/*! 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 55 The researchers decide to extend... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The researchers decide to extend the study to a 5 -year period and find that 20 of the 200 people develop a cataract over a 5 -year period. Suppose the expected incidence of cataracts among \(65-\) to 69 -year-olds in the general population is \(5 \%\) over a 5 -year period. Construct a \(95 \%\) Cl for the 5 -year true rate of cataracts among the excessive-sunlight-exposure group.

Short Answer

Expert verified
The 95% confidence interval is approximately [0.0585, 0.1415].

Step by step solution

01

Identify the Parameters

We are given that 20 of the 200 subjects developed cataracts over a 5-year period. This gives a sample proportion of \( \frac{20}{200} = 0.10 \). We need to construct a 95% confidence interval (CI) for this proportion.
02

Determine the Standard Error

The standard error (SE) of the sample proportion can be calculated using the formula: \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \), where \( \hat{p} = 0.10 \) and \( n = 200 \). Calculate \( SE = \sqrt{\frac{0.10 \times 0.90}{200}} \approx 0.0212 \).
03

Find the Critical Value

For a 95% confidence interval, use the standard normal distribution (Z-distribution). The critical value \( Z \) is approximately 1.96.
04

Calculate the Confidence Interval

The 95% CI is calculated using the formula: \( \hat{p} \pm Z \times SE \). Substitute the values: \( 0.10 \pm 1.96 \times 0.0212 \approx 0.10 \pm 0.0415 \).
05

Final Confidence Interval

The interval is thus \([0.0585, 0.1415]\), which means we are 95% confident that the true proportion of cataract development in this group is between 0.0585 and 0.1415.

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.

Sample Proportion
The sample proportion is a basic yet powerful concept in statistics. In simple terms, it is the ratio of some occurrence or outcome in a sample. For instance, in a study assessing cataract development, if 20 out of 200 people develop cataracts, the sample proportion is determined by dividing the number of cases by the total number of subjects, i.e., \(\frac{20}{200} = 0.10\). This figure represents the probability of a single individual developing a cataract in this sample.
This metric is crucial as it provides an estimate of population behavior based on sample data. It highlights the observed proportion of those affected within the group under study.
  • Calculating sample proportion: Divide the number of events by the total sample size.
  • Represents probability of an event occurring in a sample.
Understanding sample proportion is important as it forms the basis for more advanced statistical analyses, like calculating confidence intervals.
Standard Error
The standard error (SE) quantifies the variation of a sample statistic, such as the sample proportion, from the true population parameter. It tells us how much we can expect our sample proportion to fluctuate with different samples.
The formula to calculate the standard error of a sample proportion is \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \). In this exercise, where \( \hat{p} = 0.10 \) and \( n = 200 \), the SE is approximately 0.0212.
The SE helps us understand the precision of our sample proportion as an estimate of the population proportion. A smaller SE suggests the sample proportion is a more accurate estimator of the population characteristic.
  • The smaller the SE, the more precise the sample estimate.
  • Determined by both the sample size and the variability within the sample.
Overall, the standard error is key to determining the reliability of sample estimates.
Critical Value
Finding the critical value is a crucial step when constructing a confidence interval. It represents the cutoff points on a probability distribution. In the context of a 95% confidence interval, the critical value is typically derived from the standard normal distribution, also known as the Z-distribution.
For a 95% confidence level, the critical value \( Z \) is approximately 1.96. This value helps determine the margin of error for the estimate and ensures that the confidence interval truly reflects 95% of possible sample observations.
  • Used to construct confidence intervals by determining the extent of the margin of error.
  • Derived from the desired confidence level and the corresponding normal distribution.
In essence, the critical value, together with the standard error, allows us to meaningfully tell a range where the true population parameter likely falls.
Cataract Incidence
Cataract incidence refers to the rate at which new cases of cataracts develop over a certain period. In the given exercise, we are examining the occurrence of cataracts over five years in a group exposed to excessive sunlight.
The sample proportion calculated from this study is used to estimate the cataract incidence. By constructing a confidence interval, researchers could ascertain the reliability and range of this estimate, potentially leading to significant insights about the impact of sunlight exposure on cataract development.
  • Cataracts are clouding of the eye's lens, impacting vision.
  • The exercise focuses on determining how often this occurs in a specific group.
Understanding cataract incidence can contribute to public health awareness and preventive measures for groups at risk.
Normal Distribution
The normal distribution is a fundamental concept in statistics, often referred to as a "bell curve," due to its symmetrical shape. It is essential for calculating confidence intervals and understanding probability distributions.
In our exercise, when determining the standard error and critical value, the standard normal distribution is used. A normal distribution assumes that most occurrences will cluster around the mean and fewer as you move away from it.
  • Foundation for many statistical tests, including confidence intervals.
  • Reflects a situation where data tends to cluster around a central value.
Embracing the normal distribution concept helps to appreciate how statistical methods, including the estimation of confidence intervals, work. Its application extends beyond cataract studies to numerous fields where predicting future observations is key.

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

Medical errors are common in hospitals throughout the world. One possible causal factor is the long work hours of hospital personnel. In a pilot study investigating this issue, medical residents were encouraged to sleep \(6-8\) hours per night for a 3 -week period instead of their usual irregular sleep schedule. The researchers expected, given previous data, that there would be one medical error per resident per day on their usual irregular sleep schedule. Suppose two residents participate in the program (each for 3 weeks), and chart review finds a total of 20 medical errors made by the two residents combined. What test can be used to test the hypothesis that an increase in amount of sleep will change the number of medical errors per day?

Answer Problem 7.65 for the percentage of calories from fat (separately for total fat and saturated fat) as reported on the diet record and the food- frequency questionnaire. Assume there are 9 calories from fat for every gram of fat consumed.

Suppose we wish to test the hypothesis \(H_{0}: \mu=2\) vs. \(H_{1}: \mu \neq 2 .\) We find a two-sided \(p\) -value of .03 and \(a 95 \%\) Cl for \(\mu\) of \((1.5,4.0) .\) Are these two results possibly compatible? Why or why not?

Left ventricular mass (LVM) is an important risk factor for subsequent cardiovascular disease. A study is proposed to assess the relationship between childhood blood pressure levels and LVM in children as determined from echocardiograms. The goal is to stratify children into a normal bp group \((<80\) th percentile for their age, gender, and height) and an elevated bp group \((\geq 90\) th percentile for their age, gender, and height) and compare change in LVM between the 2 groups. Before this can be done, one needs to demonstrate that LVM actually changes in children over a 4-year period. To help plan the main study, a pilot study is conducted where echocardiograms are obtained from 10 random children from the Bogalusa Heart Study at baseline and after 4 years of follow-up. The data are given in Table 7.6. Since this was a pilot study, the main question of interest is how many subjects would be needed to detect an increase of \(10 \mathrm{g}\) in mean LVM over 4 years using a twosided test with \(\alpha=0.05\) and power \(=80 \%\) ? Hint. Assume that the estimated variance of change in \(L\) VM in the pilot study is the true variance of change in LVM.

Osteoporosis is an important cause of morbidity in middle-aged and elderly women. Several drugs are currently used to prevent fractures in postmenopausal women. Suppose the incidence rate of fractures over a 4 -year period is known to be \(5 \%\) among untreated postmenopausal women with no previous fractures. A pilot study conducted among 100 women without previous fractures aims to determine whether a new drug can prevent fractures. It is found that two of the women have developed fractures over a 4-year period. Suppose the new drug is hypothesized to yield a fracture rate of \(2.5 \%\) over a 4 -year period. How many subjects need to be studied to have an \(80 \%\) chance of detecting a significant difference between the incidence rate of fractures in treated women and the incidence rate of fractures in untreated women (assumed to be \(5 \%\) from Problem 7.105 )?

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.