/*! 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} Q81E Suppose the standard deviation o... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose the standard deviation of the population is knownto beσ=200 . Calculate the standard error of X for eachof the situations described in Exercise 6.80.

Short Answer

Expert verified
  1. The standard deviation is 4.898979.
  2. The standard deviation is 5.656854.
  3. The standard deviation is 6
  4. The standard deviation is 6.292853

Step by step solution

01

Given information

Assume the population standard deviation is known to beσ=200

a. n= 1,000, N= 2,500

b. n= 1,000, N= 5,000

c. n= 1,000, N= 10,000

d. n= 1,000, N= 100,000

02

Finding a standard deviation

The standard deviation is

σ^x¯=σnN-nN=20010002500-10002500=4.898979

The standard deviation is 4.898979

03

Finding a standard deviation

The standard deviation is

σ^x¯=σnN-nN=20010005000-10005000=5.656854

The standard deviation is 5.656854

04

Finding a standard deviation

The standard deviation is

σ^x¯=σnN-nN=200100010000-100010000=6

The standard deviation is 6

05

Finding a standard deviation

The standard deviation is

σ^x¯=σnN-nN=2001000100000-1000100000=6.292853

The standard deviation is 6.292853.

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

Employees with substance abuse problems. According to the New Jersey Governor’s Council for a Drug-Free Workplace Report, 50 of the 72 sampled businesses that are members of the council admitted that they had employees with substance abuse problems. At the time of the survey, 251 New Jersey businesses were members of the Governor’s Council. Use the finite population correction factor to find a 95% confidence interval for the proportion of all New Jersey Governor’s Council business members who have employees with substance abuse problems. Interpret the resulting interval.

Suppose you have selected a random sample of n = 5 measurements from a normal distribution. Compare the standard normal z-values with the corresponding t-values if you were forming the following confidence intervals.

a. 80% confidence interval

b. 90% confidence interval

c. 95% confidence interval

d. 98% confidence interval

e. 99% confidence interval

f. Use the table values you obtained in parts a–e to sketch the z- and t-distributions. What are the similarities and differences?

Suppose you wish to estimate the mean of a normal population

using a 95% confidence interval, and you know from prior information thatσ2≈1

a. To see the effect of the sample size on the width of the confidence interval, calculate the width of the confidence interval for n= 16, 25, 49, 100, and 400.

b. Plot the width as a function of sample size non graph paper. Connect the points by a smooth curve and note how the width decreases as nincreases.

The following is a 90% confidence interval for p:(0.26, 0.54). How large was the sample used to construct thisinterval?

What is the confidence level of each of the following confidence intervals μ?

  1. χ¯±1.96σn
  2. χ¯±1.645σn
  3. χ¯±2,575σn
  4. χ¯±1.282σn
  5. χ¯±.99σn
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.