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Chapter 11: Problem 9

In Problems 9鈥�12, conduct each test at the a = 0.05 level of significance by determining (a) the null and alternative hypotheses, (b) the test statistic, (c) the critical value, and (d) the P-value. Assume that the samples were obtained independently using simple random sampling. Test whether \(p_{1}>p_{2}\). Sample data: \(x_{1}=368, n_{1}=541\), \(x_{2}=351, n_{2}=593\)

Short Answer

Expert verified
Reject the null hypothesis because the P-value (0.0066) is less than 0.05, suggesting \( p_1 > p_2 \).

Step by step solution

01

Identify Null and Alternative Hypotheses

Define the null hypothesis and alternative hypothesis. Let鈥檚 denote the proportions as follows: - Null hypothesis, \( H_0 \): \( p_1 \le p_2 \)- Alternative hypothesis, \( H_a \): \( p_1 > p_2 \)
02

Calculate the sample proportions

The sample proportions can be calculated as follows: \( \hat{p}_1 = \frac{x_1}{n_1} \) and \( \hat{p}_2 = \frac{x_2}{n_2} \). - \( \hat{p}_1 = \frac{368}{541} \approx 0.680 \)- \( \hat{p}_2 = \frac{351}{593} \approx 0.592 \)
03

Determine the test statistic

The formula for the test statistic when comparing two proportions is given by \[ z = \frac{(\hat{p}_1 - \hat{p}_2)}{\sqrt{ \hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right) }} \]where \( \hat{p} \) is the pooled sample proportion: \( \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} \).First, calculate the pooled sample proportion: - \( \hat{p} = \frac{368 + 351}{541 + 593} = \frac{719}{1134} \approx 0.634 \)Next, plug in the values to get the test statistic: - \( z = \frac{(0.680 - 0.592)}{\sqrt{0.634 \times (1 - 0.634) \times \left(\frac{1}{541} + \frac{1}{593}\right)}} \approx 2.48 \)
04

Determine the critical value

For a significance level of \( \alpha = 0.05 \) for a one-tailed test (right-tail), refer to the z-table to find that the critical value \( z_{\alpha} = 1.645 \).
05

Calculate the P-value

Using the z-table, find the P-value corresponding to the test statistic \( z = 2.48 \). The P-value is the probability that a standard normal variable is greater than 2.48. From the z-table, P-value \( \approx 0.0066 \).
06

Compare P-value to significance level

Compare the P-value to the significance level \( \alpha = 0.05 \). Since \(0.0066 < 0.05\), we reject the null hypothesis.
07

Conclusion

There is sufficient evidence to conclude that \( p_1 > p_2 \) at the 0.05 level of significance.

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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, often denoted as \(H_0\), is a statement that there is no effect or no difference. In hypothesis testing, it's what we assume to be true until we have evidence to suggest otherwise. For the given exercise, the null hypothesis is \(p_1 \leq p_2\). This means that we initially assume the proportion of success in the first sample (\(p_1\)) is less than or equal to the proportion of success in the second sample (\(p_2\)). We need significant evidence from the data to reject this null hypothesis.
alternative hypothesis
Opposite to the null hypothesis, the alternative hypothesis, denoted as \(H_a \), proposes what we are trying to demonstrate. For our problem, the alternative hypothesis is \(p_1 > p_2\). This suggests that the proportion of success in the first sample is greater than in the second sample. In hypothesis testing, proving the alternative hypothesis means we have strong evidence that there is a significant effect or difference.
test statistic
A test statistic is a standardized value used to decide whether to reject the null hypothesis. It's calculated from your sample data. In our example, the test statistic (z) is determined using a specific formula that incorporates the sample proportions and the pooled proportion. The formula for comparing two proportions is: \[ z = \frac{(\hat{p}_1 - \hat{p}_2)}{\sqrt{ \hat{p}(1 - \hat{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right) }} \] Here, we first calculate the pooled proportion (\(\hat{p}\)), then substitute the values into the formula to get the test statistic. In our case, it鈥檚 about 2.48.
critical value
The critical value is a threshold that the test statistic must exceed to reject the null hypothesis. It's determined by the significance level (alpha) and the type of test (one-tailed or two-tailed). For a significance level of 0.05 and a one-tailed test, the critical value for z is 1.645. If our test statistic (z) is greater than this critical value, we reject the null hypothesis. Since our calculated test statistic is 2.48, which is greater than 1.645, we have enough evidence against the null hypothesis.
P-value
The P-value represents the probability of obtaining a test statistic at least as extreme as the one computed, assuming the null hypothesis is true. In simpler terms, it helps us measure the strength of the evidence against the null hypothesis. For our exercise, the P-value corresponding to the test statistic (z = 2.48) is approximately 0.0066. We compare this P-value with our significance level (0.05). Since 0.0066 is much smaller than 0.05, we reject the null hypothesis. This small P-value indicates strong evidence that \(p_1 > p_2\).

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Most popular questions from this chapter

Clifford Adelman, a researcher with the Department of Education, followed a cohort of students who graduated from high school in \(1992,\) monitoring the progress the students made toward completing a bachelor's degree. One aspect of his research was to compare students who first attended a community college to those who immediately attended and remained at a four-year institution. The sample standard deviation time to complete a bachelor's degree of the 268 students who transferred to a four-year school after attending community college was \(1.162 .\) The sample standard deviation time to complete a bachelor's degree of the 1145 students who immediately attended and remained at a four-year institution was \(1.015 .\) Assuming the time to earn a bachelor's degree is normally distributed, does the evidence suggest the standard deviation time to earn a bachelor's degree is different between the two groups? Use the \(\alpha=0.05\) level of significance.

Explain the difference between an independent and dependent sample.

Walking in the Airport, Part I Do people walk faster in the airport when they are departing (getting on a plane) or when they are arriving (getting off a plane)? Researcher Seth B. Young measured the walking speed of travelers in San Francisco International Airport and Cleveland Hopkins International Airport. His findings are summarized in the table. $$ \begin{array}{lcc} \text { Direction of Travel } & \text { Departure } & \text { Arrival } \\ \hline \text { Mean speed (feet per minute) } & 260 & 269 \\ \hline \begin{array}{l} \text { Standard deviation } \\ \text { (feet per minute) } \end{array} & 53 & 34 \\ \hline \text { Sample size } & 35 & 35 \\ \hline \end{array} $$ (a) Is this an observational study or a designed experiment? Why? (b) Explain why it is reasonable to use Welch's \(t\) -test. (c) Do individuals walk at different speeds depending on whether they are departing or arriving at the \(\alpha=0.05\) level of significance?

Walking in the Airport, Part II Do business travelers walk at a different pace than leisure travelers? Researcher Seth B. Young measured the walking speed of business and leisure travelers in San Francisco International Airport and Cleveland Hopkins International Airport. His findings are summarized in the table. $$ \begin{array}{lcc} \text { Type of Traveler } & \text { Business } & \text { Leisure } \\ \hline \text { Mean speed (feet per minute) } & 272 & 261 \\ \hline \begin{array}{l} \text { Standard deviation (feet per } \\ \text { minute) } \end{array} & 43 & 47 \\ \hline \text { Sample size } & 20 & 20 \\ \hline \end{array} $$ (a) Is this an observational study or a designed experiment? Why? (b) What must be true regarding the populations to use Welch's \(t\) -test to compare the means? (c) Assuming that the requirements listed in part (b) are satisfied, determine whether business travelers walk at a different speed from leisure travelers at the \(\alpha=0.05\) level of significance.

On April \(12,1955,\) Dr. Jonas Salk released the results of clinical trials for his vaccine to prevent polio. In these clinical trials, 400,000 children were randomly divided in two groups. The subjects in group 1 (the experimental group) were given the vaccine, while the subjects in group 2 (the control group) were given a placebo. Of the 200,000 children in the experimental group, 33 developed polio. Of the 200,000 children in the control group, 115 developed polio. (a) What type of experimental design is this? (b) What is the response variable? (c) What are the treatments? (d) What is a placebo? (e) Why is such a large number of subjects needed for this study? (f) Does it appear to be the case that the vaccine was effective?
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