Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Q76E (page 452)

Random samples of size n1 = 21 and n2 = 31 were drawn from populations 1 and 2, respectively. For each of the following cases, determine the test statistic and appropriate decision for a two-tailed test of the null hypothesis \({H_0}:\sigma _1^2 = \sigma _2^2\) against \({H_a}:\sigma _1^2 \ne \sigma _2^2\,at\,\alpha = 0.05\)

  1. \(s_1^2 = 500\,and\;s_2^2 = 900\)

Short Answer

Expert verified
  1. The test statistic is 1.8.
    We fail to reject the null hypothesis.

Step by step solution

01

Given Information

The hypothesis are given by

\(\begin{aligned}{H_0}:\sigma _1^2 = \sigma _2^2\\{H_0}:\sigma _1^2 \ne \sigma _2^2\end{aligned}\)

The level of significance is 0.05.

02

Compute the test statistic

The test statistic is calculated as

\(\begin{aligned}F &= \frac{{s_2^2}}{{s_1^2}}\\ &= \frac{{900}}{{500}}\\ &= \frac{9}{5}\\ &= 1.8\end{aligned}\)

Therefore, the test statistic is 1.8.

03

Decision-rule

The rejection region for two tailed test is

\(F > {F_{\frac{\alpha }{2}}}\)

For

\(\begin{aligned}{F_{\frac{\alpha }{2}}}\left( {{n_1} - 1,{n_2} - 1} \right)\\ = {F_{0.025}}\left( {20,30} \right)\end{aligned}\)

Critical value: 0.4258 and 2.1952

Hence, the statistic doesn’t fall in rejection region.

Therefore, we fail to reject the null hypothesis.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Is honey a cough remedy? Refer to the Archives of Pediatrics and Adolescent Medicine (December 2007) study of honey as a remedy for coughing, Exercise 2.31 (p. 86). Recall that the 105 ill children in the sample were randomly divided into groups. One group received a dosage of an over-the-counter cough medicine (DM); another group received a dosage of honey (H). The coughing improvement scores (as determined by the children’s parents) for the patients in the two groups are reproduced in the accompanying table. The pediatric researchers desire information on the variation in coughing improvement scores for each of the two groups.

a. Find a 90% confidence interval for the standard deviation in improvement scores for the honey dosage group.

b. Repeat part a for the DM dosage group.

c. Based on the results, parts a and b, what conclusions can the pediatric researchers draw about which group has the smaller variation in improvement scores? (We demonstrate a more statistically valid method for comparing variances in Chapter 8.)

Honey Dosage

11 12 15 11 10 13 10 13 10 4 15 16 9 14 10 6 10 11 12 12 8 12 9 11 15 10 15 9 13 8 12 10 9 5 12

DM Dosage

4 6 9 4 7 7 7 9 12 10 11 6 3 4 9 12 7 6 8 12 12 4 12 13 7 10 13 9 4 4 10 15 9

The “last name” effect in purchasing. The Journal of Consumer Research (August 2011) published a study demonstrating the “last name” effect—i.e., the tendency for consumers with last names that begin with a later letter of the alphabet to purchase an item before consumers with last names that begin with earlier letters. To facilitate the analysis, the researchers assigned a number, x, to each consumer based on the first letter of the consumer’s last name. For example, last names beginning with “A” were assigned x = 1; last names beginning with “B” were assigned x = 2; and last names beginning with “Z” were assigned x = 26.

a. If the first letters of consumers’ last names are equally likely, find the probability distribution for x.

b. Find E (x) using the probability distribution, part a. If possible, give a practical interpretation of this value.?

c. Do you believe the probability distribution, part a, is realistic? Explain. How might you go about estimating the true probability distribution for x

Salmonella in yield. Salmonella infection is the most common bacterial foodborne illness in the United States. How current is Salmonella in yield grown in the major agricultural region of Monterey, California? Experimenters from the U.S. Department of Agriculture (USDA) conducted tests for Salmonella in yield grown in the region and published their results in Applied and Environmental Microbiology (April 2011). In a sample of 252 societies attained from water used to wash the region, 18 tested positive for Salmonella. In an independent sample of 476 societies attained from the region's wildlife (e.g., catcalls), 20 tested positive for Salmonella. Is this sufficient substantiation for the USDA to state that the frequency of Salmonella in the region's water differs from the frequency of Salmonella in the region's wildlife? Use a = .01 to make your decision

Corporate sustainability of CPA firms. Refer to the Business and Society (March 2011) study on the sustainability behaviors of CPA corporations, Exercise 2.23 (p. 83). Recall that the level of support for corporate sustainability (measured on a quantitative scale ranging from 0 to 160 points) was obtained for each of 992 senior managers at CPA firms. The accompanying Minitab printout gives the mean and standard deviation for the level of support variable. It can be shown that level of support is approximately normally distributed.

a. Find the probability that the level of support for corporate sustainability of a randomly selected senior manager is less than 40 points.

b. Find the probability that the level of support for corporate sustainability of a randomly selected senior manager is between 40 and 120 points.

c. Find the probability that the level of support for corporate sustainability of a randomly selected senior manager is greater than 120 points.

d. One-fourth of the 992 senior managers indicated a level of support for corporate sustainability below what value?

Descriptive Statistics: Support

Variables

N

Mean

StDev

Variance

Minimum

Maximum

Range

Support

992

67.755

26.871

722.036

0.000

155.000

155.000

Oil content of fried sweet potato chips. Refer to theJournal of Food Engineering (September 2013) study of the characteristics of fried sweet potato chips, Exercise 7.90 (p. 431). Recall that a sample of 6 sweet potato slices fried at 130° using a vacuum fryer yielded the following statistics on internal oil content (measured in gigagrams [Gg]): x1 = .178 Gg and s1 = .011 Gg. A second sample of 6 sweet potato slices was obtained, but these were subjected to a two-stage frying process (again, at 130°) in an attempt to
improve texture and appearance. Summary statistics on internal oil content for this second sample follows: x2 = .140 Gg and s2 = .002 Gg. Using a t-test, the researchers want to compare the mean internal oil content of sweet potato chips fried with the two methods. Do you recommend the researchers carry out this analysis? Explain.
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and 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.

Sign-up for free