Testing Claims About Proportions. In Exercises 7–22, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim.

Overlap of Confidence Intervals In the article “On Judging the Significance of Differences by Examining the Overlap Between Confidence Intervals,” by Schenker and Gentleman (American Statistician, Vol. 55, No. 3), the authors consider sample data in this statement: “Independent simple random samples, each of size 200, have been drawn, and 112 people in the first sample have the attribute, whereas 88 people in the second sample have the attribute.”

a. Use the methods of this section to construct a 95% confidence interval estimate of the difference $p_1-p_2$. What does the result suggest about the equality of $p_1$and$p_2$?

b. Use the methods of Section 7-1 to construct individual 95% confidence interval estimates for each of the two population proportions. After comparing the overlap between the two confidence intervals, what do you conclude about the equality of$p_1$and$p_2$?

c. Use a 0.05 significance level to test the claim that the two population proportions are equal. What do you conclude?

d. Based on the preceding results, what should you conclude about the equality of$p_1$and$p_2$? Which of the three preceding methods is least effective in testing for the equality of$p_1$and$p_2$?

Two independent samples are considered, each of size 200. In the first sample, 112 people have the attribute. In the second sample, 88 people have the attribute. The significance level is $\alpha =0.05$.

The first random sample is of size $\left(n_1\right)$equal to 200.

The second random sample is of size $\left(n_2\right)$equal to 200.

Let ${\hat{p}}_1$denote the sampleproportion of people who have attributes in the first sample.

$$\begin{gathered} {\hat{p}}_1=\frac{x_1}{n_1} \\ =\frac{112}{200} \\ =0.56 \end{gathered}$$

Let ${\hat{p}}_2$denote the sampleproportion of people who have attributes in the second sample

$$\begin{gathered} {\hat{p}}_2=\frac{x_2}{n_2} \\ =\frac{88}{200} \\ =0.44 \end{gathered}$$

The value of the pooled sample proportion is computed as follows:

$$\begin{gathered} \overline{p}=\frac{\left(x_1+x_2\right)}{\left(n_1+n_2\right)} \\ =\frac{\left(112+88\right)}{\left(200+200\right)} \\ =0.5 \end{gathered}$$

$$\begin{gathered} \overline{q}=1-\overline{p} \\ =1-0.5 \\ =0.5 \end{gathered}$$

a.

The general formula for the confidence interval of the difference of proportions is written below:

$\text{Confidence} \text{Interval}=\left(\left({\hat{p}}_1-{\hat{p}}_2\right)-E , \left({\hat{p}}_1-{\hat{p}}_2\right)+E\right) ...\left(1\right)$

Where, E is the margin of error and has the following formula:

$E=z_{\frac{\alpha }{2}}\times \sqrt{\left(\frac{{\hat{p}}_1\times {\hat{q}}_1}{n_1}+\frac{{\hat{p}}_2\times {\hat{q}}_2}{n_2}\right)}$

For computing the confidence interval, first find the critical value $z_{\frac{\alpha }{2}}$.

The confidence level is 95%; thus, the value of the level of significance for the confidence interval becomes $\alpha =0.05$.

Hence,

$$\begin{gathered} \frac{\alpha }{2}=\frac{0.05}{2} \\ =0.025 \end{gathered}$$

The value of $z_{\frac{\alpha }{2}}$form the standard normal table is equal to 1.96

Now, the margin of error (E) is equal to:

$$\begin{gathered} E=z_{\frac{\alpha }{2}}\times \sqrt{\left(\frac{{\hat{p}}_1\times {\hat{q}}_1}{n_1}+\frac{{\hat{p}}_2\times {\hat{q}}_2}{n_2}\right)} \\ =1.96\times \sqrt{\left(\frac{0.56\times 0.44}{200}+\frac{0.44\times 0.56}{200}\right)} \\ =0.0973 \end{gathered}$$

Substitute the value of E in equation (1) as:

$$\begin{gathered} \text{Confidence} \text{Interval}=\left(\left({\hat{p}}_1-{\hat{p}}_2\right)-E , \left({\hat{p}}_1-{\hat{p}}_2\right)+E\right) \\ =\left(\left(0.56-0.44\right)-0.0973 , \left(0.56-0.44\right)+0.0973\right) \\ =\left(0.0227, 0.2173\right) \end{gathered}$$

Thus, the 95% confidence interval for the difference in the two population proportions is .

b.

Confidence interval for has the following expression:

$\left({\hat{p}}_1-E , {\hat{p}}_1+E\right) ...\left(2\right)$

Now, the margin of error (E) is equal to:

$$\begin{gathered} E=z_{\frac{\alpha }{2}}\times \sqrt{\frac{{\hat{p}}_1\times {\hat{q}}_1}{n}} \\ =1.96\times \sqrt{\frac{0.56\times 0.44}{200}} \\ =0.0688 \end{gathered}$$

Substitute the values in equation (2):

$$\begin{gathered} \left({\hat{p}}_1-E , {\hat{p}}_1+E\right)=(0.56-0.0688 , 0.56-0.0688) \\ =(0.491 , 0.629) \end{gathered}$$

Thus, the 95% confidence interval for $p_1$is $(0.491 , 0.629)$.

The confidence interval for $p_2$has the given expression:

$\left({\hat{p}}_2-E , {\hat{p}}_2+E\right) ...\left(3\right)$

Now, the margin of error (E) is equal to:

$$\begin{gathered} E=z_{\frac{\alpha }{2}}\times \sqrt{\frac{{\hat{p}}_2\times {\hat{q}}_2}{n}} \\ =1.96\times \sqrt{\frac{0.44\times 0.56}{200}} \\ =0.0688 \end{gathered}$$

Substitute the values in equation (3):

$$\begin{gathered} \left({\hat{p}}_2-E , {\hat{p}}_2+E\right)=(0.44-0.0688 , 0.44+0.0688) \\ =(0.371 , 0.509) \end{gathered}$$

Thus, the 95% confidence interval for $p_2$is $(0.371 , 0.509)$.

It can be observed that there are values that occur in both the confidence intervals. Thus, the two confidence intervals overlap. This suggests that there are possibilities where the two population proportions become equal.

Therefore, there is not sufficient evidence to conclude thattheproportionof people who have the attribute corresponding to the first population is not equal to the proportion of people who have the attribute corresponding to the second population.

c.

It is claimed that the two population proportions are equal.

The following hypotheses are set up:

Null Hypothesis:Theproportionof people who have the attribute corresponding to the first population is equal to the proportion of people who have the attribute corresponding to the second population.

$H_0:p_1=p_2$

Alternative Hypothesis:Theproportionof people who have the attribute corresponding to the first population is not equal to the proportion of people who have the attribute corresponding to the second population.

$H_1:p_1\ne p_2$

The test is two-tailed.

The value test statistic is computed as:

$$\begin{gathered} z=\frac{\left({\hat{p}}_1-{\hat{p}}_2\right)-\left(p_1-p_2\right)}{\sqrt{\left(\frac{\overline{p}\overline{q}}{n_1}+\frac{\overline{p}\overline{q}}{n_2}\right)}} \\ =\frac{\left(0.56-0.44\right)-0}{\sqrt{\left(\frac{0.5\times 0.5}{200}+\frac{0.5\times 0.5}{200}\right)}} \\ =2.4 \end{gathered}$$

Thus, the value of the test statistic is 2.4.

Referring to the standard normal distribution table, the critical values of z corresponding to$\alpha =0.05$ for a two-tailed test are -1.96 and 1.96.

Referring to the standard normal distribution table, the corresponding p-value is equal to 0.0164.

Since the p-value is less than 0.05, the null hypothesis is rejected.

There is enough evidence to reject the claim thattheproportionof people who have the attribute corresponding to the first population is equal to the proportion of people who have the attribute corresponding to the second population.

d.

Based on the above results, it is concluded that the proportionof people who have the attribute corresponding to the first population is not equal to the proportion of people who have the attribute corresponding to the second population.

It can be seen that examining the overlapping confidence intervals does not produce accurate results regarding the decision of the claim.

Further, the confidence interval estimate of the difference of the two proportions as well as the hypothesis test produces similar results which appear to be correct.

Therefore, the method in part (b) is the least effective method to test the equality of the two population proportions.