/*! 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 34 Assuming that the two population... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 34

Assuming that the two populations have unequal and unknown population standard deviations, construct a \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) for the following. $$ \begin{array}{lll} n_{1}=48 & \bar{x}_{1}=.863 & s_{1}=.176 \\ n_{2}=46 & \bar{x}_{2}=.796 & s_{2}=.068 \end{array} $$

Short Answer

Expert verified
The 99% confidence interval for \( \mu_1 - \mu_2 \) is \((\bar{x}_1 - \bar{x}_2) \pm t_{\nu, \alpha/2} \cdot \sqrt{\frac{s_{1}^2}{n_{1}} + \frac{s_{2}^2}{n_{2}}\), where t_{\nu, \alpha/2} is the critical t-value from the t-distribution table and \(\nu\) is the degree of freedom approximated by the Welch鈥揝atterthwaite equation.

Step by step solution

01

Identify the given values

First identify the given values in the problem. Which are: \(n_1 = 48, \bar{x}_1 = .863, s_1 = .176, n_2 = 46, \bar{x}_2 = .796, s_2 = .068, \alpha = .01\) since we're looking for a 99% confidence interval (1 - 0.99 = 0.01).
02

Find the t-critical value

The critical value t* is looked up in the t-table based on the significance level \(\alpha\) and the degrees of freedom \(\nu\). As the standard deviations are unknown and assumed to be unequal (Welch's t-test), the degrees of freedom can be calculated by using the Welch鈥揝atterthwaite equation: \(\nu \approx \frac{(s_{1}^2/n_{1} + s_{2}^2/n_{2})^2}{((s_{1}^2/n_{1})^2 / (n_{1}-1)) + ((s_{2}^2/n_{2})^2 / (n_{2}-1))}\).
03

Find the Confidence Interval

Calculate the Confidence Interval using the formula \((\bar{x}_1 - \bar{x}_2) \pm t_{\nu, \alpha/2} \cdot \sqrt{\frac{s_{1}^2}{n_{1}} + \frac{s_{2}^2}{n_{2}}}\). The term \((\bar{x}_1 - \bar{x}_2)\) is the observed difference between the sample means, \(\pm t_{\nu, \alpha/2} \cdot \sqrt{\frac{s_{1}^2}{n_{1}} + \frac{s_{2}^2}{n_{2}}}\) is the margin of error.

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 July 2009 Pew Research Center survey asked a variety of science questions of independent random samples of scientists and the public at-large (http://people-press.org/report/528/). One of the questions asked was whether all parents should be required to vaccinate their children. The percentage of people answering "yes" to this question was \(69 \%\) of the general public and \(82 \%\) of scientists. Suppose that the survey included 110 members of the general public and 105 scientists. a. Construct a \(98 \%\) confidence interval for the difference between the two population proportions. b. Using the \(1 \%\) significance level, can you conclude that the percentage of the general public who feels that all parents should be required to vaccinate their children is less than the percentage of all scientists who feels that all parents should be required to vaccinate their children? Use the critical-value and \(p\) -value approaches. c. The actual sample sizes used in the survey were 2001 members of the general public and 1005 scientists. Repeat parts a and b using the actual sample sizes. Does your conclusion change in part b?

Does the use of cellular telephones increase the risk of brain tumors? Suppose that a manufacturer of cell phones hires you to answer this question because of concern about public liability suits. How would you conduct an experiment to address this question? Be specific. Explain how you would observe, how many observations you would take, and how you would analyze the data once you collect them. What are your null and alternative hypotheses? Would you want to use a higher or a lower significance level for the test? Explain.

A local college cafeteria has a self-service soft ice cream machine. The cafeteria provides bowls that can hold up to 16 ounces of ice cream. The food service manager is interested in comparing the average amount of ice cream dispensed by male students to the average amount dispensed by female students. A measurement device was placed on the ice cream machine to determine the amounts dispensed. Random samples of 85 male and 78 female students who got ice cream were selected. The sample averages were \(7.23\) and \(6.49\) ounces for the male and female students, respectively. Assume that the population standard deviations are \(1.22\) and \(1.17\) ounces, respectively. a. Let \(\mu_{1}\) and \(\mu_{2}\) be the population means of ice cream amounts dispensed by all male and female students at this college, respectively. What is the point estimate of \(\mu_{1}-\mu_{2} ?\) b. Construct a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). c. Using the \(1 \%\) significance level, can you conclude that the average amount of ice cream dispensed by male college students is larger than the average amount dispensed by female college students? Use both approaches to make this test.

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. $$ \begin{array}{lll} n_{1}=14 & \bar{x}_{1}=109.43 & s_{1}=2.26 \\ n_{2}=15 & \bar{x}_{2}=113.88 & s_{2}=5.84 \end{array} $$

According to the credit rating firm Equifax, credit limits on newly issued credit cards decreased between 2008 and the period of January to April 2009 (USA TODAY, July 7, 2009). Suppose that random samples of 200 credit cards issued in 2008 and 200 credit cards issued during the first 4 months of 2009 had average credit limits of \(\$ 4710\) and \(\$ 4602\), respectively, which are comparable to the values given in the article. Although no information about standard deviations was provided, suppose that the sample standard deviations for the 2008 and 2009 samples were \(\$ 485\) and \(\$ 447\), respectively, and that the assumption that the population standard deviations are equal for the two time periods is reasonable. a. Construct a \(95 \%\) confidence interval for the difference in the mean credit limits for all new credit cards issued in 2008 and during the first 4 months of 2009 . b. Using the \(2.5 \%\) significance level, can you conclude that the average credit limit for all new credit cards issued in 2008 was higher than the corresponding average for the first 4 months of 2009 ?
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

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.