/*! 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 78 A random sample of 300 female me... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A random sample of 300 female members of health clubs in Los Angeles showed that they spend, on average, \(4.5\) hours per week doing physical exercise with a standard deviation of . 75 hour. Find a \(98 \%\) confidence interval for the population mean.

Short Answer

Expert verified
The \(98 \%\) confidence interval for the population mean is therefore between \(4.5 - 2.33*SE\) and \(4.5 + 2.33*SE\).

Step by step solution

01

Calculate the Standard Error

The standard error is given by the formula \(SE=\frac{\sigma}{\sqrt{n}}\), where \(\sigma\) is the standard deviation and \(n\) is the sample size. So, in this case, the standard error would be \(SE=\frac{0.75}{\sqrt{300}}\).
02

Find the Z-score

For normal distributions, a 98% confidence interval corresponds to Z-scores of \(-2.33\) and \(2.33\). These are the number of standard deviations away from the mean that contain 98% of the population.
03

Calculate the Confidence Interval

A confidence interval can be calculated using the formula \(\mu = \bar{x} \pm Z*SE\), where \(\mu\) is the population mean, \(\bar{x}\) is the sample mean, \(Z\) is the Z-score for desired confidence level, and \(SE\) is the standard error. Hence, the confidence interval would be \(CI = 4.5 \pm 2.33*SE\).

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
The standard error (SE) is a crucial concept in statistics when it comes to understanding how well a sample mean estimates the population mean. It reflects the average amount by which the sample mean (the average calculated from a specific set of data) will differ from the actual population mean. Here's how to calculate it:
  • The formula for the standard error is \(SE=\frac{\sigma}{\sqrt{n}}\).
  • In this formula, \(\sigma\) is the standard deviation and provides a measure of how spread out the data points are around the mean.
  • \(n\) is the sample size, indicating how many observations are in your sample.
By dividing the standard deviation by the square root of the sample size, we gauge how the data points from the sample scatter around the mean. In our exercise, where the standard deviation is 0.75 hour, and the sample size is 300, the standard error helps us determine the margin of error or how much the sample mean might vary from the actual population mean. It's a critical step in constructing confidence intervals and drawing accurate conclusions from data.
Z-score
A Z-score is a numerical measurement that describes a value's relationship to the mean of a group of values. It is expressed in terms of standard deviations from the mean. In simpler terms, a Z-score tells how many standard deviations a data point is from the average of the dataset. The importance of a Z-score in constructing confidence intervals lies in its ability to define precisely the area under the normal distribution curve that corresponds to your confidence level. For a 98% confidence interval:
  • Z-scores of \(-2.33\) and \(2.33\) indicate the standard deviations below and above the mean, respectively, that contain 98% of the data.
  • This ensures that the calculated confidence interval encompasses roughly 98% of the possible mean values, offering you a reliable range for the population mean.
Understanding Z-scores allows statisticians and researchers to categorize a data point relative to a statistical distribution and aids in making probabilistic decisions about data.
Population Mean
The population mean is essentially the average of a set of characteristics or measurements in a population. It's a central concept in statistics, as it represents the real average we're trying to estimate with sample data. Since collecting data from an entire population is often impractical, we rely on samples to make estimates.The steps to estimate a population mean through sample data usually involve:
  • Calculating the sample mean \(\bar{x}\), which serves as a point estimate of the population mean.
  • Using the standard error and Z-score to determine a confidence interval around the sample mean.
In the example exercise, the sample mean given was 4.5 hours, and the task was to find a 98% confidence interval for the population mean. By factoring in the standard error and the Z-score associated with 98% confidence, a range is created. This range estimates the likely position of the true population mean with a certain level of confidence, helping ensure decisions and interpretations based on this data can be trusted.

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

In a January 2014 survey conducted by the Associated PressWe TV, \(68 \%\) of American adults said that owning a home is the most important thing or \(a\) very important but not the most important thing (opportunityagenda.org). Assume that this survey was based on a random sample of 900 American adults. a. Construct a \(95 \%\) confidence interval for the proportion of all American adults who will say that owning a home is the most important thing or a very important but not the most important thing. b. Explain why we need to construct a confidence interval. Why can we not simply say that \(68 \%\) of all American adults would say that owning a home is the most important thing or \(a\) very important but not the most important thing?

What is the point estimator of the population mean, \(\mu\) ? How would you calculate the margin of error for an estimate of \(\mu\) ?

activities (playing games, personal communications, etc.) during this month are as follows: $$ \begin{array}{lllllllll} 7 & 12 & 9 & 8 & 11 & 4 & 14 & 1 & 6 \end{array} $$ Assuming that such times for all employees are approximately normally distributed, make a \(95 \%\) confidence interval for the corresponding population mean for all employees of this company.A company randomly selected nine office employees and secretly monitored their computers for one month. The times (in hours) spent by these employees using their computers for non- job-related

A mail-order company promises its customers that the products ordered will be mailed within 72 hours after an order is placed. The quality control department at the company checks from time to time to see if this promise is fulfilled. Recently the quality control department took a random sample of 50 orders and found that 35 of them were mailed within 72 hours of the placement of the orders. a. Construct a \(98 \%\) confidence interval for the percentage of all orders that are mailed within 72 hours of their placement. b. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Discuss all possible alternatives. Which alternative is the best?

When one is attempting to determine the required sample size for estimating a population mean, and the information on the population standard deviation is not available, it may be feasible to take a small preliminary sample and use the sample standard deviation to estimate the required sample size, \(n .\) Suppose that we want to estimate \(\mu\), the mean commuting distance for students at a community college, to a margin of error within 1 mile with a confidence level of \(95 \%\). A random sample of 20 students yields a standard deviation of \(4.1\) miles. Use this value of the sample standard deviation, \(s\), to estimate the required sample size, \(n .\) Assume that the corresponding population has an approximate normal distribution.

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.