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- 1. Multiple Choice
The National Longitudinal Study of Adolescent Health interviewed a random sample of 4877 teens (grades 7 to 12). One question asked was "What do you think are the chances you will be married in the next ten years?" Above is a two-way table of the responses by gender.
Which of the following would be the most appropriate type of graph for these data?A bar chart showing the marginal distribution of opinion about marriage
bar chart showing the marginal distribution of gender
A bar chart showing the conditional distribution of gender for each opinion about marriage
A bar chart showing the conditional distribution of opinion about marriage for each gender
Dotplots that display the number in each opinion category for each gender
- 2. Multiple Choice
The National Longitudinal Study of Adolescent Health interviewed a random sample of 4877 teens (grades 7 to 12). One question asked was "What do you think are the chances you will be married in the next ten years?" Above is a two-way table of the responses by gender.
The appropriate null hypothesis for performing a chi-square test is thatequal proportions of female and male teenagers are almost certain they will be married in 10 years
there is no difference between the distributions of female and male teenagers' opinions about marriage in this sample
there is no difference between the distributions of female and male teenagers' opinions about marriage in the population
there is no association between gender and opinion about marriage in the sample
there is no association between gender and opinion about marriage in the population
- 3. Multiple Choice
The National Longitudinal Study of Adolescent Health interviewed a random sample of 4877 teens (grades 7 to 12). One question asked was "What do you think are the chances you will be married in the next ten years?" Above is a two-way table of the responses by gender.
The expected count of females who respond "almost certain" is487.7
525
965
1038.8
1174
- 4. Multiple Choice
The National Longitudinal Study of Adolescent Health interviewed a random sample of 4877 teens (grades 7 to 12). One question asked was "What do you think are the chances you will be married in the next ten years?" Above is a two-way table of the responses by gender.
The degrees of freedom for the chi-square test for this two-way table are:4
8
10
20
4876
- 5. Multiple Choice
The National Longitudinal Study of Adolescent Health interviewed a random sample of 4877 teens (grades 7 to 12). One question asked was "What do you think are the chances you will be married in the next ten years?" Above is a two-way table of the responses by gender.
For these data, χ2 = 69.8 with a P-value of approximately 0. Assuming that the researchers used a significance level of 0.05, which of the following is true?A type I error is possible
A type II error is possible
Both a type I and type II error are possible
There is no chance of making a type I or type II error because the P-value is approximately 0
There is no chance of making a type I and type II error because the calculations are correct
- 6. Multiple Choice
When analyzing survey results from a two-way table, the main distinction between a test for independence and a test for homogeneity is
how the degrees of freedom are calculated
how the expected counts are calculated
the number of samples obtained
the number of rows in the two-way table
the number of columns in the two-way table
- 7. Multiple Choice
A Chi-square goodness-of-fit test is used to test whether a 0 - 9 spinner is “fair” (that is, the outcomes are all equally likely). The spinner is spun 100 times and the results are recorded. The degrees of freedom for the test will be
8
9
10
99
none of these
- 8. Multiple Choice
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 or not they were strongly opposed to the increase, given that it was the minimum increase necessary to maintain the university’s budget at the current levels. The results are given in the table.
Which hypotheses would be appropriate for performing a chi-square test?
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 opinions.
The null hypothesis is that the mean number of students who are strongly opposed is the same for each of the four years. The alternative is that the mean is different for at least two of the four years.
The null hypothesis is that the distribution of student opinion about the proposed tuition increase is the same for each of the four years at this university. The alternative is that the distribution is different for at least two of the four 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.
The null hypothesis is that there is an association between year in school and opinion about the tuition increase at this university. The alternative hypothesis is that these variables are not associated.
- 9. Multiple Choice
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 or not they were strongly opposed to the increase, given that it was the minimum increase necessary to maintain the university’s budget at the current levels. The results are given in the table.
The conditions for carrying out the chi-square test is:
I. Separate random samples from the populations of interest
II. Expected counts large enough
III. The samples themselves and the individual observations in each sample are independent.
Which of the conditions is (are) satisfied in this case?
I only
II only
I and II only
II and III only
I, II and III
- 10. Multiple Choice
A random sample of traffic tickets given to motorists in a large city is examined. The tickets are classified according to the race of the driver. The results are summarized in the table.
We wish to test H0: The racial distribution of traffic tickets in the city is the same as the racial distribution of the city’s population.
Assuming H0 is true, the expected number of Hispanic drivers who would receive a ticket is
8
10.36
11
11.84
12
- 11. Multiple Choice
A random sample of traffic tickets given to motorists in a large city is examined. The tickets are classified according to the race of the driver. The results are summarized in the table.
We wish to test H0: The racial distribution of traffic tickets in the city is the same as the racial distribution of the city’s population.
We compute the value of X2 to be 6.58. Assuming the conditions for inference are met, the p-value of our test is
greater than 0.2
between 0.10 and 0.20
between 0.05 and 0.10
between 0.01 and 0.05
less than 0.01
- 12. Multiple Choice
A random sample of traffic tickets given to motorists in a large city is examined. The tickets are classified according to the race of the driver. The results are summarized in the table.
We wish to test H0: The racial distribution of traffic tickets in the city is the same as the racial distribution of the city’s population.
The category that contributes the largest component to the X2 statistic is
White
Black
Hispanic
Other
The answer cannot be determined since this is only a sample
- 13. Multiple Choice
All current-carrying wires produce electromagnetic (EM) radiation, including the electrical wiring running into, through, and out of our homes. High-frequency EM radiation is thought to be a cause of cancer. The lower frequencies associated with household current are generally assumed to be harmless. To investigate this, researchers visited the addresses of a random sample of children who had died of some form of cancer (leukemia, lymphoma, or some other type) and classified the wiring configuration outside the dwelling as either a high-current configuration (HCC) or a low-current configuration (LCC). Data is given in the table.
Computer software was used to analyze the data.
X2 = 0.082 + 0.170 + 0.023 + 0.048 + 0.099 + 0.013 = 0.435
The appropriate degrees of freedom for the statistic is
1
2
3
4
5
- 14. Multiple Choice
All current-carrying wires produce electromagnetic (EM) radiation, including the electrical wiring running into, through, and out of our homes. High-frequency EM radiation is thought to be a cause of cancer. The lower frequencies associated with household current are generally assumed to be harmless. To investigate this, researchers visited the addresses of a random sample of children who had died of some form of cancer (leukemia, lymphoma, or some other type) and classified the wiring configuration outside the dwelling as either a high-current configuration (HCC) or a low-current configuration (LCC). Data is given in the table.
Computer software was used to analyze the data.
X2 = 0.082 + 0.170 + 0.023 + 0.048 + 0.099 + 0.013 = 0.435
The appropriate count of cases with lymphoma in homes with an HCC is
None of these
- 15. Multiple Choice
All current-carrying wires produce electromagnetic (EM) radiation, including the electrical wiring running into, through, and out of our homes. High-frequency EM radiation is thought to be a cause of cancer. The lower frequencies associated with household current are generally assumed to be harmless. To investigate this, researchers visited the addresses of a random sample of children who had died of some form of cancer (leukemia, lymphoma, or some other type) and classified the wiring configuration outside the dwelling as either a high-current configuration (HCC) or a low-current configuration (LCC). Data is given in the table.
Computer software was used to analyze the data.
X2 = 0.082 + 0.170 + 0.023 + 0.048 + 0.099 + 0.013 = 0.435
Which of the following may we conclude, based on the test results?
There is strong 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.
Leukemia is the most common type of cancer among children.
There is not convincing evidence of an association between wiring configuration and the type of cancer that caused the deaths of children in the study.
There is a weak evidence that HCC causes cancer in children.
- 16. Multiple Choice
All current-carrying wires produce electromagnetic (EM) radiation, including the electrical wiring running into, through, and out of our homes. High-frequency EM radiation is thought to be a cause of cancer. The lower frequencies associated with household current are generally assumed to be harmless. To investigate this, researchers visited the addresses of a random sample of children who had died of some form of cancer (leukemia, lymphoma, or some other type) and classified the wiring configuration outside the dwelling as either a high-current configuration (HCC) or a low-current configuration (LCC). Data is given in the table.
Computer software was used to analyze the data.
X2 = 0.082 + 0.170 + 0.023 + 0.048 + 0.099 + 0.013 = 0.435
A Type I error would occur if we conclude that
HCC wiring caused cancer when it actually didn’t.
HCC wiring didn’t cause cancer when it actually did.
There is no association between the type of wiring and the form of cancer when there actually is an association.
There is an association between the type of wiring and the form of cancer when there actually is no association.
The type of wiring and the form of cancer have a positive correlation when they actually don’t.
- 17. Multiple Choice
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 and 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?" The table shows the data.
An appropriate null hypothesis to test whether the food choices are equally popular is:
H0 : μ = 25, where μ = the mean number of students that prefer each type of food
H0 : p = .25, where p = the proportion of all students that prefer Asian food.
H0 : nA = nM = nP = nH = 25, where nA is the number of students in the school who would choose Asian food, and so on.
H0 : pA = pM = pP = pH = .25, where pA is the proportion of students in the school who would choose Asian food, and so on.
H0 : p̂A = p̂M = p̂P = p̂H = .25, where p̂A is the number of students in the sample who would choose Asian food, and so on.
- 18. Multiple Choice
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 and 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?" The table shows the data.
The chi-square statistic is
- 19. Multiple Choice
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 and 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?" The table shows the 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?Because 0.0129 is less than α = 0.05, reject Ho. There is convincing evidence that the food choices are equally popular.
Because 0.0129 is less than α = 0.05, reject Ho. There is not convincing evidence that the food choices are equally popular.
Because 0.0129 is less than α = 0.05, reject Ho. There is convincing evidence that the food choices are not equally popular.
Because 0.0129 is less than α = 0.05, fail to reject Ho. There is not convincing evidence that the food choices are equally popular.
Because 0.0129 is less than α = 0.05, fail to reject Ho. There is convincing evidence that the food choices are equally popular.
- 20. Multiple Choice
Which of the following is false?
A chi-square 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 degrees of freedom for a chi-square test is determined by the sample size
P (X2 > 10) is greater when df = k+1 than when df = k
The area under a chi-square density curve is always equal to 1
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