/*! 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. 12.104 The Quinnipiac University Poll c... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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?

Short Answer

Expert verified

We know, chi-square is 7.05,df=6, p-value is0.3162.
Also, x2=7.05, where 7.05<10.6446rejecting the null hypothesis.

On concluding, we can say that the satisfaction levels are not differing among the three categories of people.

Step by step solution

01

Step 1. Given information.

Consider the given question,

02

Step 2. Consider the null and alternative hypotheses.

Null Hypothesis is the satisfaction levels are not differing among the three categories of people, H0:R=B=P

Alternate Hypothesis is the satisfaction levels are not differing among the three categories of people, H1:R≠B≠P

According to the decision rule,

When P-value is less than the Level of significance then it results in the rejection of the null hypothesis.

When the test statistics value is greater than the tabulated value then reject the null hypothesis.

03

Step 3. Calculate the x2 test for the given data.

On calculating the x2test,

Chi-square=7.05,

df=6,

p-value=0.3162

04

Step 4. Calculate the test statistics value.

Consider the above table,

x2=∑O-E2E=7.05

Hence, we know x2=7.05,P-value=0.3162.

Compare x2-test value and critical value with r-1×c-1=4-1×3-1=6at 10%level of significance is role="math" localid="1651939512728" 10.6446,7.05<10.6446, fail to reject null hypothesis.

Therefore, on concluding, we can say that the satisfaction levels are not differing among the three categories of people.

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 91Ó°ÊÓ!

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

Presidential Election. According to Dave Leip's Atlas of U.S. Presidential Elections, in the 2012 presidential election, 51.01%of those voting voted for the Democratic candidate (Barack H. Obama), whereas 57.50%of those voting who lived in Illinois did so. For that presidential election, does an association exist between the variables "party of presidential candidate voted for" and "state of residence" for those who voted? Explain your answer.

To decide whether two variables of a population are associated, we usually need to resort to inferential methods such as the chi-square independence test. Why?

HPV Vaccine. In the article "Correlates for Completic 3-dose Regimen of HPV Vaccine in Female Members of a Man Care Organization" (Mayo Clinical Proceedings, Vol. 84, Pp. 870 ), C. Chao et al. examined factors that may influence wh young female patients complete a three-injection sequence of the dasil quadrivalent human papillomavirus vaccine (HPV4). HPV virus that has been linked to the development of cervical cancer. following contingency table summarizes the data obtained for ' pletion of treatment versus practice type.

In each of Exercises 12.18-12.23, we have provided a distribution and the observed frequencies of the values of a variable from a simple random sample of a population. In each case, use the chi-square goodness-of-fit test to decide, at the specified significance level, whether the distribution of the variable differs from the given distribution.

Distribution: 0.2,0.1,0.1,0.3,0.3

Observed frequencies: 29,13,5,25,28

Significance level =0.10

US. Hospitals. Refer to Exercise 12.50.


24 or fewer25-7475 or moreTotal
General260158635575403
Psychiatric24242471737
Chronic132226
Tuberculosis0224
Other25177208410
Total310201042606580

a. Determine the conditional distribution of number of beds within each facility type.

b. Does an association exist between facility type and number of beds for U.S. hospitals? Explain your answer.

c. Determine the marginal distribution of number of beds for U.S. hospitals.

d. Construct a segmented bar graph for the conditional distributions and marginal distribution of number of beds. Interpret the graph in light of your answer to part (b),

e. Without doing any further calculations, respond true or false to the following statement and explain your answer. "The conditional distributions of facility type within number-of-beds categories are identical.

f. Determine the marginal distribution of facility type and the conditional distributions of facility type within number-of-beds categories.

g. What percentage of hospitals are general facilities?

h. What percentage of hospitals that have at least 75 beds are general facilities?

i. What percentage of general facilities have at least 75 beds?

See all solutions

Recommended explanations on Math Textbooks

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.