91Ó°ÊÓ

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,

$x_1=30,n_1=80,x_2=15,n_2=20,\mathrm{Î\pm }=0.05$

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, $x_1=30,n_1=80,x_2=15,n_2=20,\mathrm{Î\pm }=0.05,\text{and}95\mathrm{\%}$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$z_{\mathrm{Î\pm }}$from table-IV (bottom).

$$\begin{gathered} z_{\mathrm{Î\pm }}=z_{0.05/2} \\ =Â\pm 1.960 \end{gathered}$$

$z>1.960$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 $1-\mathrm{Î\pm }$

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

Compute the value of $\mathrm{Î\pm }$as follow:

$$\begin{gathered} 95=100(1−\mathrm{Î\pm }) \\ \mathrm{Î\pm }=0.05 \end{gathered}$$

The value of $z$in the $z$-score table at $\mathrm{Î\pm }/2$is $1.960$.

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

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

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