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For Exercises 7 through \(27,\) perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Married People In a specific year \(53.7 \%\) of men in the United States were married and \(50.3 \%\) of women were married. Two independent random samples of 300 men and 300 women found that 178 men and 139 women were married (not to each other). At the 0.05 level of significance, can it be concluded that the proportion of men who were married is greater than the proportion of women who were married?

Short Answer

Expert verified
The proportion of married men is significantly greater than that of women at the 0.05 level.

Step by step solution

01

State the Hypotheses

To perform the hypothesis test, begin by stating the null and alternative hypotheses. Let \( p_1 \) be the proportion of married men, and \( p_2 \) be the proportion of married women.- Null Hypothesis \( H_0 \): \( p_1 \leq p_2 \)- Alternative Hypothesis \( H_a \): \( p_1 > p_2 \)The claim in this context is that the proportion of married men is greater than the proportion of married women.
02

Find the Critical Value(s)

Since this is a one-tailed test at a significance level of \( \alpha = 0.05 \), we find the critical value for a right-tailed test using a standard normal distribution (Z-distribution). The critical z-value corresponding to \( \alpha = 0.05 \) is approximately 1.645.
03

Compute the Test Value

Calculate the test statistic using the formula for the difference in proportions:\[ z = \frac{(\hat{p_1} - \hat{p_2}) - 0}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}}\]Where \( \hat{p_1} = \frac{178}{300} = 0.5933 \) and \( \hat{p_2} = \frac{139}{300} = 0.4633 \).Pooled proportion \( \hat{p} = \frac{178 + 139}{600} = 0.5283 \).Calculate:\[ z = \frac{0.5933 - 0.4633}{\sqrt{0.5283(1 - 0.5283)(\frac{1}{300} + \frac{1}{300})}} \]This simplifies to:\[ z \approx \frac{0.13}{0.0402} \approx 3.23\]
04

Make the Decision

Compare the computed test statistic \( z \approx 3.23 \) with the critical value 1.645. Since 3.23 is greater than 1.645, we reject the null hypothesis \( H_0 \).
05

Summarize the Results

At the 0.05 level of significance, we have enough evidence to support the claim that the proportion of men who were married is greater than the proportion of women who were married.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis is a statement that indicates no effect or no difference between populations. It is usually denoted as \(H_0\). You can think of the null hypothesis as the status quo or a baseline assumption that researchers aim to test against. It serves as the base for determining whether there is enough statistical evidence to support a different claim. In real-world studies, the null hypothesis might say that two groups have the same average characteristics, like the same mean or proportion. For example, in our exercise, the null hypothesis \( H_0\) suggests that the proportion of married men is less than or equal to the proportion of married women, meaning there's no true difference.One important point to remember is that the null hypothesis is never "proved"; it is either rejected or not rejected based on the data collected. Rejection of the null hypothesis suggests that there is sufficient evidence to consider the alternative hypothesis as the more likely scenario.
Alternative Hypothesis
Contrary to the null hypothesis, the alternative hypothesis represents a new claim or theory that researchers want to test. It is denoted as \(H_a\) or sometimes \(H_1\). The alternative hypothesis suggests that there is a statistically significant effect or difference between groups. It is basically what the researcher aims to prove. If evidence supports the alternative hypothesis, it indicates that there's a change from the status quo outlined by the null hypothesis.In the current exercise, the alternative hypothesis \( H_a \) states that the proportion of married men \( p_1\) is greater than the proportion of married women \( p_2\). This is a one-tailed hypothesis as we are only interested in one direction of difference. The researcher wants to gather evidence to support that more men are married compared to women.
Z-distribution
The Z-distribution, also known as the standard normal distribution, is a fundamental concept in statistics used extensively in hypothesis testing. It has a mean of 0 and a standard deviation of 1. This distribution allows us to understand where our test statistic lies in relation to a "normal" distribution, which is essential in making statistical decisions. When conducting hypothesis tests on population proportions, like the one in our exercise, the test statistic is often converted into a z-value. This conversion uses a formula that includes the sample data to standardize the test results. We use the z-distribution because it helps us find critical values and determine probabilities, allowing us to conclude whether to reject or accept the null hypothesis.In the given exercise, the z-value is calculated using the difference between sample proportions and pooled variance, resulting in \( z \approx 3.23 \). We then compare this value to the critical z-value from the distribution to make our decision.
Level of Significance
The level of significance, denoted by \( \alpha \), is a threshold set by researchers to determine the cutoff point for deciding whether to reject the null hypothesis. The level of significance is usually expressed as a decimal or percentage, and a common choice is 0.05, denoting a 5% risk of Type I error, which is the error of falsely rejecting the null hypothesis when it is actually true.In essence, the level of significance helps to decide the critical value or the cutoff point in the z-distribution which the calculated test statistic must exceed in order to indicate a significant result. For our exercise, we use a 0.05 level of significance. This means there is a 5% chance that we would incorrectly reject the null hypothesis due to sampling variability. It ensures that our test remains robust and careful, providing high confidence in the results. The critical z-value computed based on \( \alpha = 0.05 \) is approximately 1.645, which is compared against our test statistic \( z \approx 3.23 \). Since our test statistic exceeds this critical value, we decide to reject \( H_0 \).

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