Question

Refer to Exercise $15$.

(a) Carry out a significance test at the $\alpha =0.05$level.

(b) Construct and interpret a $95\%$confidence interval for the difference between the population proportions. Explain how the confidence interval is consistent with the results of the test in part (a).

Short answer

(a) There is sufficient evidence to support the claim of a difference between the population proportions.

(b) We are $95\%$confident that the proportion difference is between $0.066$and $0.174$.

Step-by-step solution

Part(a) Step 1: Given Information

Given

${\hat{p}}_1=79\%=0.79$

$n_1=800$

${\hat{p}}_2=67\%=0.67$

$n_2=400$

Determine the hypothesis

$H_0:p_1-p_2=0$

$H_a:p_1-p_2\ne 0$

Part(a) Step 2: Explanation

The sample proportion is the number of successes divided by the sample size:

$$\begin{gathered} {\hat{p}}_1=\frac{x_1}{n_1} \\ =\frac{632}{800} \\ =0.79 \end{gathered}$$

$$\begin{gathered} {\hat{p}}_2=\frac{x_2}{n_2} \\ =\frac{268}{400} \\ =0.67 \end{gathered}$$

$$\begin{gathered} {\hat{p}}_p=\frac{x_1+x_2}{n_1+n_2} \\ =\frac{632+268}{800+400} \\ =\frac{900}{1200} \\ =0.75 \end{gathered}$$

Determine the value of the test statistic:

$$\begin{gathered} z=\frac{{\hat{p}}_1-{\hat{p}}_2}{\sqrt{{\hat{p}}_p\left(1-{\hat{p}}_p\right)}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \\ =\frac{0.79-0.67}{\sqrt{0.75(1-0.75)}\sqrt{\frac{1}{800}+\frac{1}{400}}} \\ \approx 4.53 \end{gathered}$$

The $p$-value is the probability of obtaining the value of the test statistic, or a value more extreme. Determine the$p$-value using table A:
localid="1650450174625" $$\begin{gathered} P=P(Z<-4.53\text{or}Z>4.53) \\ =2\times P(Z<-4.53) \\ =2\times 0.0001 \\ =0.0002 \end{gathered}$$

If the P-value is smaller than the significance level, reject the null hypothesis:

$P<0.05\Rightarrow \text{Reject}{H}_0$

Part(b) Step 1: Given Information

Given

${\hat{p}}_1=79\%=0.79$

$n_1=800$

${\hat{p}}_2=67\%=0.67$

$n_2=400$

Determine the hypothesis

$H_0:p_1-p_2=0$

$H_a:p_1-p_2\ne 0$

Part(b) Step 2: Explanation

The sample proportion is the number of successes divided by the sample size:

$$\begin{gathered} {\hat{p}}_1=\frac{x_1}{n_1} \\ =\frac{632}{800} \\ =0.79 \end{gathered}$$

$$\begin{gathered} {\hat{p}}_2=\frac{x_2}{n_2} \\ =\frac{268}{400} \\ =0.67 \end{gathered}$$

For confidence level $1-\alpha =0.95$, determine $z_{\alpha /2}=z_{0.025}$Using table II (look up x$z_{\alpha /2}=z_{0.025}$in the table, the $z$-score is then the found $z$-score with opposite sign):

$z_{\alpha /2}=1.96$

The endpoints of the confidence interval for $p_{1}-p_{2}$ are then:

localid="1650450195998" $$\begin{gathered} \left({\hat{p}}_1-{\hat{p}}_2\right)-z_{\alpha /2}·\sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}=(0.79-0.67)-1.96\sqrt{\frac{0.79(1-0.79)}{800}+\frac{0.67(1-0.67)}{400}} \\ \approx 0.066 \end{gathered}$$

localid="1650450210366" $$\begin{gathered} \left({\hat{p}}_1-{\hat{p}}_2\right)+z_{\alpha /2}·\sqrt{\frac{{\hat{p}}_1\left(1-{\hat{p}}_1\right)}{n_1}+\frac{{\hat{p}}_2\left(1-{\hat{p}}_2\right)}{n_2}}=(0.79-0.67)+1.96\sqrt{\frac{0.79(1-0.79)}{800}+\frac{0.67(1-0.67)}{400}} \\ \approx 0.174 \end{gathered}$$