/*! 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 47 A sample of 11 observations take... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 47

A sample of 11 observations taken from a normally distributed population produced the following data. \(\begin{array}{lllllllllll}-7.1 & 10.3 & 8.7 & -3.6 & -6.0 & -7.5 & 5.2 & 3.7 & 9.8 & -4.4 & 6.4\end{array}\) a. What is the point estimate of \(\mu\) ? b. Make a \(95 \%\) confidence interval for \(\mu\). c. What is the margin of error of estimate for \(\mu\) in part b?

Short Answer

Expert verified
The point estimate of \(\mu\) is the sample mean from step 1. The 95% confidence interval for \(\mu\) is the result from step 4. The margin of error of the estimate for \(\mu\) is the result from step 5.

Step by step solution

01

Calculate the sample mean

First, add all the 11 data numbers together and then divide by 11 to find the mean of the sample. The sample mean will be the point estimate of \(\mu\).
02

Calculate the standard deviation

To calculate the standard deviation, subtract the mean from each data point, square the result, sum them up, divide the sum by the number of data points minus 1 (i.e., 10), and then take the square root.
03

Calculate the z-score

The z-score for a 95% confidence level in a normal distribution is approximately 1.96, determined from the z-table or from statistical software or calculators.
04

Calculate the confidence interval

To calculate the 95% confidence interval for \(\mu\), subtract and add the product of the z-score and the standard deviation divided by the square root of the sample size to the sample mean. The result will be the lower and upper boundaries of the confidence interval.
05

Determine the margin of error

The margin of error of the estimate for \(\mu\) in part b is half of the width of the confidence interval. Therefore, subtract the lower boundary of the confidence interval from the upper boundary and divide by 2.

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.

Point Estimate
The point estimate is a single value used to estimate a population parameter. In this exercise, we're interested in estimating the population mean, denoted by \( \mu \). To find the point estimate for \( \mu \), we use the sample mean. The sample mean is calculated by adding all the observations in the sample and dividing by the number of observations. This calculated mean serves as our point estimate for \( \mu \), providing a simple but powerful summary of the sample data.
Sample Mean
The sample mean is a foundational concept in statistics, representing the average value of a set of data points. To find it, add together all the values in your sample, and then divide by the number of values. This value is often symbolized by \( \bar{x} \).- **Formula**: \[ \bar{x} = \frac{\sum x_i}{n} \] where \( x_i \) represents each value in the sample and \( n \) is the number of values. The sample mean is used in various calculations, including forming the basis for the confidence interval and point estimate.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation means that the values tend to be close to the mean of the set, while a high standard deviation means that the values are spread out.Here鈥檚 how you calculate the standard deviation:1. Subtract the sample mean from each data point.2. Square the result for each calculation. 3. Sum all the squared results.4. Divide by \( n - 1 \) (where \( n \) is the number of data points).5. Take the square root of the result to get the standard deviation.This tells us how much the individual data points deviate, on average, from the mean.
Margin of Error
The margin of error is a crucial component of a confidence interval. It provides a range that the true population parameter is likely to fall within.To find the margin of error, you need:- **Z-score**: For a 95% confidence level, this is approximately 1.96.- **Standard deviation**: Dispersion measure of your sample.- **Sample size**: The number of observations. The formula to calculate the margin of error is: \( \text{Margin of Error} = z \times \left(\frac{\text{standard deviation}}{\sqrt{n}}\right) \)It is half the width of the confidence interval, thereby indicating the uncertainty level in your point estimate for \( \mu \).

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

A company that produces detergents wants to estimate the mean amount of detergent in 64 -ounce jugs at a \(99 \%\) confidence level. The company knows that the standard deviation of the amounts of detergent in all such jugs is \(.20\) ounce. How large a sample should the company select so that the estimate is within \(.04\) ounce of the population mean?

A researcher wants to determine a \(99 \%\) confidence interval for the mean number of hours that adults spend per week doing community service. How large a sample should the researcher select so that the estimate is within \(1.2\) hours of the population mean? Assume that the standard deviation for time spent per week doing community service by all adults is 3 hours.

At Farmer's Dairy, a machine is set to fill 32 -ounce milk cartons. However, this machine does not put exactly 32 ounces of milk into each carton; the amount varies slightly from carton to carton. It is known that when the machine is working properly, the mean net weight of these cartons is 32 ounces. The standard deviation of the amounts of milk in all such cartons is always equal to \(.15\) ounce. The quality control department takes a sample of 25 such cartons every week, calculates the mean net weight of these cartons, and makes a \(99 \%\) confidence interval for the population mean. If either the upper limit of this confidence interval is greater than \(32.15\) ounces or the lower limit of this confidence interval is less than \(31.85\) ounces, the machine is stopped and adjusted. A recent sample of 25 such cartons produced a mean net weight of \(31.94\) ounces. Based on this sample, will you conclude that the machine needs an adjustment? Assume that the amounts of milk put in all such cartons have a normal distribution.

For a data set obtained from a sample, \(n=20\) and \(\bar{x}=24.5 .\) It is known that \(\sigma=3.1\). The population is normally distributed. a. What is the point estimate of \(\mu ?\) b. Make a \(99 \%\) confidence interval for \(\mu\). c. What is the margin of error of estimate for part b?

\(8.104\) A random sample of 25 life insurance policyholders showed that the average premium they pay on their life insurance policies is \(\$ 685\) per year with a standard deviation of \(\$ 74\). Assuming that the life insurance policy premiums for all life insurance policyholders have a normal distribution, make a \(99 \%\) confidence interval for the population mean, \(\mu\).
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Calculus

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.