91Ó°ÊÓ

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. In a randomized controlled trial in Kenya, insecticide-treated bednets were tested as a way to reduce malaria. Among 343 infants using bednets, 15 developed malaria. Among 294 infants not using bednets, 27 developed malaria (based on data from "Sustainability of Reductions in Malaria Transmission and Infant Mortality in Western Kenya with Use of Insecticide-Treated Bednets," by Lindblade et al., Journal of the American Medical Association, Vol. 291, No. 21 ). We want to use a \(0.01\) significance level to test the claim that the incidence of malaria is lower for infants using bednets. a. Test the claim using a hypothesis test. b. Test the claim by constructing an appropriate confidence interval. c. Based on the results, do the bednets appear to be effective?

Step by step solution

01

- Identify the Hypotheses

State the null hypothesis (H_0) and the alternative hypothesis (H_a). The null hypothesis typically represents a statement of no effect or no difference. Here:\[ H_0: p_1 = p_2 \](This means that the proportion of infants developing malaria is the same for those using bednets and those not using bednets.)The alternative hypothesis reflects the claim we are testing:\[ H_a: p_1 < p_2 \](This means that the proportion of infants developing malaria is lower for those using bednets.)

02

- Calculate Proportions

Calculate the proportions of infants who developed malaria in both groups.For infants using bednets:\[ \hat{p}_1 = \frac{15}{343} \approx 0.044 \]For infants not using bednets:\[ \hat{p}_2 = \frac{27}{294} \approx 0.092 \]

03

- Calculate the Test Statistic

To perform a two-proportion z-test, use the formula for the test statistic:\[ z = \frac{(\hat{p}_1 - \hat{p}_2)}{\sqrt{\hat{p}(1 - \hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}} \]where \( \hat{p} \) is the pooled sample proportion:\[ \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{15 + 27}{343 + 294} \approx 0.066 \]Substitute the values:\[ z = \frac{0.044 - 0.092}{\sqrt{0.066(1 - 0.066)(\frac{1}{343} + \frac{1}{294})}} \approx -2.45 \]

04

- Determine the P-value

Using the standard normal distribution, find the P-value for the test statistic. For \( z = -2.45 \):Check the value in the standard normal (z) table or use a calculator. The P-value is approximately 0.0071.

05

- Compare P-value to Significance Level

The significance level is given as 0.01. If the P-value is less than the significance level, we reject the null hypothesis:\( 0.0071 < 0.01 \)So, reject the null hypothesis.

06

- Construct the Confidence Interval

Compute the confidence interval for the difference in proportions using the following formula for a 99% confidence level:\[ (\hat{p}_1 - \hat{p}_2) \pm z_{\alpha/2} \sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}} \]where \( z_{\alpha/2} \) for 99% confidence level is approximately 2.576.Insert the values:\[ (0.044 - 0.092) \pm 2.576 \sqrt{\frac{0.044(1 - 0.044)}{343} + \frac{0.092(1 - 0.092)}{294}} \approx -0.048 \pm 0.036 \]Thus, the confidence interval is approximately [-0.084, -0.012].

07

- State Final Conclusions

Since the confidence interval does not include 0 and the null hypothesis was rejected, we conclude that there is significant evidence to support the claim that the incidence of malaria is lower for infants using bednets. Bednets appear to be effective in reducing the incidence of malaria in infants.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Null Hypothesis

A null hypothesis, often represented by \(H\text{_}0\), is a statement that there is no effect or no difference in the context of the problem being tested. It is the hypothesis that researchers try to disprove. For the given exercise, the null hypothesis is that the proportion of infants developing malaria is the same whether they use bednets or not. Mathematically, it is expressed as \[ H_0: p_1 = p_2 \] where \(p_1\) is the proportion of infants using bednets who develop malaria and \(p_2\) is the proportion of infants not using bednets who develop malaria. By starting with the assumption that there is no difference, we aim to find evidence against this claim.

Alternative Hypothesis

The alternative hypothesis, represented by \(H\text{_}a\), is a statement that indicates the presence of an effect or difference. It is what we aim to support with our data. In the given exercise, the alternative hypothesis is that the proportion of infants developing malaria is lower for those using bednets compared to those who do not use them. Mathematically, it is expressed as \[ H_a: p_1 < p_2 \] This alternative hypothesis reflects the original claim that insecticide-treated bednets are effective in reducing the incidence of malaria among infants.

Test Statistic

To test the hypotheses, we calculate a test statistic, which helps us determine how far our sample data is from the null hypothesis. In this exercise, we use a two-proportion z-test, where the test statistic is calculated using the formula: \[ z = \frac{(\hat{p}_1 - \hat{p}_2)}{\sqrt{\hat{p}(1 - \hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}} \] Here, \(\hat{p}\) is the pooled sample proportion obtained by combining the data from both groups. It is calculated as: \[ \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} \] where \(x_1\) and \(x_2\) are the number of malaria cases in each group, and \(n_1\) and \(n_2\) are the total number of infants in each group. Using the values given, we obtain a test statistic of approximately \(z = -2.45\).

P-value

The P-value helps us determine the significance of our test statistic. It represents the probability that we would observe a test statistic as extreme as the one calculated, assuming the null hypothesis is true. For a test statistic \(z = -2.45\), the P-value is approximately 0.0071. To make a decision, we compare the P-value with our given significance level (\(\alpha = 0.01\)): \[ \text{If } P\text{-value} < \alpha , \text{ reject } H_0 \] In this case, 0.0071 is less than 0.01, so we reject the null hypothesis. This means there is strong evidence to support the alternative hypothesis that bednets reduce the incidence of malaria among infants.

Confidence Interval

A confidence interval provides a range of values within which we expect the true difference in proportions to lie, with a certain degree of confidence. For a 99% confidence level, the confidence interval is calculated using the formula: \[ (\hat{p}_1 - \hat{p}_2) \pm z_{\alpha/2} \sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2 (1 - \hat{p}_2)}{n_2}} \] where \(z_{\alpha/2}\) is the critical value for the standard normal distribution at the desired confidence level. Plugging in the values, we get an interval of approximately [-0.084, -0.012]. Because this interval does not contain 0, it supports the conclusion that there is a significant difference between the two proportions. Thus, the evidence supports that bednets are indeed effective in reducing malaria incidence.