/*! 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 12 Example \(16.1\) (page 377) assu... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Example \(16.1\) (page 377) assumed that the body mass index (BMI) of all American young women follows a Normal distribution with standard deviation \(\sigma=7.5\). How large a sample would be needed to estimate the mean BMI \(\mu\) in this population to within \(\pm 1\) with \(95 \%\) confidence?

Short Answer

Expert verified
A sample size of 217 is needed.

Step by step solution

01

Identify the known parameters

We are given that the standard deviation \(\sigma\) is 7.5 and we need to estimate the mean BMI to within \(\pm 1\) with 95% confidence. This means the margin of error \(E\) is 1.
02

Understanding the concept of Margin of Error

The formula for the margin of error \(E\) at 95% confidence for a Normal distribution is given by \(E = z^* \cdot \frac{\sigma}{\sqrt{n}}\), where \(z^*\) is the critical value from the standard Normal distribution corresponding to 95% confidence, \(\sigma\) is the standard deviation, and \(n\) is the sample size.
03

Finding the critical value

For a 95% confidence level, the critical value \(z^*\) is approximately 1.96 because 95% corresponds to the distribution tail probabilities of 0.025 each (as derived from Normal distribution tables).
04

Solve for the sample size \(n\)

Substitute the known values into the margin of error formula: \(1 = 1.96 \cdot \frac{7.5}{\sqrt{n}}\). Rearranging and solving for \(n\), we find: \[ n = \left(\frac{1.96 \cdot 7.5}{1}\right)^2 = \left(14.7\right)^2 = 216.09 \].
05

Round up the sample size

Since the sample size \(n\) must be a whole number and it is customary to round up when dealing with sample sizes to ensure the desired precision, we round 216.09 up to 217.

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.

Margin of Error
The margin of error is an important concept when determining how precise an estimate of a population parameter, like the mean BMI, is likely to be. It tells us the range within which the true population parameter is expected to fall. In our exercise, we want to estimate BMI with a margin of error of 1 unit with 95% confidence.

The margin of error is calculated using the formula:

\[ E = z^* \cdot \frac{\sigma}{\sqrt{n}} \]

where:
  • \( E \) is the margin of error
  • \( z^* \) is the critical value from the standard normal distribution
  • \( \sigma \) is the standard deviation of the population
  • \( n \) is the sample size
For a given confidence level, the margin of error decreases as the sample size increases. Thus, if you want a smaller margin of error, you will generally need a larger sample size.
Confidence Interval
A confidence interval provides a range of values, derived from the sample data, that is believed to contain the true population parameter. When we talk about a 95% confidence interval, we mean that if we were to take 100 different samples and compute the interval estimate for each sample, approximately 95 of those intervals would contain the population mean.

In our exercise context, we are interested in constructing a confidence interval to estimate the mean of the BMI for American young women. The endpoint of a confidence interval is calculated as follows:

\[ \bar{x} \pm z^* \frac{\sigma}{\sqrt{n}} \]

where:
  • \( \bar{x} \) is the sample mean
  • \( \pm \) signifies that we are calculating both the upper and lower bounds of the interval
  • \( z^* \frac{\sigma}{\sqrt{n}} \) is the margin of error
Knowing the confidence interval helps us understand the range in which the true population mean is likely to exist with the specified level of confidence.
Normal Distribution
Normal distribution is a continuous probability distribution characterized by a symmetric bell-shaped curve. It's important in statistics because it describes how data values are typically distributed, thus allowing us to make inferences about a population from our sample data.

In the context of the exercise, it's assumed that the BMI of American young women follows a normal distribution. This assumption allows us to apply certain statistical methods, like calculating the margin of error and constructing a confidence interval.

Here are some properties of normal distribution:
  • The mean, median, and mode of a normal distribution are all equal.
  • The distribution is symmetric around the mean.
  • The area under the curve corresponds to the probability, with a total area (probability) of 1.
  • The empirical rule states that approximately 68% of data falls within one standard deviation (\(\sigma\)), 95% within two \(\sigma\), and 99.7% within three \(\sigma\) of the mean.
Applications of normal distribution make it easier to perform statistical tests and determine probabilities for real-world scenarios, such as predicting population characteristics based on sample data.

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 most important condition for sound conclusions from statistical inference is usually that (a) the data can be thought of as a random sample from the population of interest. (b) the population distribution is exactly Normal. (c) the data contain no outliers.

Mortality rates vary from city to city in the United States. We have lots of data on many U.S. cities. Statistical software makes it easy to perform dozens of significance tests on dozens of variables to see which ones best predict mortality rate. One interesting finding is that those cities with major league ballparks tend to have significantly higher mortality rates than other cities. To improve your chances of a long life, should you use this "significant" variable to decide where to live? Explain your answer.

Many sample surveys use well-designed random samples, but half or more of the original sample can't be contacted or refuse to take part. Any errors due to this nonresponse (a) have no effect on the accuracy of confidence intervals. (b) are included in the announced margin of error. (c) are in addition to the random variation accounted for by the announced margin of error.

A college administrator questions the first 50 students he meets on campus the day after final exams are over. He asks them whether they had positive, neutral, or negative overall feelings about the term that had just ended. Suggest some reasons it may be risky to act as if the first 50 students at this particular time are an SRS of all students at this college.

Which of the following questions does a test of significance answer? Briefly explain your replies. (a) Is the sample or experiment properly designed? (b) Is the observed effect due to chance? (c) Is the observed effect important?

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.