/*! 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 13 Arthritis The Centers for Diseas... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 21: Problem 13

Arthritis The Centers for Disease Control and Prevention reported a survey of randomly selected Americans age 65 and older, which found that 411 of 1012 men and 535 of 1062 women suffered from some form of arthritis. a) Are the assumptions and conditions necessary for inference satisfied? Explain. b) Create a \(95 \%\) confidence interval for the difference in the proportions of senior men and women who have this disease. c) Interpret your interval in this context. d) Does this confidence interval suggest that arthritis is more likely to afflict women than men? Explain.

Short Answer

Expert verified
Yes, necessary assumptions are satisfied. The 95% CI for the difference is approximately \([-0.133, -0.057]\), suggesting arthritis is more prevalent in women.

Step by step solution

01

Check Sample Size Condition

For both samples of men and women, verify that the sample size is large enough. We need at least 10 expected successes and 10 expected failures in each group. Let's calculate this: - Men: Expected successes = 411, failures = 1012 - 411 = 601 - Women: Expected successes = 535, failures = 1062 - 535 = 527 Both conditions are satisfied as we have more than 10 in successes and failures for each group.
02

Independence Assumption

Check if the samples are independent. Both samples of men and women are randomly selected. There is no indication of overlap or bias in how the samples were chosen, so we can assume that the samples are independent.
03

Calculate the Proportions

Compute the sample proportions for men and women:- For men: \( \hat{p}_1 = \frac{411}{1012} \approx 0.4063 \)- For women: \( \hat{p}_2 = \frac{535}{1062} \approx 0.5038 \)
04

Find the Standard Error

Calculate the standard error for the difference between the proportions with the formula:\[ SE = \sqrt{\frac{\hat{p}_1 (1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2 (1 - \hat{p}_2)}{n_2}} \]Where:- \( n_1 = 1012, \hat{p}_1 = 0.4063 \)- \( n_2 = 1062, \hat{p}_2 = 0.5038 \)Substitute these values to calculate the standard error.
05

Compute the Confidence Interval

For a 95% confidence interval, use the critical value of approximately 1.96 (from the standard normal distribution). The confidence interval for the difference in proportions, \( \hat{d} = \hat{p}_1 - \hat{p}_2 \) is:\[ \hat{d} \pm Z^* \times SE \]Substitute the values calculated to find the interval.
06

Interpret the Confidence Interval

If the confidence interval does not include 0, it suggests a significant difference between the two proportions. Find the numerical result of the confidence interval and determine whether 0 is included, providing insight into the difference in arthritis prevalence between men and women.
07

Conclusion on More Likely Affliction

Evaluate the confidence interval to see if it suggests that arthritis afflicts a higher proportion of women than men. If the confidence interval is entirely above 0, it affirms the hypothesis that women are more likely than men to have arthritis.

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 91影视!

Key Concepts

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

Confidence Intervals
When we're exploring differences between two groups, like senior men and women in this exercise, confidence intervals are incredibly useful. A confidence interval gives us a range of values鈥攂ased on our sample data鈥攖hat is likely to contain the true difference in population proportions.
For a 95% confidence interval, we are saying that if we took 100 different samples, approximately 95 of them would contain the true difference in proportions within this interval.
This particular exercise required creating a 95% confidence interval for the difference in proportions of senior men and women who have arthritis. To do this, we started by calculating the sample proportions for both men and women, then looked at the difference. With these, and the critical value from a standard normal distribution (z-value) of 1.96, we calculated the range using the formula for confidence intervals.
This range hints at how different the arthritis rates are between these two groups in the population.
Sample Proportions
Sample proportions give us a snapshot of a characteristic within a group鈥攊n this case, how many senior men and women reported having arthritis. These are calculated by dividing the number of favorable outcomes by the total number of people studied in each group.
  • For men, the sample proportion (\( \hat{p}_1 \)) was calculated as \( \frac{411}{1012} \approx 0.4063 \).
  • For women, it was \( \frac{535}{1062} \approx 0.5038 \).

These proportions are vital because they represent our best estimate of the actual proportion in the population based on the sample we have. In real-world problems like this one, calculating these proportions is the first step toward making broader statistical inferences.
Standard Error
The standard error tells us how much the sample proportion is expected to vary from sample to sample. It acts as a margin of error reflecting the fact that any one sample might just by chance be a bit different from others.
To find it, we use the formula:
\[ SE = \sqrt{\frac{\hat{p}_1 (1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2 (1 - \hat{p}_2)}{n_2}} \]where \( \hat{p}_1 \) and \( \hat{p}_2 \) are the sample proportions and \( n_1 \) and \( n_2 \) are the total number of individuals in each group.
By applying these values, we calculate a measure of variability in the difference between men's and women's arthritis rates. Standard error is crucial because it forms the basis of our confidence interval calculation, influencing the margin around our sample estimate.
Independence Assumption
In statistical inference, especially when comparing two groups, assuming independence between samples means that what happens in one group doesn't affect the results in the other.
For our study of arthritis in senior men and women, assuming independence is essential. The samples were randomly selected without any overlap or connection, satisfying this independence.
We must assume independence to confidently apply formulas for standard error and confidence intervals. If samples aren't independent, our confidence interval may tell us little about the true relationship between groups.
In many real-world scenarios, verifying this assumption is a key part of validating the reliability of your statistical conclusions.

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

Buy it again? A consumer magazine plans to poll car owners to see if they are happy enough with their vehicles that they would purchase the same model again. They'll randomly select 450 owners of American-made cars and 450 owners of Japanese models. Obviously, the actual opinions of the entire population couldn't be known, but suppose \(76 \%\) of owners of American cars and \(78 \%\) of owners of Japanese cars would purchase another. a) What sampling design is the magazine planning to use? b) What difference would you expect their poll to show? c) Of course, sampling error means the poll won't reflect the difference perfectly. What's the standard deviation for the difference in the proportions? d) Sketch a sampling model for the difference in proportions that might appear in a poll like this. e) Could the magazine be misled by the poll, concluding that owners of American cars are much happier with their vehicles than owners of Japanese cars? Explain.

Revealing information 886 randomly sampled teens were asked which of several personal items of information they thought it okay to share with someone they had just met. \(44 \%\) said it was okay to share their e-mail addresses, but only \(29 \%\) said they would give out their cell phone numbers. A researcher claims that a twoproportion \(z\) -test could tell whether there was a real difference among all teens. Explain why that test would not be appropriate for these data.

Gender gap Candidates for political office realize that different levels of support among men and women may be a crucial factor in determining the outcome of an election. One candidate finds that \(52 \%\) of 473 men polled say they will vote for him, but only \(45 \%\) of the 522 women in the poll express support. a) Write a \(95 \%\) confidence interval for the percent of male voters who may vote for this candidate. Interpret your interval. b) Write and interpret a \(95 \%\) confidence interval for the percent of female voters who may vote for him. c) Do the intervals for males and females overlap? What do you think this means about the gender gap? d) Find a \(95 \%\) confidence interval for the difference in the proportions of males and females who will vote for this candidate. Interpret your interval. e) Does this interval contain zero? What does that mean? f) Why do the results in parts c and e seem contradictory? If we want to see if there is a gender gap among voters with respect to this candidate, which is the correct approach? Why?

\- Online activity checks Are more parents checking up on their teen's online activities? A Pew survey in 2004 found that \(33 \%\) of 868 randomly sampled teens said that their parents checked to see what Web sites they visited. In 2006 the same question posed to 811 teens found \(41 \%\) reporting such checks. Do these results provide evidence that more parents are checking?

Twins In \(2001,\) one county reported that, among 3132 white women who had babies, 94 were multiple births. There were also 20 multiple births to 606 black women. Does this indicate any racial difference in the likelihood of multiple births? a) Test an appropriate hypothesis and state your conclusion in context. b) If your conclusion is incorrect, which type of error did you commit?
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
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.