/*! 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 17 Population requirement for prima... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Population requirement for primary hip-replacement surgery: a cross-sectional study" was conducted by the University of Bristol in the United Kingdom. The findings resulted in the following statement: "The prevalence of self reported hip pain was 107 per 1000 (95\% CI \(101-113\) ) for men and 173 per \(1000(166-180)\) for women." a. Explain the meaning of the confidence interval, \(95 \%\) CI \(101-113\) b. Find the standard error for the men's self-reported hip pain \(95 \%\) confidence interval. c. Assuming the women's data was also from a \(95 \%\) confidence interval, find the standard error.

Short Answer

Expert verified
a) 95% confidence interval (CI) of 101-113 indicates that we can be 95% confident that the mean self-reported hip pain for men falls between 101 and 113 per 1000. b) The standard error for men's self-reported hip pain is approximately 3.06. c) The standard error for women's self-reported hip pain is approximately 3.57.

Step by step solution

01

Explanation of Confidence Interval

A 95% confidence interval of 101-113 indicates that, based on the data collected in this study, one can be 95% confident that the true mean of self-reported hip pain in men falls between 101 and 113 per 1000. This essentially means if the experiment were conducted 100 times, the average result would fall in this range 95 times.
02

Calculate the standard error for men

To calculate the standard error, the width of the confidence interval is divided by the number of standard deviations that define the interval. For a 95% confidence interval, this value is approximately 1.96. Here, the width of the confidence interval is \(113 - 101 = 12\), so the standard error = \((113 - 101) / (2 \times 1.96) = 12 / 3.92 = 3.06\). This means the standard deviation of men's reported hip pain is approximately 3.06 per 1000.
03

Calculate the standard error for women

Following the same process as above, the width of the women's confidence interval is calculated as \(180 - 166 = 14\). Hence, the standard error = \((180 - 166) / (2 \times 1.96) = 14 / 3.92 = 3.57\). Therefore, the standard deviation of women's reported hip pain is approximately 3.57 per 1000.

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.

Standard Error
Understanding the standard error is crucial when interpreting statistical data. It measures the variability of sample means around the population mean. In simpler terms, if you were to take multiple samples from a population and calculate the mean for each sample, the standard error tells you how much those sample means would typically differ from the actual population mean.

When we look at the given exercise, the standard error for men's self-reported hip pain was calculated using the range of the confidence interval (101-113) and the z-score (1.96 for 95% confidence). The computation resulted in a standard error of 3.06 per 1000, offering insight into the precision of the estimated population mean. A smaller standard error represents a more precise estimate, enabling researchers to make more reliable inferences about the population.
Statistical Significance
The term statistical significance is a measure of whether the results observed in a study or experiment are unlikely to have occurred by chance. It gives an indication of the trustworthiness of the findings and whether they reflect a genuine effect or relationship in the population being studied.

Likewise, the confidence interval plays a role here—when it does not cross a value of interest (like zero in a null hypothesis), the result is usually considered statistically significant. In the given study, the self-reported hip pain among men and women had a 95% confidence interval that didn't include zero, implying that the reported pain prevalence is a significant finding for the population.
Population Study Statistics
In population study statistics, researchers aim to make inferences about an entire population based on a sample. This practice is common because it's typically impractical or impossible to study every individual in a population. Key metrics include population mean, variance, standard deviation, and the standard error mentioned earlier.

In our exercise's context, the conclusions drawn about the prevalence of hip pain are based on the statistics calculated from samples of the population. By constructing confidence intervals, which are influenced by the sample size and standard error, researchers infer the prevalence rate of hip pain in the broader population with a stated level of confidence—95% in this case.

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

The null hypothesis, \(H_{o}: \mu=250,\) was tested against the alternative hypothesis, \(H_{a}: \mu<250 .\) A sample of \(n=85\) resulted in a calculated test statistic of \(z \star=-1.18 .\) If \(\sigma=22.6,\) find the value of the sample mean, \(\bar{x}\). Find the sum of the sample data, \(\Sigma x\)

Suppose you want to test the hypothesis that "the mean salt content of frozen 'lite' dinners is more than \(350 \mathrm{mg}\) per serving." An average of \(350 \mathrm{mg}\) is an acceptable amount of salt per serving; therefore, you use it as the standard. The null hypothesis is "The average content is not more than \(350 \mathrm{mg} "(\mu=350) .\) The alternative hypothesis is "The average content is more than \(350 \mathrm{mg} "\) \((\mu > 350)\) a. Describe the conditions that would exist if your decision results in a type I error. b. Describe the conditions that would exist if your decision results in a type II error.

A random sample of the scores of 100 applicants for clerk-typist positions at a large insurance company showed a mean score of \(72.6 .\) The preparer of the test maintained that qualified applicants should average \(75.0 .\) a. Determine the \(99 \%\) confidence interval for the mean score of all applicants at the insurance company. Assume that the standard deviation of test scores is 10.5 b. Can the insurance company conclude that it is getting qualified applicants (as measured by this test)?

Waiting times (in hours) at a popular restaurant are believed to be approximately normally distributed with a variance of 2.25 during busy periods. a. A sample of 20 customers revealed a mean waiting time of 1.52 hours. Construct the \(95 \%\) confidence interval for the population mean. b. Suppose that the mean of 1.52 hours had resulted from a sample of 32 customers. Find the \(95 \%\) confidence interval. c. What effect does a larger sample size have on the confidence interval?

Consider the null hypothesis in Applied Example \(8.11, " H_{o}:\) Teaching techniques have no significant effect on students' exam scores." Describe the actions that would result in a type I and a type II error if \(H_{o}\) were tested.

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.