Question

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. The percentage of males 18 years and older who have never married is \(30.4 .\) For females the percentage is \(23.6 .\) Looking at the records in a particular populous county, a random sample of 250 men showed that 78 had never married and 58 of 200 women had never married. At the 0.05 level of significance, is the proportion of men greater than the proportion of women? Use the \(P\) -value method.

Short answer

We do not reject the null hypothesis; there's not enough evidence that the proportion of men who never married is greater than women.

Step-by-step solution

State the Hypotheses and Identify the Claim

Let \( p_1 \) be the proportion of never-married men, and \( p_2 \) be the proportion of never-married women. We want to test whether the proportion of men who never married is greater than that of women. The null hypothesis \( H_0 \) states that \( p_1 \leq p_2 \), and the alternative hypothesis \( H_a \) states that \( p_1 > p_2 \). Our claim is the alternative hypothesis \( H_a \).

Calculate Sample Proportions

Calculate the sample proportion for men and women. For men, \( \hat{p}_1 = \frac{78}{250} = 0.312 \). For women, \( \hat{p}_2 = \frac{58}{200} = 0.29 \).

Find Critical Value

Using a significance level \( \alpha = 0.05 \) for a one-tailed test, find the critical value from the standard normal distribution table. The critical value \( z_c \) is approximately 1.645.

Compute the Test Value

Compute the test statistic for comparing two proportions. The formula is:\[ z = \frac{(\hat{p}_1 - \hat{p}_2)}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}} \]where \( \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} \) is the pooled sample proportion. With \( x_1 = 78, x_2 = 58, n_1 = 250, \) and \( n_2 = 200 \), calculate \( \hat{p} = \frac{78 + 58}{250 + 200} = \frac{136}{450} \approx 0.3022 \).The test statistic is then:\[ z = \frac{(0.312 - 0.29)}{\sqrt{0.3022 \times (1 - 0.3022) \times (\frac{1}{250} + \frac{1}{200})}} \approx 0.61 \]

Make the Decision

Compare the computed test statistic \( z = 0.61 \) to the critical value \( z_c = 1.645 \). Since the test statistic is less than the critical value, we do not reject the null hypothesis \( H_0 \).

Summarize the Results

We have insufficient evidence to support the claim that the proportion of never-married men is greater than the proportion of never-married women in the given county.

Key concepts

Critical Value

In hypothesis testing, the critical value is a crucial concept. It helps determine the threshold where we either reject or fail to reject the null hypothesis. For our exercise on the differences in marital status among men and women, the critical value helps us decide if the observed difference in proportions is significant. The critical value depends on the significance level and whether the test is one-tailed or two-tailed.

In this case, we have a one-tailed test with a significance level of 0.05. For such a test, we use the standard normal distribution (Z-distribution) to find the critical value. The chosen critical value is 1.645. This number marks the point beyond which we would reject the null hypothesis. If your test statistic exceeds this critical value, you could potentially have enough evidence to support the alternative hypothesis.

Test Statistic

The test statistic is like a yardstick. It measures how far our sample data deviates from the null hypothesis. For the proportions of never-married individuals, we use the formula for test statistics of two proportions:

\[ z = \frac{(\hat{p}_1 - \hat{p}_2)}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}}\]
Where:

• \( \hat{p_1} \) is the sample proportion of men
• \( \hat{p_2} \) is the sample proportion of women
• \( \hat{p} \) is the pooled sample proportion

Plugging the appropriate values into the formula gives us a test statistic of approximately 0.61. By comparing this test statistic to the critical value, we decide whether to reject the null hypothesis.

Significance Level

The significance level, denoted as \( \alpha \), is akin to a safety margin. It indicates the probability of making a Type I error, which involves rejecting the null hypothesis when it is actually true. For this exercise, the significance level is set at 0.05 or 5%.

This 5% threshold means that if our results are within the extreme 5% of the distribution (in the direction of the alternative hypothesis), we consider the findings statistically significant. The critical value aligns with this significance level to help guide our decision process in the test.
Choosing a smaller significance level (e.g., 0.01) would make our test more stringent, reducing the risk of falsely rejecting the null hypothesis but also requiring stronger evidence to do so.

Sample Proportion

Sample proportions are used to estimate population proportions in hypothesis testing. For the case of never-married individuals, we calculated separate sample proportions for men and women. Here's a quick breakdown of what sample proportions tell us:

• For men, the sample proportion \(\hat{p}_1\) was calculated as \( \frac{78}{250} = 0.312 \), indicating that about 31.2% of the sample of men had never married.
• For women, the sample proportion \(\hat{p}_2\) was \( \frac{58}{200} = 0.29 \), showing that 29% of the sampled women had never married.

Sample proportions allow us to conduct precise comparisons and calculate test statistics for hypothesis testing. They provide an evidence-based estimate of population parameters and serve as the groundwork for making statistical inferences.