/*! 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 59 When are the samples considered ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 59

When are the samples considered large enough for the sampling distribution of the difference between two sample proportions to be (approximately) normal?

Short Answer

Expert verified
The samples are considered large enough for the sampling distribution of the difference between two sample proportions to be approximately normal when the sample size and proportions satisfy the following conditions: n1*p1 > 5, n1*(1 - p1) > 5, n2*p2 > 5, and n2*(1 - p2) > 5.

Step by step solution

01

Understanding the Condition for Approximate Normality

The sampling distribution of the difference between two independent proportions is approximately normal when the sample size is large enough. The generally accepted rule for large enough sample sizes involves checking the following conditions for both groups being sampled: n*p > 5 and n*(1 - p) > 5. Here, n is the sample size and p is the proportion.
02

Applying the Rule to Samples

In practice, if both of the following conditions hold: n1*p1 > 5, n1*(1 - p1) > 5, n2*p2 > 5, and n2*(1 - p2) > 5, where n1 and n2 are the sample sizes for first and second group, and p1 and p2 are the proportions in the first and second group respectively. Then the samples are large enough for the distribution of the difference between them to be approximately normal.
03

Explanation

These rules 鈥� i.e., n*p > 5 and n*(1 - p) > 5 for both population samples 鈥� are usually used to ensure that the sample size n is large enough so that the binomial distribution can be approximated using the normal distribution. If these inequalities hold for both the sample proportions, then the difference between them will also follow an approximate normal distribution.

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

Construct a \(95 \%\) confidence interval for \(p_{1}-p_{2}\) for the following. $$ n_{1}=100 \quad \hat{p}_{1}=.81 \quad n_{2}=150 \quad \hat{p}_{2}=.77 $$

Find the following confidence intervals for \(\mu_{d}\), assuming that the populations of paired differences are normally distributed. a. \(n=12, \bar{d}=17.5, \quad s_{d}=6.3, \quad\) confidence level \(=99 \%\) b. \(n=27, \quad \bar{d}=55.9, \quad s_{d}=14.7\), confidence level \(=95 \%\) c. \(n=16, \bar{d}=29.3, \quad s_{d}=8.3, \quad\) confidence level \(=90 \%\)

What is the shape of the sampling distribution of \(\hat{p}_{1}-\hat{p}_{2}\) for two large samples? What are the mean and standard deviation of this sampling distribution?

The manager of a factory has devised a detailed plan for evacuating the building as quickly as possible in the event of a fire or other emergency. An industrial psychologist believes that workers actually leave the factory faster at closing time without following any system. The company holds fire drills periodically in which a bell sounds and workers leave the building according to the system. The evacuation time for each drill is recorded. For comparison, the psychologist also records the evacuation time when the bell sounds for closing time each day. A random sample of 36 fire drills showed a mean evacuation time of \(5.1\) minutes with a standard deviation of \(1.1\) minutes. A random sample of 37 days at closing time showed a mean evacuation time of \(4.2\) minutes with a standard deviation of \(1.0\) minute. a. Construct a \(99 \%\) confidence interval for the difference between the two population means. b. Test at the \(5 \%\) significance level whether the mean evacuation time is smaller at closing time than during fire drills.

The following information was obtained from two independent samples selected from two normally distributed populations with unknown but equal standard deviations. $$ \begin{array}{lll} n_{1}=21 & \bar{x}_{1}=13.97 & s_{1}=3.78 \\ n_{2}=20 & \bar{x}_{2}=15.55 & s_{2}=3.26 \end{array} $$ a. What is the point estimate of \(\mu_{1}-\mu_{2} ? \quad\) b. Construct a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\).
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

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.