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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. Never Married People 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

Expert verified
We do not reject the null hypothesis; there's not enough evidence to show a higher proportion of never married men than women.

Step by step solution

01

State the Hypotheses

We are testing if the proportion of never married men is greater than that of never married women. Let \( p_m \) be the proportion of never married men, and \( p_w \) be the proportion of never married women. The null and alternative hypotheses are:- Null Hypothesis \( (H_0) \): \( p_m \leq p_w \)- Alternative Hypothesis \( (H_1) \): \( p_m > p_w \)The claim is that the proportion of never married men is greater than the proportion of never married women.
02

Find the Critical Value

For a right-tailed test at the 0.05 level of significance, we need the z-critical value. Since the significance level is 0.05 for a right-tailed test, the critical value of \( z \) is 1.645.
03

Compute the Test Value

First, calculate the sample proportions: \[ \hat{p}_m = \frac{78}{250} = 0.312 \] and \[ \hat{p}_w = \frac{58}{200} = 0.29 \].Next, calculate the pooled proportion \( \hat{p} \):\[ \hat{p} = \frac{78 + 58}{250 + 200} = \frac{136}{450} \approx 0.302\].Now compute the standard error \( SE \):\[ SE = \sqrt{\hat{p}(1 - \hat{p})\left(\frac{1}{250} + \frac{1}{200}\right)} \]\[ SE = \sqrt{0.302 \times 0.698 \left(\frac{1}{250} + \frac{1}{200}\right)} \approx 0.042\].Finally, compute the z-test value:\[ z = \frac{\hat{p}_m - \hat{p}_w}{SE} = \frac{0.312 - 0.29}{0.042} \approx 0.524\].
04

Make the Decision

Compare the computed z-test value with the critical value: - Computed z-value: 0.524 - Critical z-value: 1.645 Since the computed z-value (0.524) is less than the critical value (1.645), we fail to reject the null hypothesis.
05

Summarize the Results

At the 0.05 level of significance, we do not have enough evidence to support the claim that the proportion of never married men is greater than the proportion of never married women in this sample.

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Key Concepts

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

Significance Level
In hypothesis testing, the significance level is a critical threshold that helps us decide whether to accept or reject the null hypothesis. It is denoted by \( \alpha \), and it represents the probability of committing a Type I error, which is rejecting a true null hypothesis. In this context, a 0.05 level of significance means that there is a 5% risk of concluding that the proportion of never married men is greater than that of women when that may not actually be the case. This is often chosen as a conventional cut-off point.
When setting up hypothesis tests, it is crucial to choose an appropriate significance level as it affects the sensitivity of the test. A lower significance level reduces Type I error risk but increases the chance of a Type II error, i.e., failing to reject a false null hypothesis.
In our exercise, the significance level is set at 0.05, which means we have a reasonable balance between being too lenient and too strict. If your test statistic shows results that are inside this critical range determined by the significance level, it suggests evidence against the null hypothesis being sufficient enough to support the alternative hypothesis.
Critical Value
The critical value is the boundary or threshold of the confidence interval that dictates whether a test statistic is in the rejection zone or not. It is like a 1c cutoff number1d that helps determine statistical significance.
The critical value is reliant on the significance level; for our scenario with a right-tailed test, the critical value was found to be 1.645. This means that for the test statistic to show that the proportion of never married men is significantly greater than that of women, the computed z-value must exceed 1.645.
Where this value precisely lands on the distribution curve allows for demonstrating whether observed differences (if any) between proportions in your sample are unlikely under the null hypothesis. Thus, it helps in making a data-informed judgment about the hypotheses, aiding in deciding whether to support or refute the null or alternative claims you've set.
Z-test
The z-test in hypothesis testing is a statistical method used to determine if there is a significant difference between sample and population means or proportions. In our context, it's used to compare two proportions from independent populations.
This type of test assumes that the data is normally distributed, which allows the use of the critical value table to determine statistical significance. The z-statistic itself is calculated using the difference between sample proportions, the standard error, and often a pooled proportion as part of the formula.
For this exercise:
  • Sample proportions for both men and women are found, and a pooled proportion is calculated.
  • Then, the standard error (SE) is determined using the pooled proportion and the sample sizes.
  • Finally, the z-test value is computed showing how far (in standard errors) the actual sample difference is from the expected difference under the null hypothesis.
This practical approach provides a clear view of whether to accept or reject the null hypothesis based on how extreme the z-calculated value is in relation to known normal distribution curves. The test outcome enables students and analysts to conclude if the observed data reflects a statistically supported deviation from an initial hypothesis.
Proportion Comparison
Proportion comparison, a key objective in this exercise, involves inspecting two sample proportions to see if one is statistically larger than the other. This kind of analysis is common in statistical practice when looking into differences between categorical data.
In this exercise, we compare the proportion of never married men and women in a given county. The process involves several steps:
  • Identify each group19s sample proportion by dividing the number successful by the total number surveyed, for both men and women.
  • Calculate a pooled proportion that essentially combines both samples, giving a weighted average.
  • Using these proportions, determine the standard error, an indicator of variability in the data.
  • Perform a z-test to reveal if there is statistical evidence supporting a significant difference between the two proportions.
Successful comparison might show a significant difference, or it might not, guiding decisions in practical research fields such as social science or market research. Understanding the technical foundation lays ground for interpreting survey or study findings reliably, thus supporting better-informed conclusions based on quantitative assessments.

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Most popular questions from this chapter

For Exercises 2 through \(12,\) perform each of these steps. Assume that all variables are normally or approximately normally distributed. 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. Mistakes in a Song A random sample of six music students played a short song, and the number of mistakes in music each student made was recorded. After they practiced the song 5 times, the number of mistakes each student made was recorded. The data are shown. At \(\alpha=0.05,\) can it be concluded that there was a decrease in the mean number of mistakes? $$ \begin{array}{l|cccccc}{\text { Student }} & {\mathrm{A}} & {\mathrm{B}} & {\mathrm{C}} & {\mathrm{D}} & {\mathrm{E}} & {\mathrm{F}} \\ \hline \text { Before } & {10} & {6} & {8} & {8} & {13} & {8} \\ \hline \text { After } & {4} & {2} & {2} & {7} & {8} & {9}\end{array} $$

Weights of Running Shoes The weights in ounces of a sample of running shoes for men and women are shown. Test the claim that the means are different. Use the \(P\) -value method with \(\alpha=0.05 .\) $$ \begin{array}{ll|ccc}{\text { Men }} & {} & {\text { Women }} & {} \\ \hline 10.4 & {12.6} & {10.6} & {10.2} & {8.8} \\ {1.1} & {14.7} & {9.6} & {9.5} & {9.5} \\ {10.8} & {12.9} & {10.1} & {11.2} & {9.3} \\ {11.7} & {13.3} & {9.4} & {10.3} & {9.5} \\ {12.8} & {14.5} & {9.8} & {10.3} & {11.0} \\\ \hline\end{array} $$

If there is a significant difference between \(p_{1}\) and \(p_{2}\) and between \(p_{2}\) and \(p_{3},\) can you conclude that there is a significant difference between \(p_{1}\) and \(p_{3} ?\)

Commuting Times The U.S. Census Bureau reports that the average commuting time for citizens of both Baltimore, Maryland, and Miami, Florida, is approximately 29 minutes. To see if their commuting times appear to be any different in the winter, random samples of 40 drivers were surveyed in each city and the average commuting time for the month of January was calculated for both cities. The results are shown. At the 0.05 level of significance, can it be concluded that the commuting times are different in the winter? $$ \begin{array}{lcc}{} & {\text { Miami }} & {\text { Baltimore }} \\ \hline \text { Sample size } & {40} & {40} \\ {\text { Sample mean }} & {28.5 \min } & {35.2 \mathrm{min}} \\ {\text { Population standard deviation }} & {7.2 \mathrm{min}} & {9.1 \mathrm{min}}\end{array} $$

When a researcher selects all possible pairs of samples from a population in order to find the difference between the means of each pair, what will be the shape of the distribution of the differences when the original distributions are normally distributed? What will be the mean of the distribution? What will be the standard deviation of the distribution?

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