91Ó°ÊÓ

Given in the question that,

$x_1=14,n_1=20,x_2=8,n_2=20$;

right-tailed test, $\mathrm{Î\pm }=0.05;90\%$confidence interval

we need to determine the sample proportions.

The given values are, $x_1=14,n_1=20,x_2=8,n_2=20,\mathrm{Î\pm }=0.05$, and $90\%$confidence interval.

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

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

Substitute $x_1=14,n_1=20$

role="math" localid="1651491983163" ${\tilde{p}}_1=\frac{14}{20}$

role="math" localid="1651492005630" $=0,7$

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

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

Substitute $x_2=8,n_2=20$

$=\frac{8}{20}$

$=0.4$

Therefore, the sample proportions are $=0.4$and $0.4$.

Given in the question that,

$x_1=14,n_1=20,x_2=8,n_2=20$

right-tailed test, $\mathrm{Î\pm }=0.05;90\%$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, $x_1=14,n_1=20,x_2=8,n_2=20,\mathrm{Î\pm }=0.05$, and $90\%$confidence interval.

To begin, calculate$n_1-x_1$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 $5$.

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

$n_1-x_1=20-14$

$=16$

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

$n_2-x_2=20-8$

$=12$

The two-proportion z-Procedure 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,

$x_1=14,n_1=20,x_2=8,n_2=20$;

right-tailed test, $\mathrm{Î\pm }=0.05;90\%$confidence interval

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

The given values are, $x_1=14,n_1=20,x_2=8,n_2=20,\mathrm{Î\pm }=0.05$, and $90\%$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 $x_1=14,n_1=20,x_2=8,n_2=20$

role="math" localid="1651493218114" ${\tilde{p}}_p=\frac{14+8}{20+20}$

$=\frac{22}{40}$

$0.55$

The value of $z$is calculated as,

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

$=\frac{0.7-0.4}{\sqrt{0.55(1-0.55)}\left(\sqrt{\frac{1}{20}+\frac{1}{20}}\right)}$

$=\frac{0.3}{0.157}$

$=1.907$

Perform the test at $5\%$level of significance that is $\mathrm{Î\pm }=0.05$from table-IV (at the bottom) the value of

$z_{\mathrm{Î\pm }}=z_{0.05}=1.645\text{.}$

$z>1.645$is the rejected region. Since then, the test static has fallen into the reject zone. As a result, the hypothesis $H_o$is rejected, and the test results at the $5\%$level are statistically significant.

As a result, the data give adequate evidence to reject the null hypothesis at a level of significance of .

Given in the question that,

$x_1=14,n_1=20,x_2=8,n_2=20$;

right-tailed test, $\mathrm{Î\pm }=0.05;90\%$confidence interval

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

The given values are, $x_1=14,n_1=20,x_2=8,n_2=20,\mathrm{Î\pm }=0.05$, and $90\%$confidence interval.

For confidence level of $(1-\mathrm{Î\pm })$the confidence interval for $p_1-p_2$are

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

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

$90=100(1-\mathrm{Î\pm })$

$\mathrm{Î\pm }=0.1$

The value of $z$at $\mathrm{Î\pm }/2$from the $z$-score table is $1.645$.

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

$\left({\tilde{p}}_1-{\tilde{p}}_2\right)Â\pm z_{\mathrm{Î\pm }/2}·\sqrt{\frac{{\tilde{p}}_1\left(1-{\tilde{p}}_1\right)}{n_1+2}+\frac{{\tilde{p}}_2\left(1-{\tilde{p}}_2\right)}{n_2+2}}=(0.7-0.4)Â\pm 1.645$

$.\sqrt{\frac{0.7(1-0.7)}{20}+\frac{0.4(1-0.4)}{20}}$

$=0.3Â\pm 0.247$

$=0.053$ to $0.547$

Therefore, the difference between the percentage of the adult-Americans is $0.053$to $0.547$.