Question

a. Determine the sample proportions.

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

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

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

$x_1=15$

$n_1=20$

$x_2=18$

$n_2=30$

Right tailed test, $\alpha =0.05$

Confidence interval is$90\%$

Short answer

Part (a) The sample proportions are $0.75$and $0.6$

Part (b) The two-proportion $z-$test procedure is appropriate.

Part (c) The hypothesis $H_0$is rejected, and the test results at the $5\%$level are statistically significant.

Part (d) The difference between the percentage of the adult-Americans is $-0.067to0.367$.

Step-by-step solution

Part (a) Step 1: Given information

The given values are

$x_1=15$

$n_1=20$

$x_2=18$

$n_2=30$

Right tailed test, $\alpha =0.05$

Confidence interval is $90\%$

Part (a) Step 2: Explanation

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

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

$=\frac{15}{20}$

$=0.75$

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

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

$=\frac{18}{30}$

$=0.6$.

Part (b) Step 1: Given information

The given values are

$x_1=15$

$n_1=20$

$x_2=18$

$n_2=30$

Right tailed test, $\alpha =0.05$

Confidence interval is $90\%$

Part (b) Step 2: Explanation

Calculate $n_1-x_1$and $n_2-x_2$first.

After that, compare the outcome to $5$.

The two-proportion $z-$test technique is appropriate if it is more than or equal to $5$.

$n_1-x_1=20-15$

$=5$

$n_2-x_2=30-18$

$=12$

Since the values are greater than or equal to$5$ hence, the two-proportion z-test procedure is appropriate.

Part (c) Step 1: Given information

The given values are

$x_1=15$

$n_1=20$

$x_2=18$

$n_2=30$

Right tailed test, $\alpha =0.05$

Confidence interval is$90\%$

Part (c) Step 2: Explanation

Using the given values,

The $z$value is given as

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

The formula for${\tilde{p}}_p$

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

$=\frac{15+18}{20+30}$

$=\frac{33}{50}$

$=0.66$

From this, the value of $z$is

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

$=\frac{0.75-0.6}{\sqrt{0.66(1-0.66)}\left(\sqrt{\frac{1}{20}+\frac{1}{30}}\right)}$

$=\frac{0.15}{0.137}$

$=1.097$

Perform the test at $5\%$level of significance that is $\alpha =0.05$from table-IV (at the bottom) the value of

$z_{\alpha }=z_{0.05}$

$=1.645$

This$z>1.645$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 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 $5\%$.

Part (d) Step 1: Given information

The given values are

$x_1=15$

$n_1=20$

$x_2=18$

$n_2=30$

Right tailed test, $\alpha =0.05$

Confidence interval is$90\%$

Part (d) Step 2: Explanation

For the confidence level $(1-\alpha )$

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

Calculate the value of $a$

$90=100(1-\alpha )$

$=0.1$

The value of $z$at $a/2$from the $z-$score table is$1.645$

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

$\left({\tilde{p}}_1-{\tilde{p}}_2\right)\pm z_{\alpha /2}\times \sqrt{\frac{{\tilde{p}}_1\left(1-{\tilde{p}}_1\right)}{n_1+2}+\frac{{\tilde{p}}_2\left(1-{\tilde{p}}_2\right)}{n_2+2}}=(0.75-0.6)\pm 1.645$

$\times \sqrt{\frac{0.75(1-0.75)}{20}+\frac{0.6(1-0.6)}{30}}$

$=0.15\pm 0.217$

$=-0.067to0.367$.