/*! 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} Q 8.15. In each of Exercises 8.11-8.16, ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In each of Exercises 8.11-8.16, we provide a sample mean, sample size, and population standard deviation. In each case, perform the following Lacks.

a. Find a 95%confidence interval for the population means. (Note: You may want to review Example 8.2, which begins on page 3.16)

b. Identify and interpret the margin of error.

c. Express the endpoints of the confidence interval in terms of the point estimate and the margin of error.

x¯=50,n=16,σ=5

Short Answer

Expert verified

Part (a) The 95%confidence interval for the population mean is from 47.5to 52.5

Part (b) The population mean 95%to within of our sample mean with 2.5confidence.

Part (c) Point estimate ±Margin of error =50±2.5

Step by step solution

01

Part (a) Step 1: Given information

x¯=50,n=16,σ=5

02

Part (a) Step 2: Concept

The formula used:x¯-2σntox¯+2σn

03

Part (a) Step 3: Calculation

Find a 95%confidence interval for the population mean.

Consider x¯=50,n=16, and σ=5

Empirical rule:

Property 1: Around 68%of the data set is located between (x¯-s,x¯+s)

Property 2: Approximately 95%of the data set is contained within the range (x¯-2s,x¯+2s)

Property $3: Approximately 99.7%of the data set is located between (x¯-3s,x¯+3s)

Using Property 2, 95%of all observations fall within two standard deviations of the mean on either side.

The 95%confidence interval for the population mean is,

x¯-2σntox¯+2σn=50-2(5)16to50+2(5)16=50-104to50+104=50-2.5to50+2.5=47.5to52.5

Thus, the 95% confidence interval for the population mean is from 47.5 to 52.5

04

Part (b) Step 1: Explanation

Identify the margin of error.

From part a., the margin of error is 2.5

Interpretation:

With 95% confidence, the population mean (μ) can be projected to be within 2.5 of our sample mean.

05

Part (c) Step 1: Calculation

In terms of the point estimate and the margin of error, express the confidence interval's endpoints.

The point estimate and margin of error endpoints of the confidence interval are as follows: Point estimate ± Margin of error =50±2.5

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!

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

Prices of New Mobile Homes. Recall that a simple random sample of 36 new mobile homes yielded the prices, in thousands of dollars, shown in Table 8.1on page 315 . We found the mean of those prices to be \(63.28thousand.

a. Use this information and Procedure 8.1on page 322 to find a 95%confidence interval for the mean price of all new mobile homes. Recall that σ=\)7.2thousand.

b. Compare your 95%confidence interval in part (a) to the one found in Example 8.2(c) on page 317 and explain any discrepancy that you observe.

American Alligators. Refer to Exercise 8.78.

a. Determine the margin of error for the 95%confidence interval.

b. Determine the margin of error for the 99%confidence interval.

c. Compare the margins of error found in parts (a) and (b).

d. What principle is being illustrated?

"Chips Ahoy! 1,000 Chips Challenge." As reported by B. Warner and J. Rutledge in the paper "Checking the Chips Ahoy! Guarantee" (Chanee, Vol. 12. Issue 1. pp. 10-14), a random sample of forty-two 18-ounce bags of Chips Ahoy! cookies yielded a mean of 1261.6 chips per bag with a standard deviation of 117.6 chips per bug.

a. Determine a 95% confidence interval for the mean number of chips per bag for all 18-ounce bags of Chips Ahoy! cookies, and interpret your result in words.

b. Can you conclude that the average 18-ounce bag of Chips Ahoy! cookies contain at least 1000 chocolate chips? Explain your answer.

A simple random sample is taken from a population and yields the following data for a variable of the population:

find a point estimate for the population standard deviation (i.e., the standard deviation of the variable).

What is meant by saying that a 1-αconfidence interval is

a. exact?

b. approximately correct?

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.