/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 Marriage and status "Would you m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Marriage and status "Would you marry a person from a lower social class than your own?" Researchers asked this question of a random sample of 385 black, never-married college students. Of the 149 men in the sample, 91 said "Yes." Among the 236 women, 117 said "Yes." 15 Did a significantly higher proportion of the men than the women who were surveyed say "Yes"? Give appropriate evidence to justify your answer.

Short Answer

Expert verified
Yes, significantly more men said "Yes" than women.

Step by step solution

01

Define the Hypotheses

We want to determine if a significant difference exists in the proportion of men and women who said "Yes" to marrying someone from a lower social class. Therefore, set up the null hypothesis and alternative hypothesis: \(H_0: p_m = p_w\) (The proportions of men and women saying "Yes" are the same). \(H_a: p_m > p_w\) (The proportion of men saying "Yes" is greater).
02

Calculate Proportions

Next, calculate the sample proportions for men and women. The proportion of men saying "Yes" is \( \hat{p}_m = \frac{91}{149} \approx 0.6107 \). The proportion of women saying "Yes" is \( \hat{p}_w = \frac{117}{236} \approx 0.4966 \).
03

Determine the Pooled Proportion

The pooled proportion combines the successes from both groups over the total sample size. It is calculated as \( \hat{p} = \frac{91 + 117}{149 + 236} = \frac{208}{385} \approx 0.5403 \).
04

Compute the Standard Error

The standard error for the difference between the two proportions is calculated as follows:\[SE = \sqrt{\hat{p}(1 - \hat{p}) \left(\frac{1}{n_m} + \frac{1}{n_w}\right)} = \sqrt{0.5403 \cdot 0.4597 \left(\frac{1}{149} + \frac{1}{236}\right)} \approx 0.0642\]
05

Find the Test Statistic

The test statistic is calculated using the formula:\[Z = \frac{\hat{p}_m - \hat{p}_w}{SE} = \frac{0.6107 - 0.4966}{0.0642} \approx 1.776\]
06

Compare to Critical Value

Assuming a significance level \(\alpha = 0.05\) and using a Z-table, the critical value for a one-tailed test is 1.645. Compare our calculated Z value to this critical value.
07

Make a Decision

Since the calculated Z value (1.776) is greater than the critical value (1.645), we reject the null hypothesis.
08

Conclusion

There is significant evidence at the \(\alpha = 0.05\) level that a higher proportion of men than women are willing to marry someone from a lower social class.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Null Hypothesis
When we conduct statistical hypothesis tests, the null hypothesis plays a crucial role. It is the starting point for any test, representing a statement of no effect or no difference. In the context of the marriage and status study, the null hypothesis (\(H_0\): \(p_m = p_w\)) suggests that there is no difference between the proportions of men and women willing to marry someone from a lower social class.

This assumption gives us a baseline to compare against and is presumed true unless evidence suggests otherwise. By initially assuming the null hypothesis is true, we can objectively assess the data and determine if there is sufficient statistical evidence to support the alternative hypothesis.

In simple terms, the null hypothesis asks, "Is there really no difference between the two groups in question?" The subsequent steps of hypothesis testing help us answer this fundamental question.
Standard Error
The standard error is an important concept in statistics, providing an estimate of the variability of a statistic; in this case, the difference between two sample proportions. It measures how much the sample proportion may vary from the true population proportion due to random sampling.

In our example, the standard error allows us to understand how much the observed difference between men's and women's willingness to marry from a lower class could vary naturally, without any actual difference in the population.

To calculate the standard error, we use the formula:
\[SE = \sqrt{\hat{p}(1 - \hat{p}) \left(\frac{1}{n_m} + \frac{1}{n_w}\right)}\]
here \(\hat{p}\) is the pooled proportion, which aggregates the successes across both groups. The standard error is crucial for calculating the test statistic and ultimately influences decisions about rejecting or not rejecting the null hypothesis.
Test Statistic
A test statistic is a standardized value derived from sample data used to decide whether to reject the null hypothesis. It helps compare the observed data against a theoretical distribution assuming the null hypothesis is true.

In this exercise, the test statistic is a Z-score calculated with the formula:
\[Z = \frac{\hat{p}_m - \hat{p}_w}{SE}\]
The Z-score quantifies the difference between the sample proportions of men and women, taking into account the variability that standard error represents. In this case, a Z value of approximately 1.776 indicates that the difference between sample proportions is 1.776 standard deviations away from the null hypothesis's expected value (zero difference).

We use this score to determine if the observed data are statistically significant compared to what we'd expect if the null hypothesis were true.
Significance Level
Setting a significance level, denoted by \(\alpha\), is a pivotal step in hypothesis testing. The significance level represents the probability of rejecting the null hypothesis when it is actually true. In simpler terms, it is the threshold for determining whether the test results are due to chance or if there is a real effect.

In practical terms, a common choice for the significance level is 0.05, indicating a 5% risk of concluding that a difference exists when it doesn't. In the context of the marriage and status study, an \(\alpha\) of 0.05 means that only if the probability of observing the test statistic under the null hypothesis is less than 5%, we reject the null hypothesis.

We compare our test statistic (Z-score) to the critical value from the Z-table. If the test statistic exceeds the critical value for our chosen significance level, it means the difference is statistically significant, leading us to reject the null.

This process helps ensure that the results are not just random noise, making our conclusions more robust and trustworthy.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Shrubs and fire Fire is a serious threat to shrubs in dry climates. Some shrubs can resprout from their roots after their tops are destroyed. One study of resprouting took place in a dry area of Mexico. The investigators randomly assigned shrubs to treatment and control groups. They clipped the tops of all the shrubs. They then applied a propane torch to the stumps of the treatment group to simulate a fire. All 12 of the shrubs in the treatment group resprouted. Only 8 of the 12 shrubs in the control group resprouted.

Who owns iPods? As part of the Pew Internet and American Life Project, researchers surveyed a random sample of 800 teens and a separate random sample of 400 young adults. For the teens, \(79 \%\) said that they own an iPod or MP3 player. For the young adults, this figure was \(67 \%\). Do the data give convincing evidence of a difference in the proportions of all teens and young adults who would say that they own an iPod or MP3 player? State appropriate hypotheses for a test to answer this question. Define any parameters you use.

Which of the following is the correct margin of error for a \(99 \%\) confidence interval for the difference in the proportion of male and female college students who worked for pay last summer? (a) \(2.576 \sqrt{\frac{0.851(0.149)}{550}+\frac{0.851(0.149)}{500}}\) (b) \(2.576 \sqrt{\frac{0.851(0.149)}{1050}}\) (c) \(2.576 \sqrt{\frac{0.880(0.120)}{550}+\frac{0.820(0.180)}{500}}\) (d) \(1.960 \sqrt{\frac{0.851(0.149)}{550}+\frac{0.851(0.149)}{500}}\) (e) \(1.960 \sqrt{\frac{0.880(0.120)}{550}+\frac{0.820(0.180)}{500}}\)

Driving school A driving school owner believes that Instructor \(\mathrm{A}\) is more effective than Instructor \(\mathrm{B}\) at preparing students to pass the state's driver's license exam. An incoming class of 100 students is randomly assigned to two groups, each of size \(50 .\) One group is taught by Instructor \(A ;\) the other is taught by Instructor B. At the end of the course, 30 of Instructor A's students and 22 of Instructor \(\mathrm{B}\) 's students pass the state exam. (a) Do these results give convincing evidence at the \(\alpha=0.05\) level that Instructor \(\mathrm{A}\) is more effective? (b) Describe a Type I and a Type II error in this setting. Which error could you have made in part (a)?

Information online (8.2,10.1) A random digit dialing sample of 2092 adults found that 1318 used the Internet. \({ }^{42}\) Of the users, 1041 said that they expect businesses to have Web sites that give product information; 294 of the 774 nonusers said this. (a) Construct and interpret a \(95 \%\) confidence interval for the proportion of all adults who use the Internet. (b) Construct and interpret a \(95 \%\) confidence interval to compare the proportions of users and nonusers who expect businesses to have Web sites that give product information.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.