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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; there is a significant difference between the proportions.

Step by step solution

01

Formulate the Decision Rule

To test the hypothesis at a 0.05 significance level, we use a two-tailed test for the difference between proportions. The decision rule is determined by comparing the test statistic to the critical value. For a 0.05 significance level in a two-tailed test, the critical value of Z is approximately ±1.96. If the test statistic falls outside this range, we reject the null hypothesis.
02

Compute the Pooled Proportion

The pooled proportion is calculated using both the sample proportions. First, find the sample proportions:\( \hat{p}_1 = \frac{X_1}{n_1} = \frac{170}{200} = 0.85 \) and \( \hat{p}_2 = \frac{X_2}{n_2} = \frac{110}{150} = \frac{11}{15} \approx 0.7333 \). Now, the pooled proportion is given by \( \hat{p} = \frac{X_1 + X_2}{n_1 + n_2} = \frac{170 + 110}{200 + 150} = \frac{280}{350} = 0.8 \).
03

Compute the Standard Error for Proportion Difference

The standard error of the difference between the two proportions is calculated as:\( SE = \sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{0.8 \times 0.2 \left(\frac{1}{200} + \frac{1}{150}\right)} \). Simplifying gives:\( SE = \sqrt{0.8 \times 0.2 \left(0.005 + 0.0067\right)} = \sqrt{0.8 \times 0.2 \times 0.0117} = \sqrt{0.001872} \approx 0.0433 \).
04

Compute the Z Test Statistic

The Z test statistic for comparing two proportions is given by:\( Z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.85 - 0.7333}{0.0433} \approx \frac{0.1167}{0.0433} \approx 2.69 \).
05

Make a Decision on the Null Hypothesis

Since the computed Z value of 2.69 exceeds the critical Z value of ±1.96, we reject the null hypothesis \( H_0 : \pi_1 = \pi_2 \). This indicates that there is a significant difference between the proportions of the two populations.

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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 key concept in hypothesis testing. It is a statement asserting that there is no effect or no difference between groups or conditions under study. In the given exercise, the null hypothesis is \( \pi_1 = \pi_2 \), indicating that the proportions of two populations are equal.

We rely on statistical evidence from the data to evaluate this hypothesis. If the data provide sufficient evidence against the null hypothesis, we might reject it, suggesting that a difference likely exists between the groups being compared. Conversely, if the data do not offer strong evidence against the null hypothesis, we fail to reject it, implying the absence of enough evidence to conclude a difference exists. This approach ensures an objective assessment of the data, based on the significance level and statistical test used.
Alternate Hypothesis
The alternate hypothesis, represented as \( H_1 \), stands in opposition to the null hypothesis. It suggests that there is an effect or a difference between the groups or conditions studied. For the exercise, the alternate hypothesis \( \pi_1 eq \pi_2 \) proposes that the two populations have different proportions.

This hypothesis is crucial because it embodies the alternative scenario that researchers aim to prove through their data analysis. When testing, we examine if the data provide enough evidence to reject the null hypothesis in favor of the alternate hypothesis.

In hypothesis testing, rejecting the null hypothesis is synonymous with supporting the alternate hypothesis, implying a significant difference or effect that wasn't assumed initially.
Pooled Proportion
In comparing two proportions, the pooled proportion is a critical calculation. It combines data from both groups to provide a single estimate, which serves to stabilize variance. The exercise computes the pooled proportion as \( \hat{p} = \frac{X_1 + X_2}{n_1 + n_2} = \frac{280}{350} = 0.8 \).

This calculation aggregates the success counts from both populations divided by their total instances, offering an average proportion. By using a pooled proportion, we ensure that the subsequent calculations, such as the standard error, are more accurate and balanced. This step is essential, especially when the sample sizes are different, as it provides a single estimate to guide the hypothesis testing process.
Significance Level
The significance level, commonly denoted by \( \alpha \), is a threshold used in hypothesis testing to determine statistical significance. In the exercise, a significance level of 0.05 is utilized, meaning there's a 5% risk of concluding that a difference exists when there is none.

This value is crucial as it defines the critical region where we would reject the null hypothesis. A lower significance level means stricter criteria for rejection, requiring stronger evidence to conclude a difference. In practice, it is often chosen based on the desired balance between the Type I error (false positive) and Type II error (false negative) risks.

The significance level helps set the decision rule: if the test statistic falls beyond the critical value, we reject the null hypothesis, implying statistical significance.

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Most popular questions from this chapter

A sample of 65 observations is selected from one population. The sample mean is 2.67 and the sample standard deviation is \(0.75 .\) A sample of 50 observations is selected from a second population. The sample mean is 2.59 and the sample standard deviation is \(0.66 .\) Conduct the following test of hypothesis using the .08 significance level. $$\begin{array}{l}H_{0}: \mu_{1} \leq \mu_{2} \\\H_{1}: \mu_{1}>\mu_{2}\end{array}$$ a. Is this a one-tailed or a two-tailed test? b. State the decision rule. c. Compute the value of the test statistic. d. What is your decision regarding \(H_{0} ?\) e. What is the \(p\) -value? Compute and interpret the \(p\) -value.

The USA Today Attp://www.usatoday.com/sports/baseball/front.htm) and Major League Baseball's fittp://www.majorleaguebaseball.com) websites regularly report information on individual player salaries in the American League and the National League. Go to one of these sites and find the individual salaries for your favorite team in each league. Compute the mean and the standard deviation. Is it reasonable to conclude that there is a difference in the salaries of the two teams?

A cell phone company offers two plans to it subscribers. At the time new subscribers sign up, they are asked to provide some demographic information. The mean yearly income for a sample of 40 subscribers to Plan \(\mathrm{A}\) is \(\$ 57,000\) with a standard deviation of \(\$ 9,200 .\) This distribution is positively skewed; the actual coefficient of skewness is 2.11. For a sample of 30 subscribers to Plan B the mean income is \(\$ 61,000\) with a standard deviation of \(\$ 7,100 .\) The distribution of Plan B subscribers is also positively skewed, but not as severely. The coefficient of skewness is 1.54 . At the .05 significance level, is it reasonable to conclude the mean income of those selecting Plan \(\mathrm{B}\) is larger? What is the \(p\) -value? Do the coefficients of skewness affect the results of the hypothesis test? Why?

Lester Hollar is Vice President for Human 91Ó°ÊÓ for a large manufacturing company. In recent years he has noticed an increase in absenteeism that he thinks is related to the general health of the employees. Four years ago, in an attempt to improve the situation, he began a fitness program in which employees exercise during their lunch hour. To evaluate the program, he selected a random sample of eight participants and found the number of days each was absent in the six months before the exercise program began and in the last six months. Below are the results. At the .05 significance level, can he conclude that the number of absences has declined? Estimate the \(p\) -value.

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 three-year period?

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