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Sorry, no chi-square How do U.S. residents who travel overseas for leisure differ from

those who travel for business? The following is the breakdown by occupation.

Occupation
Leisure travelers (%)
Business travelers (%)
Professional/technical
36
39
Manager/executive
23
48
Retired
14
3
Student
7
3
Other
20
7
Total
100
100

Explain why we can’t use a chi-square test to learn whether these two distributions differ

significantly.

Short Answer

Expert verified

We can’t use a chi-square test to learn whether these two distributions differ significantly because the information given to us is in percentage and no information is given about total number of travellers.

Step by step solution

01

Step 1:Given Information 

We have been given that,

Percentage of U.S. residents who travel overseas for leisure,and

Percentage of U.S. residents who travel overseas for business.

Occupation
Leisure travelers (%)
Business travelers (%)
Professional/technical
36
39
Manager/executive
23
48
Retired
14
3
Student
7
3
Other
20
3
Total
100
100
02

Step 2:Explanation 

The information given to us is in percentage and no information is given about total number of travellers.

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Most popular questions from this chapter

The manager of a high school cafeteria is planning to offer several new types of food for student lunches in the new school year. She wants to know if each type of food will be equally popular so she can start ordering supplies and making other plans. To find out, she selects a random sample of 100students and asks them, “Which type of food do you prefer: Ramen, tacos, pizza, or hamburgers?” Here are her data:

The P-value for a chi-square test for goodness of fit is 0.0129. Which of the following is the most appropriate conclusion at a significance level of 0.05?

a. Because 0.0129 is less than α=0.05 reject H0 . There is convincing evidence that the food choices are equally popular.

b. Because 0.0129 is less than α=0.05 reject H0 There is not convincing

evidence that the food choices are equally popular.

c. Because 0.0129 is less than α=0.05 reject H0 . There is convincing evidence that the food choices are not equally popular.

d. Because 0.0129 is less than α=0.05 fail to reject H0 There is not convincing evidence that the food choices are equally popular.

e. Because 0.0129 is less than α=0.05 fail to reject H0 There is convincing

evidence that the food choices are equally popular.

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a. State appropriate hypotheses for a chi-square test.

b. Here is Minitab output for a chi-square test. Interpret the P-value. What conclusion would you draw?

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A random sample of traffic tickets given to motorists in a large city is examined. The tickets are classified according to the race or ethnicity of the driver. The results are summarized in the following table.

The proportion of this city's population in each of the racial/ethnic categories listed is as follows.

We wish to test H0: The racial/ethnic distribution of traffic tickets in the city is the same as the racial/ethnic distribution of the city's population.

Assuming H0is true, what is the expected number of Hispanic drivers who would receive a ticket?

a.8

b.10.36

c.11

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a. Should we use a chi-square test for homogeneity or a chi-square test for independence in this setting? Justify your answer.

b. State appropriate hypotheses for performing the type of test you chose in part (a). Here is Minitab output from a chi-square test.

c. Check that the conditions for carrying out the test are met.

d. Interpret the P-value. What conclusion would you draw?

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a. Is there a relationship between attendance at religious services and alcohol consumption? A random sample of 1000adults was asked whether they regularly attend religious services and whether they drink alcohol daily.

b. Separate random samples of 75 college students and 75 high school students were asked how much time, on average, they spend watching television each week. We want to estimate the difference in the average amount of TV watched by high school and college students.

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