/*! 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 114 What sample size is needed to gi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 114

What sample size is needed to give the desired margin of error in estimating a population mean with the indicated level of confidence? A margin of error within 卤12 with \(95 \%\) confidence, assuming we estimate that \(\sigma \approx 125\)

Short Answer

Expert verified
The required sample size to estimate the population mean with a margin of error 卤12 and a level of confidence of 95% is 107.

Step by step solution

01

Identify known values

First, we confirm the values provided in the problem. From the problem, it鈥檚 known that the margin of error \(E=12\), the standard deviation \(\sigma = 125\), and the value of z for a 95% confidence level is approximately 1.96 (using a Z-table or similar resource).
02

Apply the formula for the margin of error

The formula to calculate the margin of error when the population standard deviation is known can be expressed as \(E = Z * \frac{蟽}{\sqrt{n}}\). We can rearrange this formula to isolate \(n\), getting \(\sqrt{n} = \frac{Z * 蟽}{E}\) or \(n = (\frac{Z * 蟽}{E})^2\). Plug in the known values: \(n = (\frac{1.96 * 125}{12})^2\).
03

Calculate Sample Size

Perform the calculation inside the parenthesis first then square the result. It gets approximately 106.72. Since we cannot have a fraction of a participant, we should always round up to ensure the sample size is sufficiently large. Therefore, the smallest necessary sample size is 107.

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

Use StatKey or other technology to generate a bootstrap distribution of sample means and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample standard deviation as an estimate of the population standard deviation. Mean price of used Mustang cars online (in \(\$ 1000\) s) using the data in MustangPrice with \(n=25,\) \(\bar{x}=15.98,\) and \(s=11.11\)

In Exercises 6.1 to \(6.6,\) if random samples of the given size are drawn from a population with the given proportion: (a) Find the mean and standard error of the distribution of sample proportions. (b) If the sample size is large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 1000 from a population with proportion 0.70

In the mid-1990s a Nabisco marketing campaign claimed that there were at least 1000 chips in every bag of Chips Ahoy! cookies. A group of Air Force cadets collected a sample of 42 bags of Chips Ahoy! cookies, bought from locations all across the country, to verify this claim. \({ }^{41}\) The cookies were dissolved in water and the number of chips (any piece of chocolate) in each bag were hand counted by the cadets. The average number of chips per bag was \(1261.6,\) with standard deviation 117.6 chips. (a) Why were the cookies bought from locations all over the country? (b) Test whether the average number of chips per bag is greater than 1000 . Show all details. (c) Does part (b) confirm Nabisco's claim that every bag has at least 1000 chips? Why or why not?

A sample with \(n=12, \bar{x}=7.6,\) and \(s=1.6\)

We see } in the AllCountries dataset that the percent of the population living in rural areas is 8.0 in Argentina and 34.4 in Bolivia. Suppose we take random samples of size 200 from each country, and compute the difference in sample proportions \(\hat{p}_{A}-\hat{p}_{B},\) where \(\hat{p}_{A}\) represents the sample proportion living in rural areas in Argentina and \(\hat{p}_{B}\) represents the proportion of the sample that lives in rural areas in Bolivia. (a) Find the mean and standard deviation of the distribution of differences in sample proportions, \(\hat{p}_{A}-\hat{p}_{B}\) (b) If the sample sizes are large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. (c) Using the graph drawn in part (b), are we likely to see a difference in sample proportions as large in magnitude as -0.4 ? As large as \(-0.3 ?\) Explain.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

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.