\%$, or $a=0.05$, using the value of$z_{\alpha }$from table-IV (bottom).

$$\begin{gathered} z_{\alpha }=z_{0.05/2} \\ =\pm 1.960 \end{gathered}$$

$z>1.960$is the rejected region. Since then, the test static has fallen into the reject zone. As a result, the hypothesis$H_0$is rejected, and the test findings at the$5\%$level are not statistically significant.

To determine the confidence interval that has been specified

The confidence interval for $p1-p2$is for a confidence level of $1-\alpha$

$\left({\widehat{p}}_1-{\widehat{p}}_2\right)\pm z1/2\times \sqrt{{\widehat{p}}_1\left(1-{\widehat{p}}_1\right)/n_1+{\widehat{p}}_2\left(1-{\widehat{p}}_2\right)/n_2}$

Compute the value of $\alpha$as follow:

$$\begin{gathered} 95=100(1-\alpha ) \\ \alpha =0.05 \end{gathered}$$

The value of $z$in the $z$-score table at $\alpha /2$is $1.960$.

For the difference between the two-population proportion, the needed confidence interval is determined as,

$\left({\tilde{p}}_1-{\tilde{p}}_2\right)\pm z\alpha /2·\sqrt{\frac{{\tilde{p}}_1\left(1-{\tilde{p}}_1\right)}{n_1+2}+\frac{{\tilde{p}}_2\left(1-{\tilde{p}}_2\right)}{n_2+2}}=(0.375-0.75)\pm 1.960\times \sqrt{\frac{0.375(1-0.375)}{80}+\frac{0.75(1-0.75)}{20}}$

$$\begin{gathered} =0.375\pm 0.217 \\ =-0.592\text{to}-0.158 \end{gathered}$$