/*! 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} Q. 12.78 We have presented a contingency ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Q. 12.78 (page 510)

We have presented a contingency table that gives a cross-classification of a random sample of values for two variables x and y, of a population.

Perform the following tasks

a. Find the expected frequencies Note: You will first need to compute the row totals, column totals, and grand total.

b. Determine the value of the chi-square statistic

c. Decide at the 5% significance level whether the data provide sufficient evidence to conclude that the two variables are associated.

Short Answer

Expert verified

The data do not provide sufficient evidence to conclude that the two variables are associated at the 5% significance level.

Step by step solution

01

Step 1. Given

02

Step 2. Solution a). Find the Expected frequencies using MINITAB

MINITAB procedure:

Step 1: Choose Stat > Tables > Chi-Square test (Two-Way Table in Worksheet).

Step 2: In Columns containing the table, enter the columns of AandB.

Step 3: Click OK.

03

Step 3. MINITAB output

04

Step 4. Row total, column total and grand total

yABTotal
a53540
b2080100
c3585110
Total50200250
05

Step 5. Expected Frequencies

yABTotal
a83240
b2080100
c2288110
Total50200250
06

Step 6. Solution b)

Determine the value of chi-squared statistic

From the MINITAB output, the value of chi-squared statistic is1.918

07

Step 7. Solution c) 

Check whether or not the data provide sufficient evidence to conclude that the two variables are associated at the 5% significance level

The hypotheses are given below

Null hypothesis:

: The two variables are not associated

Alternative hypothesis:

: The two variables are associated

08

Step 8. Conclusion for 5% significance level

From the output, the value of test statistic is 1.918and the p-value is 0.383.

Here, the p-value is lesser than the level of significance

That is, p-value(=0.383)>伪(=0.05).

Therefore, the null hypothesis is not rejected at 5% level

Thus, the data do not provide sufficient evidence to conclude that the two variables are associated at the 5% significance level.

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

In given exercise use either the critical-value approach or the P-value approach to perform a chi square independence lest. provided the conditions for using the test are met.

Siskel In the classic TV show Sneak Previews, originally hosted by the late Gene Siskel and Roger Ebert, the two Chicago movie critics reviewed the weeks new movie releases and

then rated them thumbs up (positive), mixed, thumbs down (negative). These two critics often saw the merits of a movie differently. In general, however, were ratings given by Siskel and Ebert

associated? The answer to this question was the focus of the paper "Evaluating Agreement and Disagreement Among Movie Reviewers" by A. Agresti and L. Winner that appeared in Chance

(Vol- 10(2), pp-10-14). The following contingency table summarizes ratings by Siskel and Ebert for 160 movies.

At the 1% significance level, do the data provide sufficient evidence to conclude that an association exists between ratings of Siskel and Ebert?

The Quinnipiac University Poll conducts nationwide surveys as a public service and of research. This problem is based on the results of once such poll. Independent simple random samples of 300residents each in read (predominantly Republiclian), blue (predominantly Democratic) and purple (mixed) states were asked how satisfied they were with the way things are going today. The following table summarizes the responses.

At the 10%significance level, do the data provide sufficient evidence to conclude that the satisfaction-level distributions differ among residents of red, blue and purple states?

In each of the given Exercises, we have presented a contingency table that gives a cross-classification of a random sample of values for two variables, x, and y, of a population. For each exercise, perform the following tasks.

a. Find the expected frequencies. Note: You will first need to compute the row totals, column totals, and grand total.

b. Determine the value of the chi-square statistic.

c. Decide at the 5% significance level whether the data provide sufficient evidence to conclude that the two variables are associated.

The U.S. Census Bureau publishes census data on the resident population of the United States in Current Population Reports. According to that document, 7.3% of male residents are in the age group20鈭�24 years.
a. If there is no association exists between age group and gender, what percentage of the resident population would be in the age group20鈭�24 years? Explain your answer.
b. If there is no association exists between age group and gender, what percentage of female residents would be in the age group20鈭�24 years? Explain your answer.
c. There are about 158million female residents of the United States. If no association exists between age group and gender, how many female residents would there be in the age group20鈭�24 years?
d. In fact, there are some 10.2 million female residents in the age group20鈭�24 years. Given this number and your answer to part (c), what do you conclude?

Variegated Plants. Arabidopsis is a genus of flowering plants related to cabbage. A variegated mutant of the Arabidopsis has yellow streaks or marks. E. Miura et al. studied the origin of this variegated mutant in the article "The Balance between Protein Synthesis and Degradation in Chloroplasts Determines Leaf Variegation in Arabidopsis yellow variegated Mutants" (The Plant Cell, Vol. 19, No. 4, pp. 1313-1328). In a second-generation cross of variegated plants, 216 were variegated and 84 were normal. Genetics predicts that 75% of crossed variegated plants would be variegated and 25% would be normal. At the 10% significance level, do the data provide sufficient evidence to conclude that the second generation of crossed variegated plants does not follow the genetic predictions?
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Applied 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.