$x_1=18,n_1=40,x_2=30,n_2=40$; left-tailed test, $\alpha =0.10;80\%$ confidence interval

Given in the question that, $x_1=18,n_1=40,x_2=30,n_2=40$; left-tailed test, $\alpha =0.10;80\%$confidence interval.

we need to determine the sample proportions.

The given values are, $x_1=18,n_1=40,x_2=30,n_2=40,\alpha =0.10$, and $80\%$confidence interval.

The formula for ${\tilde{p}}_1$is given by,

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

Substitute $x_1=18,n_1=40$

role="math" localid="1651484766063" ${\tilde{p}}_1=\frac{18}{40}$

$=0.45$

The formula for ${\tilde{p}}_2$is given by,

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

Substitute $x_2=30,n_2=40$

width="61" height="41" role="math">p~2=3040

$=0.75$

Therefore, the sample proportions are$0.45$and $0.75$.

Given in the question that, $x_1=18,n_1=40,x_2=30,n_2=40$; left-tailed test, $\alpha =0.10;80\%$confidence interval

we need to decide that whether using the two-proportions z-procedures is appropriate. If so, also do parts (c) and (d).

The given values are, $x_1=18,n_1=40,x_2=30,n_2=40,\alpha =0.10$, and $80\%$confidence interval.

To begin, calculate role="math" localid="1651485400650" $n_1-x_1$and role="math" localid="1651485418546" $n_2-x_2$. After that, compare the outcome to 5. The two-proportion z-test technique is appropriate if it is more than or equal to 5.

The value of $n_1-x_1$is calculated as,

$n_1-x_1=40-18$

$=22$

The value of $n_2-x_2$is calculated as,

$n_2-x_2=40-30$

$=10$

The two-proportion z-test technique is appropriate because the values are more than 5.

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

Given in the question that, $x_1=18,n_1=40,x_2=30,n_2=40$; left-tailed test, $\alpha =0.10;80\%$confidence interval

we need to use the two-proportions z-test to conduct the required hypothesis test.

The given values are, $x_1=18,n_1=40,x_2=30,n_2=40,\alpha =0.10$, and $80\%$confidence interval.

The formula for $z$is given by,

$z=\frac{{\tilde{p}}_1-{\tilde{p}}_2}{\sqrt{{\tilde{p}}_p\left(1-{\tilde{p}}_p\right)}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

The formula for ${\tilde{p}}_p$is given by,

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

Substitute $x_1=18,n_1=40,x_2=30,n_2=40$

$=\frac{18+30}{40+40}$

$=\frac{48}{80}$

The value of $z$is calculated as,

$z=\frac{{\tilde{p}}_1-{\tilde{p}}_2}{\sqrt{{\tilde{p}}_p\left(1-{\tilde{p}}_p\right)}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

$=\frac{0.45-0.75}{\sqrt{0.6(1-0.6)}\left(\sqrt{\frac{1}{40}+\frac{1}{40}}\right)}$

$=\frac{-0.3}{0.11}$

$=-2.739$

Perform the test at $10\%$level of significance that is $\alpha =0.1$from table-IV (at the bottom) the value of

$z_{\alpha }=z_{0.1}=1.282\text{.}$

$z-1.282$is the rejected region. As a result, the test static does not fall into the reject zone. As a result, the hypothesis $H_o$is rejected, and the test findings at the $10\%$level are not statistically significant.

As a result, the data does not provide sufficient evidence to reject the null hypothesis at the $10\%$level of significance.

Given in the question that, $x_1=18,n_1=40,x_2=30,n_2=40$; left-tailed test, $\alpha =0.10;80\%$confidence interval

we need to find the specified confidence interval by using the two-proportions z-interval procedure

The given values are,$x_1=18,n_1=40,x_2=30,n_2=40,\alpha =0.10$, and $80\%$confidence interval.

For confidence level of $(1-\alpha )$the confidence interval for $p_1-p_2$are

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

Calculate the value of $\alpha$,

$80=100(1-\alpha )$

$\alpha =0.2$

The value of$z$at $\alpha /2$from the $z$-score table is $1.282$.

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

$\left({\tilde{p}}_1-{\tilde{p}}_2\right)\pm z_{\alpha /2}·\sqrt{\frac{{\tilde{p}}_1\left(1-{\tilde{p}}_1\right)}{n_1+2}+\frac{{\tilde{p}}_2\left(1-{\tilde{p}}_2\right)}{n_2+2}}=(0.45-0.75)\pm 1.282$

$.\sqrt{\frac{0.45(1-0.45)}{40}+\frac{0.75(1-0.75)}{40}}$

$=-0.3\pm 0.134$

$=-0.434$to $-0.166$

Therefore, the difference between the percentage of the adult-Americans is $-0.434$ to $-0.166$.