91Ó°ÊÓ

The given data is

$x_1=18$

$$\begin{gathered} n_1=30 \\ {x}_2=10 \\ {n}_2=20 \end{gathered}$$

Two-tailed test, $\mathrm{Î\pm }=0.05$

Confidence interval is$95\%$.

From the given information

The value of${\tilde{p}}_1$is

${\tilde{p}}_1=\frac{x_1}{n_1}$

${\tilde{p}}_1=\frac{18}{30}$

$=0.6$

Then,

${\tilde{p}}_2=\frac{x_2}{n_2}$

${\tilde{p}}_2=\frac{10}{20}$

$=0.5$.

The given data is

$$\begin{gathered} x_1=18 \\ {n}_1=30 \\ {x}_2=10 \\ {n}_2=20 \end{gathered}$$

Two-tailed test, $\mathrm{Î\pm }=0.05$

Confidence level is$95\%$

First calculate $n_1-x_1$and $n_2-x_2$. And then compare the result with $5$.

If it is greater than or equal to $5$that the two-proportion $z-$test procedure is appropriate.

$n_1-x_1=30-18$

$=12$

Then,

$n_2-x_2=20-10$

$=10$

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

As a result, the two-proportion$z-$test is applicable.

The given data is

$$\begin{gathered} x_1=18 \\ {n}_1=30 \\ {x}_2=10 \\ {n}_2=20 \end{gathered}$$

Two-tailed test, $\mathrm{Î\pm }=0.05$

Confidence interval is$95\%$

The value of $z$is calculated as

$z=\frac{{\tilde{p}}_1-{\tilde{p}}_2}{\sqrt{{\tilde{p}}_p\left(1-{\tilde{p}}_p\right)}\sqrt{\frac{1}{n_1}+\frac{1}{m_2}}}$

First find,

${\tilde{p}}_p=\frac{x_1+x_2}{n_1+n_2}$

$=\frac{18+10}{30+20}$

$=\frac{28}{50}$

Then the value of $z$is

$z=\frac{{\tilde{p}}_1-{\tilde{p}}_2}{\sqrt{{\tilde{p}}_p\left(1-{\tilde{p}}_p\right)}\sqrt{\frac{1}{n_1}+\frac{1}{m_2}}}$

$=\frac{0.6-0.5}{\sqrt{0.56(1-0.56)}\left(\sqrt{\frac{1}{30}+\frac{1}{20}}\right)}$

$=\frac{0.1}{0.145}$

$=0.69$

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

$z_{\mathrm{Î\pm }}=z_{0.05/2}$

$=Â\pm 1.960$

The $z<1.960$is the rejected region.

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

As a result, the data give adequate evidence to support the null hypothesis at a level of significance of $5\%$.

The given data is

$$\begin{gathered} x_1=18 \\ {n}_1=30 \\ {x}_2=10 \\ {n}_2=20 \end{gathered}$$

Two-tailed test, $\mathrm{Î\pm }=0.05$

Confidence interval is$95\%$

For confidence level $(1-\mathrm{Î\pm })$, the confidence interval is

$\left({\hat{p}}_1-{\hat{p}}_2\right)Â\pm z_{1/2}×\sqrt{{\hat{p}}_1\left(1-{\hat{p}}_1\right)/n_1+{\hat{p}}_2\left(1-{\hat{p}}_2\right)/n_2}$

Calculate the value of $\mathrm{Î\pm }$

$95=100(1-\mathrm{Î\pm })$

$\mathrm{Î\pm }=0.05$

The value of $z$at $a/2$from the $z-$score table is $1.960$.

The required confidence interval for the difference between the two-population proportion is calculated as,

$\left({\tilde{p}}_1-{\tilde{p}}_2\right)Â\pm z_{\mathrm{Î\pm }/2}×\sqrt{\frac{{\tilde{p}}_1\left(1-{\tilde{p}}_1\right)}{n_1+2}+\frac{{\tilde{p}}_2\left(1-{\tilde{p}}_2\right)}{n_2+2}}=(0.6-0.5)Â\pm 1.960$

$×\sqrt{\frac{0.6(1-0.6)}{30}+\frac{0.5(1-0.5)}{20}}$

$=0.1Â\pm 0.280$

$=-0.18to0.38$.