/*! 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 22 Find and interpret a \(95 \%\) c... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 22

Find and interpret a \(95 \%\) confidence interval for a population mean \(\mu\) for these values: a. \(n=36, \bar{x}=13.1, s^{2}=3.42\) b. \(n=64, \bar{x}=2.73, s^{2}=.1047\)

Short Answer

Expert verified
Answer: For the first set of values, the 95% confidence interval for the population mean 渭 is between 12.49 and 13.71. For the second set of values, the 95% confidence interval for the population mean 渭 is between 2.65 and 2.81. These intervals can be interpreted to mean that we are 95% confident that the true population mean falls within these ranges for each respective set of values.

Step by step solution

01

Identify the given values

In this exercise, we are given two different sets of values (a and b), which include the sample size (\(n\)), sample mean (\(\bar{x}\)), and the sample variance (\(s^2\)). First, let's identify the given values for each case: a. \(n=36, \bar{x}=13.1, s^{2}=3.42\) b. \(n=64, \bar{x}=2.73, s^{2}=.1047\)
02

Calculate the standard deviation

Now, calculate the sample standard deviation (\(s\)) by taking the square root of the sample variance (\(s^2\)) for each case: a. \(s = \sqrt{3.42} \approx 1.85\) b. \(s = \sqrt{.1047} \approx 0.32\)
03

Find the standard error

The standard error (\(SE\)) represents the uncertainty in the sample mean estimation. It can be calculated by dividing the standard deviation by the square root of the sample size: \(SE = \frac{s}{\sqrt{n}}\) Calculate the standard error for each case: a. \(SE_a = \frac{1.85}{\sqrt{36}} \approx 0.31\) b. \(SE_b = \frac{0.32}{\sqrt{64}} \approx 0.04\)
04

Determine the Confidence Interval using the z score

For a 95% confidence interval, the z score for a standard normal distribution is approximately 1.96. We can find the margin of error (\(ME\)) and compute the confidence interval using the formulas: \(ME = 1.96 \times SE\) \(CI = (\bar{x} - ME, \bar{x} + ME)\) Calculate the confidence interval for each case: a. \(ME_a = 1.96 \times 0.31 \approx 0.61\) \(CI_a = (13.1 - 0.61, 13.1 + 0.61) = (12.49, 13.71)\) b. \(ME_b = 1.96 \times 0.04 \approx 0.08\) \(CI_b = (2.73 - 0.08, 2.73 + 0.08) = (2.65, 2.81)\)
05

Interpret the Confidence Intervals

Now that we have the 95% confidence intervals, we can interpret them as follows: a. We are 95% confident that the population mean \(\mu\) is between 12.49 and 13.71 for the first set of values. b. We are 95% confident that the population mean \(\mu\) is between 2.65 and 2.81 for the second set of values.

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!

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 study of the relationship between birth order and college success, an investigator found that 126 in a sample of 180 college graduates were firstborn or only children. In a sample of 100 nongraduates of comparable age and socioeconomic background, the number of firstborn or only children was \(54 .\) Estimate the difference between the proportions of firstborn or only children in the two populations from which these samples were drawn. Use a \(90 \%\) confidence interval and interpret your results.

You want to estimate the mean hourly yield for a process that manufactures an antibiotic. You observe the process for 100 hourly periods chosen at random, with the results \(\bar{x}=34\) ounces per hour and \(s=3\). Estimate the mean hourly yield for the process using a \(95 \%\) confidence interval.

Exercise 8.106 presented statistics from a study of fast starts by ice hockey skaters. The mean and standard deviation of the 69 individual average acceleration measurements over the 6 -meter distance were 2.962 and .529 meters per second, respectively. a. Find a \(95 \%\) confidence interval for this population mean. Interpret the interval. b. Suppose you were dissatisfied with the width of this confidence interval and wanted to cut the interval in half by increasing the sample size. How many skaters (total) would have to be included in the study?

In a study to compare the effects of two pain relievers it was found that of \(n_{1}=200\) randomly selectd individuals instructed to use the first pain reliever, \(93 \%\) indicated that it relieved their pain. Of \(n_{2}=450\) randomly selected individuals instructed to use the second pain reliever, \(96 \%\) indicated that it relieved their pain. a. Find a \(99 \%\) confidence interval for the difference in the proportions experiencing relief from pain for these two pain relievers. b. Based on the confidence interval in part a, is there sufficient evidence to indicate a difference in the proportions experiencing relief for the two pain relievers? Explain.

Don't Americans know that eating pizza and french fries leads to being overweight? In the same American Demographics article referenced in Exercise \(8.98,\) a survey of women who are the main meal preparers in their households reported these results: \(\cdot$$90 \%\) know that obesity causes health problems. \(\cdot$$80 \%\) know that high fat intake may lead to health problems. \(\cdot$$86 \%\) know that cholesterol is a health problem. \(\cdot$$88 \%\) know that sodium may have negative effects on health. a. Suppose that this survey was based on a random sample of 750 women. How accurate do you expect the percentages given above to be in estimating the actual population percentages? (HINT: If these are the only four percentages for which you need a margin of error, a conservative estimate for \(p\) is \(p \approx .80 .)\) b. If you want to decrease your sampling error to \(\pm 1 \%,\) how large a sample should you take?
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Statistics

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.