/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 79 The Salk polio vaccine experimen... [FREE SOLUTION] | 91Ó°ÊÓ

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The Salk polio vaccine experiment in 1954 focused on the effectiveness of the vaccine in combating paralytic polio. Because it was felt that without a control group of children there would be no sound basis for evaluating the efficacy of the Salk vaccine, the vaccine was administered to one group, and a placebo (visually identical to the vaccine but known to have no effect) was administered to a second group. For ethical reasons, and because it was suspected that knowledge of vaccine administration would affect subsequent diagnoses, the experiment was conducted in a doubleblind fashion. That is, neither the subjects nor the administrators knew who received the vaccine and who received the placebo. The actual data for this experiment are as follows: Placebo group: \(n=201,299: 110\) cases of polio observed Vaccine group: \(n=200,745: 33\) cases of polio observed (a) Use a hypothesis-testing procedure to determine if the proportion of children in the two groups who contracted paralytic polio is statistically different. Use a probability of a type I error equal to \(0.05 .\) (b) Repeat part (a) using a probability of a type I error equal to 0.01 (c) Compare your conclusions from parts (a) and (b) and explain why they are the same or different.

Short Answer

Expert verified
The Salk vaccine significantly reduces paralytic polio incidence at both 0.05 and 0.01 levels.

Step by step solution

01

Identify Hypotheses

Let's define the null hypothesis, \( H_0 \), and the alternative hypothesis, \( H_a \). \( H_0: p_1 = p_2 \) (the proportion of polio cases in placebo and vaccine groups are equal) \( H_a: p_1 eq p_2 \) (the proportions are not equal) where \( p_1 \) is the proportion of the placebo group and \( p_2 \) is the proportion of the vaccine group.
02

Calculate Sample Proportions

Calculate the sample proportions of polio cases for both groups. For the placebo group: \( \hat{p}_1 = \frac{110}{201299} \approx 0.000546 \) For the vaccine group: \( \hat{p}_2 = \frac{33}{200745} \approx 0.000164 \)
03

Combined Sample Proportion

Calculate the combined sample proportion (\( \hat{p} \)), which is used for calculating the standard error. \( \hat{p} = \frac{110 + 33}{201299 + 200745} = \frac{143}{402044} \approx 0.000356 \)
04

Standard Error Calculation

Calculate the standard error (SE) of the difference in proportions. \[ SE = \sqrt{ \hat{p}(1 - \hat{p})(\frac{1}{n_1} + \frac{1}{n_2}) } \] Substitute values: \[ SE = \sqrt{0.000356 \times (1-0.000356) \times \left(\frac{1}{201299} + \frac{1}{200745}\right)} \approx 0.000048 \]
05

Test Statistic Calculation

Calculate the test statistic (z) using the formula: \[ z = \frac{\hat{p}_1 - \hat{p}_2}{SE} \] Insert the calculated values: \[ z = \frac{0.000546 - 0.000164}{0.000048} \approx 7.958 \]
06

Determine Critical Value at \( \\alpha = 0.05 \)

For a two-tailed test with \( \alpha = 0.05 \), the critical z-values are \( \pm 1.96 \). Compare the test statistic with these values to test the hypothesis.
07

Make Decision for \( \\alpha = 0.05 \)

Since the calculated z-value \( 7.958 \) is greater than \( 1.96 \), reject the null hypothesis \( H_0 \). There is enough evidence to say that the proportions are significantly different.
08

Determine Critical Value at \( \\alpha = 0.01 \)

For a two-tailed test with \( \alpha = 0.01 \), the critical z-values are \( \pm 2.576 \). Again, compare the test statistic with these values.
09

Make Decision for \( \\alpha = 0.01 \)

Since the calculated z-value \( 7.958 \) is greater than \( 2.576 \), reject \( H_0 \) again. There is strong evidence that the proportions are statistically different at this level as well.
10

Comparison of Part (a) and Part (b)

In both cases \((\alpha = 0.05 \text{ and } \alpha = 0.01)\), \( H_0 \) was rejected. The conclusion is consistent, indicating that the Salk vaccine significantly reduces the incidence of paralytic polio, regardless of the level of significance chosen because the p-value is very low compared to both significance levels.

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Key Concepts

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

Proportion Difference
When comparing two groups, understanding the difference in proportions is crucial. In the context of the Salk polio vaccine experiment, we are interested in the proportions of children who contracted polio in both the placebo and the vaccine groups. The proportion difference helps us measure how effective the vaccine is in reducing the incidence of polio compared to the placebo.We compute these proportions as the number of cases observed divided by the total number of participants in each group. For the placebo group, the proportion is given by \( \hat{p}_1 = \frac{110}{201299} \), and for the vaccine group, it is \( \hat{p}_2 = \frac{33}{200745} \). The difference in these proportions is critical as it provides a numerical value representing the vaccine's impact. If these proportions are significantly different, it suggests that the vaccine has a real effect.
Type I Error
In hypothesis testing, a Type I error occurs when we wrongly reject a true null hypothesis. It's crucial to understand this concept because it essentially means we're claiming there is an effect or difference when there truly isn't one.The probability of committing a Type I error is denoted by \( \alpha \), the significance level. In the polio vaccine example, tests are conducted with \( \alpha = 0.05 \) and \( \alpha = 0.01 \). This probability defines our tolerance for deciding that the vaccine has an effect when it might not be true. A smaller \( \alpha \) (like 0.01) is more stringent than larger \( \alpha \) (like 0.05), meaning we require more evidence to claim significance and reduce the chance of making a Type I error.
Standard Error
The standard error measures the variability or the "average" distance between a sample statistic and the population parameter. It helps us understand the confidence we have in the difference we see between sample proportions.For the polio vaccine experiment, the standard error of the proportion difference is calculated using:\[SE = \sqrt{ \hat{p}(1 - \hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}\]Here, \( \hat{p} \) is the combined sample proportion, representing the overall proportion of polio cases across both groups. The standard error is critical because it is used to determine the test statistic, helping us assess if the observed difference is statistically significant.
Null and Alternative Hypothesis
Setting hypotheses is the foundation of hypothesis testing. In the polio vaccine trial, we start by defining a null hypothesis \( H_0 \) and an alternative hypothesis \( H_a \).- **Null Hypothesis \( H_0 \):** States that there is no difference in the proportion of polio cases between the vaccine and placebo groups, mathematically presented as \( p_1 = p_2 \).- **Alternative Hypothesis \( H_a \):** States that there is a difference, implying \( p_1 eq p_2 \).These hypotheses guide the analysis and conclusions. Rejecting or failing to reject \( H_0 \) informs us about the vaccine's effectiveness.
Z-test
The Z-test is a statistical method to determine if there is a significant difference between two population proportions. It assesses how far away the sample proportion difference is from what the null hypothesis claims, in units of the standard error.For our case study, the Z-test statistic is calculated using:\[z = \frac{\hat{p}_1 - \hat{p}_2}{SE}\]The test statistic is then compared to critical z-values determined by the significance level \( \alpha \). For example, at \( \alpha = 0.05 \), the critical z-values are \( \pm 1.96 \). If the calculated \( z \) is beyond these values, we reject \( H_0 \). In our vaccine experiment, the calculated \( z \) indicates a significant result, as its value is much larger than the critical values, strongly suggesting that the vaccine is effective.

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Most popular questions from this chapter

A study was performed to determine whether men and women differ in their repeatability in assembling components on printed circuit boards. Random samples of 25 men and 21 women were selected, and each subject assembled the units. The two sample standard deviations of assembly time were \(s_{\text {men }}=0.98\) minutes and \(s_{\text {women }}=1.02\) minutes. (a) Is there evidence to support the claim that men and women differ in repeatability for this assembly task? Use \(\alpha=\) 0.02 and state any necessary assumptions about the underlying distribution of the data. (b) Find a \(98 \%\) confidence interval on the ratio of the two variances. Provide an interpretation of the interval.

Consider the hypothesis test \(H_{0}: \mu_{1}=\mu_{2}\) against \(H_{1}: \mu_{1}<\mu_{2} .\) Suppose that sample sizes \(n_{1}=15\) and \(n_{2}=15\) that \(\bar{x}_{1}=6.2\) and \(\bar{x}_{2}=7.8,\) and that \(s_{1}^{2}=4\) and \(s_{2}^{2}=6.25\) Assume that \(\sigma_{1}^{2}=\sigma_{2}^{2}\) and that the data are drawn from normal distributions. Use \(\alpha=0.05 .\). (a) Test the hypothesis and find the \(P\) -value. (b) Explain how the test could be conducted with a confidence interval. (c) What is the power of the test in part (a) if \(\mu_{1}\) is 3 units less than \(\mu_{2} ?\) (d) Assuming equal sample sizes, what sample size should be used to obtain \(\beta=0.05\) if \(\mu_{1}\) is 2.5 units less than \(\mu_{2} ?\) Assume that \(\alpha=0.05\)

An electrical engineer must design a circuit to deliver the maximum amount of current to a display tube to achieve sufficient image brightness. Within her allowable design constraints, she has developed two candidate circuits and tests prototypes of each. The resulting data (in microamperes) are as follows: $$\begin{array}{l|l}\text { Circuit } 1: & 251,255,258,257,250,251,254,250,248 \\\\\hline \text { Circuit } 2: & 250.253 .249 .256 .259 .252 .260 .251\end{array}$$ (a) Use the Wilcoxon rank-sum test to test \(H_{0}: \mu_{1}=\mu_{2}\) against the alternative \(H_{1}: \mu_{1}>\mu_{2} .\) Use \(\alpha=0.025 .\) (b) Use the normal approximation for the Wilcoxon rank-sum test. Assume that \(\alpha=0.05 .\) Find the approximate \(P\) -value for this test statistic.

Two machines are used to fill plastic bottles with dishwashing detergent. The standard deviations of fill volume are known to be \(\sigma_{1}=0.10\) fluid ounces and \(\sigma_{2}=0.15\) fluid ounces for the two machines, respectively. Two random samples of \(n_{1}=12\) bottles from machine 1 and \(n_{2}=10\) bottles from machine 2 are selected, and the sample mean fill volumes are \(\bar{x}_{1}=30.87\) fluid ounces and \(\bar{x}_{2}=30.68\) fluid ounces. Assume normality. (a) Construct a \(90 \%\) two-sided confidence interval on the mean difference in fill volume. Interpret this interval. (b) Construct a \(95 \%\) two-sided confidence interval on the mean difference in fill volume. Compare and comment on the width of this interval to the width of the interval in part (a). (c) Construct a \(95 \%\) upper-confidence interval on the mean difference in fill volume. Interpret this interval. (d) Test the hypothesis that both machines fill to the same mean volume. Use \(\alpha=0.05 .\) What is the \(P\) -value? (e) If the \(\beta\) -error of the test when the true difference in fill volume is 0.2 fluid ounces should not exceed \(0.1,\) what sample sizes must be used? Use \(\alpha=0.05 .\)

One of the authors travels regularly to Seattle, Washington. He uses either Delta or Alaska. Flight delays are sometimes unavoidable, but he would be willing to give most of his business to the airline with the best on-time arrival record. The number of minutes that his flight arrived late for the last six trips on each airline follows. Is there evidence that either airline has superior on-time arrival performance? Use \(\alpha=0.01\) and the Wilcoxon rank-sum test. $$ \begin{array}{l|l} \text { Delta: } & 13,10,1,-4, \quad 0,9 \text { (minutes late) } \\ \hline \text { Alaska: } & 15,8,3,-1,-2,4 \text { (minutes late) } \end{array} $$

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