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Chapter 11: Q 8. (page 716)

No chi-square The principal in Exercise 7 also asked the random sample of students to record whether they did all of the homework that was assigned on each of the five school days that week. Here are the data:

Short Answer

Expert verified

Because the observations are not independent, a chi-square test for goodness of fit is not appropriate.

Step by step solution

01

Given information

02

Explanation

The observations must be independent in order to use the chi-square test for goodness of fit. Because most kids will have the same teachers, independent observations are not possible. As a result, a chi-square test for goodness of fit would be inappropriate.

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Most popular questions from this chapter

More candy The two-way table shows the results of the experiment

described in Exercise 27.


Red SurveyBlue Survey
Control Survey
Total
Red Candy13
5
8
26
Blue Candy
7
15
12
34
Total20
20
20
60

a. State the appropriate null and alternative hypotheses.

b. Show the calculation for the expected count in the Red/Red cell. Then provide a

complete table of expected counts.

c. Calculate the value of the chi-square test statistic.

鈥淲ill changing the rating scale on a survey affect how people answer the question?鈥� To find out, the group took an SRS of 50students from an alphabetical roster of the school鈥檚 just over 1000students. The first 22students chosen were asked to rate the cafeteria food on a scale of 1(terrible) to 5(excellent). The remaining 28students were asked to rate the cafeteria food on a scale of 0(terrible) to 4(excellent). Here are the data:

a. Was this an observational study or an experiment? Justify your answer.

b. Explain why it would not be appropriate to perform a chi-square test in this setting.

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鈥檛 use a chi-square test to learn whether these two distributions differ

significantly.

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, 鈥淲hich type of food do you prefer: Ramen, tacos, pizza, or hamburgers?鈥� Here are her data:

The chi-square test statistic is

a. (18鈭�25)225+(22鈭�25)225+(39鈭�25)225+(21鈭�25)225

b. (25鈭�18)218+(25鈭�22)222+(25鈭�39)239+(25鈭�21)221

c. (18鈭�25)25+(22鈭�25)25+(39鈭�25)25+(21鈭�25)25

d. (18鈭�25)2100+(22鈭�25)2100+(39鈭�25)2100+(21鈭�25)2100

e. (0.18鈭�0.25)20.25+(0.22鈭�0.25)20.25+(0.39鈭�0.25)20.25+(0.21鈭�0.25)20.25

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, 鈥淲hich type of food do you prefer: Ramen, tacos, pizza, or hamburgers?鈥� Here are her data:

Which of the following is false?

a. A chi-square distribution with k degrees of freedom is more right-skewed than a chi square distribution with k+1 degrees of freedom.

b. A chi-square distribution never takes negative values.

c. The degrees of freedom for a chi-square test are determined by the sample size.

d. P(蠂2>10) is greater when df=k+1 than when df=k

e. The area under a chi-square density curve is always equal to 1
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