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The null and alternate hypotheses are: $$ \begin{array}{l} H_{0}: \pi_{1}=\pi_{2} \\ H_{1}: \pi_{1} \neq \pi_{2} \end{array} $$ A sample of 200 observations from the first population indicated that \(x_{1}\) is \(170 .\) A sample of 150 observations from the second population revealed \(x_{2}\) to be \(110 .\) Use the .05 significance level to test the hypothesis. a. State the decision rule. b. Compute the pooled proportion. c. Compute the value of the test statistic. d. What is your decision regarding the null hypothesis?

Short Answer

Expert verified
Reject the null hypothesis; the proportions differ significantly.

Step by step solution

01

Understanding the Decision Rule

Since the alternate hypothesis is that the proportions are not equal, \(H_1: \pi_1 eq \pi_2\), this is a two-tailed test. Using a 0.05 significance level, we look for critical values that cover a total of 5% of both tails. The critical values for a two-tailed test at 0.05 significance are typically \( \pm 1.96\) for z-scores. The decision rule is to reject \(H_0\) if \(Z < -1.96\) or \(Z > 1.96\).
02

Calculate Pooled Proportion

The pooled sample proportion \( \hat{p} \) is calculated using the formula:\[\hat{p} = \frac{x_1 + x_2}{n_1 + n_2}\]Given \(x_1 = 170\), \(x_2 = 110\), \(n_1 = 200\), and \(n_2 = 150\), the pooled proportion is:\[\hat{p} = \frac{170 + 110}{200 + 150} = \frac{280}{350} \approx 0.8\]
03

Compute the Test Statistic

The test statistic \( Z \) is computed using the formula:\[Z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}\]Where \(\hat{p}_1 = \frac{x_1}{n_1}\) and \(\hat{p}_2 = \frac{x_2}{n_2}\). First, calculate \(\hat{p}_1\) and \(\hat{p}_2\):\[\hat{p}_1 = \frac{170}{200} = 0.85, \quad \hat{p}_2 = \frac{110}{150} \approx 0.7333\]Plugging into the formula, we have:\[Z = \frac{0.85 - 0.7333}{\sqrt{0.8(1-0.8)\left(\frac{1}{200} + \frac{1}{150}\right)}}\]Calculating further gives:\[Z \approx \frac{0.1167}{\sqrt{0.8 \times 0.2 \times (0.005 + 0.0067)}} \approx \frac{0.1167}{\sqrt{0.8 \times 0.2 \times 0.0117}} \approx \frac{0.1167}{0.0365} \approx 3.195\]
04

Make the Decision

The calculated test statistic \( Z \approx 3.195 \) is greater than the critical value of 1.96. Since \( Z > 1.96 \), we reject the null hypothesis \( H_0 \). This indicates there is a significant difference between the population proportions.

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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 is a statement that suggests no effect or no difference between populations. It is the starting point for hypothesis testing, where we aim to determine if there is enough statistical evidence to reject this assumption. In hypothesis testing, researchers often assume the null hypothesis is true until proven otherwise. This forms the basis for evaluating the observed data. In the exercise, the null hypothesis is stated as \(H_0: \pi_1 = \pi_2\), meaning there is no difference between the two population proportions. This assumption simplifies analysis by providing a specific frame of reference.
Alternate Hypothesis
The alternate hypothesis (or alternative hypothesis) is the statement that contradicts the null hypothesis. It represents what the researcher aims to support or prove through data analysis. It's often denoted as \(H_1\), and in the given exercise, it is \(H_1: \pi_1 eq \pi_2\). This indicates a belief that there is a difference between the two population proportions. When conducting a hypothesis test, if the evidence strongly suggests the null hypothesis is unlikely, the conclusion is to reject the null hypothesis in favor of the alternate hypothesis. This forms the basis for the decision-making process in hypothesis testing.
Z-Test
A z-test is a statistical test used to determine if there is a significant difference between the means or proportions of two groups. It utilizes the z-distribution, which is a special case of the normal distribution, typically when sample sizes are large. Z-tests are useful for testing the hypothesis about population proportions where the sample size is sufficient for the normal approximation of the binomial distribution. In a z-test for proportions, the test statistic is calculated to compare an observed sample proportion with an expected one under the null hypothesis. In our exercise, the calculated z-statistic was approximately 3.195. It was evaluated against the critical value for a two-tailed test at a 0.05 significance level, which is ±1.96. Since the calculated statistic exceeds this threshold, it suggests that the observed difference is unlikely to have occurred by random chance.
Pooled Proportion
Pooled proportion refers to the combined proportion from two samples. It is a useful measure when comparing two population proportions, as it provides a way to standardize these proportions into a single value. The pooled proportion is calculated using the formula:
  • \( \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} \)
where \(x_1\) and \(x_2\) are the number of successes in the two samples, and \(n_1\) and \(n_2\) are their respective sample sizes. In the exercise, the pooled proportion was calculated as approximately 0.8. It is important because it allows for the estimation of the standard error and forms part of the z-test calculation. By providing an expected proportion under the null hypothesis, it helps to identify whether the observed differences between sample proportions are statistically significant.

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Most popular questions from this chapter

A recent article reported that a job awaits only one in three new college graduates. The major reasons given were an overabundance of college graduates and a weak economy. A survey of 200 recent graduates from your school revealed that 80 students had jobs. At the .01 significance level, can we conclude that a larger proportion of students at your school have jobs?

The null and alternate hypotheses are: $$ \begin{array}{l} H_{0}: \pi_{1} \leq \pi_{2} \\ H_{1}: \pi_{1}>\pi_{2} \end{array} $$ A sample of 100 observations from the first population indicated that \(x_{1}\) is \(70 .\) A sample of 150 observations from the second population revealed \(x_{2}\) to be \(90 .\) Use the .05 significance level to test the hypothesis. a. State the decision rule. b. Compute the pooled proportion. c. Compute the value of the test statistic. d. What is your decision regarding the null hypothesis?

The Consumer Confidence Survey is a monthly review that measures consumer confidence in the U.S. economy. It is based on a typical sample of 5,000 U.S. households. Last month \(9.1 \%\) of consumers said conditions were "good." In the prior month, only \(8.5 \%\) said they were "good." Use the six-step hypothesis-testing method at the .05 level of significance to see whether you can determine if there is an increase in the share asserting conditions are "good." Find the \(p\) -value and explain what it means.

The research department at the home office of New Hampshire Insurance conducts ongoing research on the causes of automobile accidents, the characteristics of the drivers, and so on. A random sample of 400 policies written on single persons revealed 120 had at least one accident in the previous three-year period. Similarly, a sample of 600 policies written on married persons revealed that 150 had been in at least one accident. At the .05 significance level, is there a significant difference in the proportions of single and married persons having an accident during a threeyear period? Determine the \(p\) -value.

The policy of the Suburban Transit Authority is to add a bus route if more than \(55 \%\) of the potential commuters indicate they would use the particular route. A sample of 70 commuters revealed that 42 would use a proposed route from Bowman Park to the downtown area. Does the Bowman-to-downtown route meet the STA criterion? Use the .05 significance level.

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