Question

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.

Is Echinacea Effective for Colds? Rhinoviruses typically cause common colds. In a test of the effectiveness of Echinacea, 40 of the 45 subjects treated with Echinacea developed rhinovirus infections. In a placebo group, 88 of the 103 subjects developed rhinovirus infections (based on data from “An Evaluation of Echinacea Angustifolia in Experimental Rhinovirus Infections,” by Turner et al., New England Journal of Medicine, Vol. 353, No. 4). We want to use a 0.05 significance level to test the claim that Echinacea has an effect on rhinovirus infections.

a. Test the claim using a hypothesis test.

b. Test the claim by constructing an appropriate confidence interval.

c. Based on the results, does Echinacea appear to have any effect on the infection rate?

Short answer

a.

The hypotheses are as follows.

$$\begin{gathered} H_0:p_1=p_2 \\ H_1:p_1\ne p_2 \end{gathered}$$

The test statistic is 0.565. The p-value is 0.569.

The null hypothesis is failed to be rejected. Thus, there is no sufficient evidence to support the claim that Echinacea has an effect.

b. The 95% confidence interval is$-0.0798<\left({\text{p}}_1-{\text{p}}_2\right)<0.149$.Thus, there is no sufficient evidence to support the claim that Echinacea has an effect, as 0 is included in the interval.

c. The result suggests thatEchinaceadoes not have any effect on reducing the infection rate.

Step-by-step solution

Given information

The test for the effectiveness of Echinacea involves two groups:

In the treatment group, 40 of 45 subjects developed infections.

In the placebo group, 88 of 103 subjects developed infections.

The significance level is $\alpha =0.05$.

State the null and alternative hypotheses

a.

To test the effectiveness, let $p_1,p_2$be the proportion of subjects who develop infections in the treatment and subject groups, respectively.

From the claim, the null and alternative hypotheses are as follows.

$$\begin{gathered} H_0:p_1=p_2 \\ H_1:p_1\ne p_2 \end{gathered}$$

Compute the proportions

From the given information, summarize the following:

$$\begin{gathered} n_1=45 \\ {x}_1=40 \\ {n}_2=103 \\ {x}_2=88 \end{gathered}$$

The sample proportions are as follows.

$$\begin{gathered} {\hat{p}}_1=\frac{x_1}{n_1} \\ =\frac{40}{45} \\ =0.8889 \end{gathered}$$

and

$$\begin{gathered} {\hat{p}}_2=\frac{x_2}{n_2} \\ =\frac{88}{103} \\ =0.8544 \end{gathered}$$.

Find the sample pooled proportion 

The sample pooled proportions are calculated as follows.

$$\begin{gathered} \overline{p}=\frac{\left(x_1+x_2\right)}{\left(n_1+n_2\right)} \\ =\frac{\left(40+88\right)}{\left(45+103\right)} \\ =0.8649 \end{gathered}$$

and

$$\begin{gathered} \overline{q}=1-\overline{p} \\ =1-0.8649 \\ =0.1351 \end{gathered}$$

Define the test statistic

To conduct a hypothesis test of two proportions, the test statistic is computed as

$z=\frac{\left({\hat{p}}_1-{\hat{p}}_2\right)-\left(p_1-p_2\right)}{\sqrt{\left(\frac{\overline{p}\overline{q}}{n_1}+\frac{\overline{p}\overline{q}}{n_2}\right)}}$

Here, $\overline{p}$ is the pooled sample proportion, and $\overline{q}=1-\overline{p}$

Substitute the values. So,

$$\begin{gathered} z=\frac{\left({\hat{p}}_1-{\hat{p}}_2\right)-\left(p_1-p_2\right)}{\sqrt{\left(\frac{\overline{p}\overline{q}}{n_1}+\frac{\overline{p}\overline{q}}{n_2}\right)}} \\ =\frac{\left(0.8889-0.8544\right)-0}{\sqrt{\left(\frac{0.8649\times 0.1351}{45}+\frac{0.8649\times 0.1351}{103}\right)}} \\ =0.565 \end{gathered}$$

The value of the test statistic is 0.565.

Find the p-value 

Referring to the standard normal table for the positive z-score of 0.7157, the cumulative probability of 0.57 is obtained from the cell intersection for row 0.5 and the column value of 0.07.

For the two-tailed test, the p-value is twice the area to the right of the test statistic, as shown below.

$$\begin{gathered} 2\times P\left(Z>0.57\right)=2\times \left(1-P\left(Z<0.57\right)\right) \\ =2\times \left(1-0.7157\right) \\ =0.5686 \end{gathered}$$

.

Thus, the p-value is 0.569.

State the decision 

As the p-value is greater than 0.05, the null hypothesis is failed to be rejected. Thus, there is not enough evidence to support the claim that Echinacea is effective on rhinovirus infections.

Describe the confidence interval

b.

The general formula for the confidence interval of the difference of proportions is as follows.

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

Here, E is the margin of error calculated as follows.

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

Find the confidence interval

For the two-tailed test with a significance level of 0.05, the confidence level will be 95%.

The critical value$z_{\frac{\alpha }{2}}$has the cumulative area to its left as $1-\frac{\alpha }{2}$.

Mathematically,

$$\begin{gathered} P\left(Z<z_{\frac{\alpha }{2}}\right)=1-\frac{\alpha }{2} \\ P\left(Z<z_{0.025}\right)=0.975 \end{gathered}$$

From the standard normal table, the area of 0.975 is observed corresponding to the intersection of the row value 1.9 and the column value 0.06 as 1.96.

The margin of error is as follows.

$$\begin{gathered} \text{E}=z_{\frac{\alpha }{2}}\times \sqrt{\left(\frac{{\hat{p}}_1\times {\hat{q}}_1}{n_1}+\frac{{\hat{p}}_2\times {\hat{q}}_2}{n_2}\right)} \\ =1.96\times \sqrt{\left(\frac{0.8889\times 0.1111}{45}+\frac{0.8544\times 0.1456}{103}\right)} \\ =0.1143 \end{gathered}$$

Substitute the value of E as follows.

$$\begin{gathered} \text{Confidence} \text{Interval}=\left(\left({\hat{p}}_1-{\hat{p}}_2\right)-E , \left({\hat{p}}_1-{\hat{p}}_2\right)+E\right) \\ =\left(\left(0.8889-0.8544\right)-0.1143 , \left(0.8889-0.8544\right)+0.1143\right) \\ =\left(-0.0798 ,\hspace{0.17em}0.149\right) \end{gathered}$$

.

Thus, the 95% confidence interval for two proportions is $-0.0798<\left(p_1-p_2\right)<0.149$.

As 0 is included in the interval, there is not sufficient evidence to support the effectiveness of Echinacea.

Conclude the results

c.

From the results, Echinacea does not appear to have a significant effect on reducing the infection rate of rhinovirus.