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The drug Prevnar is a vaccine meant to prevent certain types of bacterial meningitis, typically administered to infants around 2 months. In randomized, double-blind clinical trials of Prevnar, infants were randomly divided into two groups. Subjects in group 1 received Prevnar, while subjects in group 2 received a control vaccine. After the first dose, 107 of 710 subjects in the experimental group (group 1) experienced fever as a side effect. After the first dose, 67 of 611 of the subjects in the control group (group 2) experienced fever as a side effect. Does the evidence suggest that a higher proportion of subjects in group 1 experienced fever as a side effect than subjects in group 2 at the a = 0.05 level of significance?

Step by step solution

01

Define the Hypotheses

First, define the null and alternative hypotheses. Null hypothesis (H_0): \(p_1 = p_2\), meaning the proportion of infants experiencing fever in both groups is the same.Alternative hypothesis (H_a): \(p_1 > p_2\), meaning the proportion of infants experiencing fever in group 1 is higher than in group 2.

02

Calculate the Sample Proportions

Calculate the sample proportions for both groups.Group 1: \(\hat{p_1} = \frac{107}{710} \approx 0.1507\)Group 2: \(\hat{p_2} = \frac{67}{611} \approx 0.1096\)

03

Compute the Pooled Proportion

Calculate the pooled sample proportion:\[\hat{p} = \frac{107 + 67}{710 + 611} \approx 0.1311\]

04

Find the Standard Error

Compute the standard error (SE) for the difference in proportions using the pooled proportion.\[SE = \sqrt{\hat{p}(1 - \hat{p})(\frac{1}{n_1} + \frac{1}{n_2})} \approx 0.0202\]

05

Calculate the Test Statistic

Determine the test statistic (z) for the difference in proportions.\[z = \frac{\hat{p_1} - \hat{p_2}}{SE} = \frac{0.1507 - 0.1096}{0.0202} \approx 2.04\]

06

Determine the Critical Value and Compare

Find the critical value for \(\alpha = 0.05\) (one-tailed test). The critical value is about 1.645. Compare the test statistic to the critical value.2.04 > 1.645

07

Make a Decision

Since the test statistic is greater than the critical value, we reject the null hypothesis. This suggests that there is enough evidence at the \(\alpha = 0.05\) level to conclude that a higher proportion of subjects in group 1 experienced fever as a side effect than subjects in group 2.

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Key Concepts

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

Null Hypothesis

The null hypothesis, denoted as \(H_0\), is the statement we want to test in a hypothesis testing scenario. It represents the idea that there is no effect or no difference. In this exercise, the null hypothesis states that the proportion of infants experiencing fever is the same in both groups: \(H_0: p_1 = p_2\). This means we assume initially that the fever side effect occurs equally in both groups.

Alternative Hypothesis

The alternative hypothesis, denoted as \(H_a\), is the statement we consider if the null hypothesis is rejected. It indicates the presence of an effect or a difference. For this problem, the alternative hypothesis states that the proportion of infants with fever is higher in the group receiving Prevnar compared to the control group: \(H_a: p_1 > p_2\). This suggests that the vaccine has a higher likelihood of causing fever.

Pooled Proportion

The pooled proportion is used when comparing two proportions. It is a weighted average of the individual sample proportions, assuming the null hypothesis is true. The formula for the pooled proportion \(\hat{p}\) is:

\[ \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} \]
where \(x_1\) and \(x_2\) are the number of successes (fever cases) in each group, and \(n_1\) and \(n_2\) are the sample sizes. In this exercise, the pooled proportion comes out to approximately 0.1311, indicating the overall proportion of infants experiencing fever across both groups.

Standard Error

The standard error (SE) measures the variability of the difference between sample proportions. It indicates how much the sample proportions would fall due to random chance. The formula for SE with pooled proportion is:

\[ SE = \sqrt{\hat{p}(1 - \hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)} \]
For this exercise, we substitute the values to get approximately 0.0202. This helps us understand the expected variation in the difference between the two group proportions.

Test Statistic

The test statistic (z) evaluates how far the sample proportion difference is from the hypothesized difference, given the standard error. It's calculated using:

\[ z = \frac{\hat{p_1} - \hat{p_2}}{SE} \]
In this case, substituting the sample proportions and the calculated SE gives around 2.04. Comparing this test statistic to the critical value (for example, 1.645 for a one-tailed test with \(\alpha = 0.05\)), helps us determine if the observed difference is statistically significant. Since 2.04 > 1.645, we conclude there is significant evidence to reject the null hypothesis.