/*! 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 3 The following information is obt... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 3

The following information is obtained from two independent samples selected from two normally distributed populations. $$ \begin{array}{lll} n_{1}=18 & \bar{x}_{1}=7.82 & \sigma_{1}=2.35 \\ n_{2}=15 & \bar{x}_{2}=5.99 & \sigma_{2}=3.17 \end{array} $$ a. What is the point estimate of \(\mu_{1}-\mu_{2}\) ? b. Construct a \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). Find the margin of error for this estimate.

Short Answer

Expert verified
The point estimate of \( \mu_{1}-\mu_{2} \) is 1.83. For finding the confidence interval and margin of error, the standard error(SE) must be calculated first using \( SE = \sqrt{ \frac{{\sigma^2_1}}{{n_1}} + \frac{{\sigma^2_2}}{{n_2}} } \). Once SE is found, confidence interval could be found using \( CI = 1.83 卤 (2.576 * SE) \) and margin of error using \( MoE = 2.576 * SE \). These values could be calculated in the next steps given the information from the problem.

Step by step solution

01

Calculate the Point Estimate

First, it's required to calculate the point estimate of \( \mu_{1}-\mu_{2} \). The point estimate can be calculated as the difference between the sample means, \(\bar{x}_{1} - \bar{x}_{2}\). Plugging in the provided values from the problem:\( \bar{x}_{1} = 7.82 \) , \( \bar{x}_{2} = 5.99 \)the point estimate is \( \bar{x}_{1} - \bar{x}_{2} = 7.82 - 5.99 = 1.83 \).
02

Finding the Standard Error

Before we move forward to finding the confidence interval, we need to find the standard error for the difference of the means. Standard Error (SE) can be calculated using the formula: \[ SE = \sqrt{ \frac{{\sigma^2_1}}{{n_1}} + \frac{{\sigma^2_2}}{{n_2}} }\]Here, \( \sigma_1 = 2.35 \), \( \sigma_2 = 3.17 \), \( n_1 = 18 \), \( n_2 = 15 \).After substituting these values:\[ SE = \sqrt{ \frac{{2.35^2}}{18} + \frac{{3.17^2}}{15} } \]
03

Constructing the Confidence interval

Next, we need to construct a 99% confidence interval for \( \mu_{1}-\mu_{2} \). As per the Z-distribution table, Z value for 99% confidence interval is approximately 2.576. Using the formula:\[ CI = Point \: estimate 卤 (Z \: value * SE) \]In the case at hand: \[ CI = 1.83 卤 (2.576 * SE) \]
04

Compute the Margin of Error

Finally, find the margin of error for this estimate, which is actually the part after the 卤 sign in the confidence interval calculation, that is, \( MoE = Z \: value * SE \).So the margin error for this problem would be \( MoE = 2.576 * SE \).

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

A consulting agency was asked by a large insurance company to investigate if business majors were better salespersons than those with other majors. A sample of 20 salespersons with a business degree showed that they sold an average of 11 insurance policies per week. Another sample of 25 salespersons with a degree other than business showed that they sold an average of 9 insurance policies per week.

Assuming that the two populations are normally distributed with unequal and unknown population standard deviations, construct a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) for the following. $$ \begin{array}{lll} n_{1}=14 & \bar{x}_{1}=109.43 & s_{1}=2.26 \\ n_{2}=15 & \bar{x}_{2}=113.88 & s_{2}=5.84 \end{array} $$

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 \(=90 \%\)

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

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?
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

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.