The manager of a high school cafeteria is planning to offer several new types of food for student lunches in the following school year. She wants to know if each type of food will be equally popular so she can start ordering supplies. To find out, she selects a random sample of 100 students and asks them, "Which type of food do you prefer: Asian food, Mexican food, pizza, or hamburgers?" Here are the data in the table provided. Identify the p-value and decide which of the following is the most appropriate conclusion?

Because 0.0129 is less than α = 0.05, reject H 0 . There is convincing evidence that the food choices are not equally popular.

Because 0.0129 is less than α = 0.05, fail to reject H 0 . There is convincing evidence that the food choices are equally popular.

Because 0.0289 is less than α = 0.05, reject H 0 . There is convincing evidence that the food choices are not equally popular.

Because 0.0289 is less than α = 0.05, fail to reject H 0 . There is not convincing evidence that the food choices are equally popular.

Recent revenue shortfalls in a midwestern state led to a reduction in the state budget for higher education. To offset the reduction, the largest state university proposed a 25% tuition increase. It was determined that such an increase was needed simply to compensate for the lost support from the state. Separate random samples of 50 freshmen, 50 sophomores, 50 juniors, and 50 seniors from the university were asked whether they were strongly opposed to the increase, given that it was the minimum increase necessary to maintain the university's budget at current levels. The results are in the table provided. Why hypotheses would be appropriate for performing a chi-square test?

The null hypothesis is that the distribution of student opinion about the proposed tuition increase is the same for each of the 4 years at the university. The alternative is that the distribution is different for at least 2 of the 4 years.

The null hypothesis is that the closer students get to graduation, the less likely they are to be opposed to tuition increases. The alternative is that how close students are to graduation makes no difference in their opinion.

The null hypothesis is that the mean number of students who are strongly are opposed is the same for each of the 4 years. The alternative is that the mean is different for at least 2 of the 4 years.

The null hypothesis is that year in school and student opinion about the tuition increase in the sample are independent. The alternative is that these variables are dependent.

All current-carrying wires produce electro-magnetic (EM) radiation. High-frequency EM radiation is thought to be a cause of cancer. The lower frequencies are generally assumed to be harmless. To investigate the relationship between current configuration and type of cancer, researchers visited the addresses of a random sample of children who died of some form of cancer (leukemia, lymphoma, or some other type) and classified the wiring configuration outside the dwelling as either high-current configuration (HCC) or low-current configuration (LCC). The data is provided in a table. Which of the following may we conclude, based on the test results?

There is not convincing evidence of an association between wiring configuration and the type of cancer that cause the deaths of children in the study.

There is convincing evidence of an association between wiring configuration and the chance that a child will develop some form of cancer.

HCC either causes cancer directly or is a major contributing factor to the development of cancer in children.

There is convincing evidence that HCC does not cause cancer in children.

Which statement is NOT correct about the chi-square test statistic?

a)A value close to 0 would indicate expected counts are much different from observed counts.

b)A large value of the test statistic would be in support of the alternative hypothesis.

c)A small value of the test statistic would indicate evidence supporting the null hypothesis.

d)The test statistic is the sum of positive numbers and therefore must be positive.

Which of the following is false ?

The degrees of freedom for chi-square test is determined by sample size.

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

A chi-square distribution never takes negative values.

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

Using a computer program, the chi-square test statistic for the commuter data is c2 = 5.37 with P-value < 0.01. If we use a = 0.05, what conclusion is appropriate?

c)Reject H 0; the results are significant.

a)Don’t reject H 0 ; the results are significant.

b)Don’t reject H 0 ; the results are not significant.

d)Reject H 0 ; the results are not significant.

The manager of a high school cafeteria is planning to offer several new types of food for student lunches in the following school year. She wants to know if each type of food will be equally popular so she can start ordering supplies making other plans. To find out, she selects a random sample of 100 students and asks them, "Which type of food do you prefer: Asian food, Mexican food, pizza, or hamburgers?" Here are the data in the table provided. An appropriate null hypothesis to test whether the food choices are equally popular is:

H 0 : p A = p M = p P = p H = 0.25, where η A is the proportion of students in the sample who would choose Asian food, and so on.

H 0 : μ = 25, where μ = the mean number of students that prefer each type of food.

H 0 : p = 0.25, where p = the proportion of all students who prefer Asian food.

H 0 : η A = η M = η P = η H = 0.25, where η A is the number of students in the school who would choose Asian food, and so on.

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

(18-25) 2 /25 + (22-25) 2 /25 + (39-25) 2 /25 + (21-25) 2 /25

(25-18) 2 /18 + (25-22) 2 /22 + (25-39) 2 /39 + (25-21) 2 /21

(18-25) /25 + (22-25) /25 + (39-25) /25 + (21-25) /25

(0.18-0.25) 2 /0.25 + (0.22-0.25) 2 /0.25 + (0.39-0.25) 2 /0.25 + (0.21-0.25) 2 /0.25

The manager of a high school cafeteria is planning to offer several new types of food for student lunches in the following school year. She wants to know if each type of food will be equally popular so she can start ordering supplies making other plans. To find out, she selects a random sample of 100 students and asks them, "Which type of food do you prefer: Asian food, Mexican food, pizza, or hamburgers?" Here are the data in the table provided. The P-value for a chi-square test for goodness of fit is 0.0129. Which is the following is the most appropriate conclusion?

Because 0.0129 is less than α = 0.05, reject H 0 . There is convincing evidence that the food choices are not equally popular.

Because 0.0129 is less than α = 0.05, reject H 0 . There is convincing evidence that the food choices are equally popular.

Because 0.0129 is less than α = 0.05, reject H 0 . There is not convincing evidence that the food choices are equally popular.

Because 0.0129 is less than α = 0.05, fail to reject H 0 . There is convincing evidence that the food choices are equally popular.

Recent revenue shortfalls in a midwestern state led to a reduction in the state budget for higher education. To offset the reduction, the largest state university proposed a 25% tuition increase. It was determined that such an increase was needed simply to compensate for the lost support from the state. Separate random samples of 50 freshmen, 50 sophomores, 50 juniors, and 50 seniors from the university were asked whether they were strongly opposed to the increase, given that it was the minimum increase necessary to maintain the university's budget at current levels. The results are in the table provided. Why hypotheses would be appropriate for performing a chi-square test?

The null hypothesis is that the distribution of student opinion about the proposed tuition increase is the same for each of the 4 years at the university. The alternative is that the distribution is different for at least 2 of the 4 years.

The null hypothesis is that the closer students get to graduation, the less likely they are to be opposed to tuition increases. The alternative is that how close students are to graduation makes no difference in their opinion.

The null hypothesis is that the mean number of students who are strongly are opposed is the same for each of the 4 years. The alternative is that the mean is different for at least 2 of the 4 years.

The null hypothesis is that year in school and student opinion about the tuition increase in the sample are independent. The alternative is that these variables are dependent.

Chapter 11 AP Statistics

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11th - 12th Grade

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Practice Problem

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Hard

Barbara White

FREE Resource

30 questions

Show all answers

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Find the degrees of freedom.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

The table shows the number or babies born on each day of the week. Is there evidence that births are more likely on specific days of the week?

yes, the p value is greater than .05

no, the p value is less than .05

yes, the p value is less than .05

no, the p value is greater than .05

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What must be true about the expected values in a chi square test?

greater than or equal to 2

greater than or equal to 5

greater than or equal to 10

greater than or equal to 30

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Which test would you use to see if the distribution of music that Californians listen to is the same as that of Nevadans? 900 Californians and 600 Nevadans were surveyed.

Chi-square Goodness of Fit Test

Chi-Square Test for Homogeneity

Chi-Square Test for Independence/Association

Not a Chi-square test

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Which test would you use .... In 1974, each acre of the Tahoe basin was assessed to see what condition the soil was in. 23% of acres were of type I, 32% were of type II, 41% were of type III, and 16% were of type IV. This year the region was assessed again and the number of each type were again determined.

Chi-square Goodness of Fit Test

Chi-Square Test for Homogeneity

Chi-Square Test for Independence/Association

Not a Chi-square test

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Which test would you use to see if people on computers and people on handhelds use the same Internet search engines? 1000 web searches from a computer and 1500 web searches from a hand held were tracked to see what search engine was used.

Chi-square Goodness of Fit Test

Chi-Square Test for Homogeneity

Chi-Square Test for Independence/Association

Not a Chi-square test

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Which test would you use to see if gender is a factor choosing a person’s favorite fruit? 900 college students were surveyed and their favorite fruit recorded.

Chi-square Goodness of Fit Test

Chi-Square Test for Homogeneity

Chi-Square Test for Independence/Association

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The table shows the number or babies born on each day of the week. Is there evidence that births are more likely on specific days of the week?

What must be true about the expected values in a chi square test?

Which test would you use to see if the distribution of music that Californians listen to is the same as that of Nevadans? 900 Californians and 600 Nevadans were surveyed.

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What are the expected counts of a female who likes Pepsi?

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