Title: Relationships in Categorical Data with Intro to Probability
Course: Question Banks

Q: The U.S. government collects information on Americans who do not have health insurance. Here is the data, sorted by geographic region:

Region | Uninsured | Insured | Row Totals
Northeast | 6,782 | 47,043 | 53,825
Midwest | 7,757 | 57,135 | 64,892
South | 19,090 | 85,800 | 104,890
West | 11,676 | 55,427 | 67,103
Column Totals | 45,305 | 245,405 | 290,710

The percentage of people from the West that are insured is
- 55,427 / 245,405 which is about 22.6% (You calculated the people from the west with insurance out of the total people who have insurance. Remember that the explanatory variable determines the denominator. Since we are comparing regions, the total in the West is the denominator.)
- 55,427 / 290,710 which is about 19.1% (you correctly identified the number of insured people in the West. But you divided by the total number of people instead of the number of people in the West.)
- 67,103 / 245,405 which is about 27.3% (You divided the total number of people living in the West by the total who are insured. Start by identifying the total in the West. Now determine how many of that group are insured.)
- 55,427 / 67,103 which is about 82.6% (The total people in the West is 67,103. Out of these people we want the percentage that are insured.)
A: 55,427 / 67,103 which is about 82.6%

Q: The following table summarizes the full-time enrollment at a community college located in a West Coast city. There are a total of 12,000 full-time students enrolled at the college. The two categorical variables here are gender and program. The programs include academic and vocational programs at the college. Assume that a student can enroll in only one program.

 | Arts-Sci | Bus-Econ | Info Tech | Health Science | Graphics Design | Culinary Arts | Row Totals
Female | 4,660 | 435 | 494 | 421 | 105 | 83 | 6,198
Male | 4,334 | 490 | 564 | 223 | 97 | 94 | 5,802
Column Totals | 8,994 | 925 | 1,058 | 644 | 202 | 177 | 12,000

If we select a student at random the marginal probability that they are male, that is, P(student is male)is approximately:
- 0.02 (Did you divide 12000 by 5802? If so, you identified the correct numbers in the table, but you divided backwards.)
- 0.52 (P(female) is 6198 / 12000, which is about 0.52. We want P(male).)
- 0.48 (5802 of the 12000 students are male, so P(male) is 5802 / 12000, or about 0.48.)
A: 0.48

Q: The following table summarizes the full-time enrollment at a community college located in a West Coast city. There are a total of 12,000 full-time students enrolled at the college. The two categorical variables here are gender and program. The programs include academic and vocational programs at the college. Assume that a student can enroll in only one program.

 | Arts-Sci | Bus-Econ | Info Tech | Health Science | Graphics Design | Culinary Arts | Row Totals
Female | 4,660 | 435 | 494 | 421 | 105 | 83 | 6,198
Male | 4,334 | 490 | 564 | 223 | 97 | 94 | 5,802
Column Totals | 8,994 | 925 | 1,058 | 644 | 202 | 177 | 12,000

Suppose we select a male student. What is the probability that he is in the Graphics Design program?
- 97 out of 202 (You have misidentified the condition. The fraction you chose selects a student that is in the Graphics Design program (202) then finds the male students in that group (97).)
- 97 out of 12000 (The fraction you chose computes the probability that a randomly selected student from the entire college is both in Graphics Design and male. However, the condition that the student is a male defines the initial group as 5802.)
- 97 out of 105 (There is no probability that relates male students in the Graphics Design program to female students in that program. This is a conditional probability. What is the condition?)
- 97 out of 5802 (There are 5802 male students. 97 of the male students are in the Graphics Design program.)
A: 97 out of 5802

Q: The following table summarizes the full-time enrollment at a community college located in a West Coast city. There are a total of 12,000 full-time students enrolled at the college. The two categorical variables here are gender and program. The programs include academic and vocational programs at the college. Assume that a student can enroll in only one program.

 | Arts-Sci | Bus-Econ | Info Tech | Health Science | Graphics Design | Culinary Arts | Row Totals
Female | 4,660 | 435 | 494 | 421 | 105 | 83 | 6,198
Male | 4,334 | 490 | 564 | 223 | 97 | 94 | 5,802
Column Totals | 8,994 | 925 | 1,058 | 644 | 202 | 177 | 12,000

P(male | Culinary Arts) =
- 0.03 (Did you calculate 177 out of 5802? There is no probability that relates total male students to total Culinary Arts students. This is a conditional probability. The condition defines the initial group and is the denominator of the fraction.)
- 0.01 (Did you calculate 94 out of 12000? The fraction you chose computes the probability that a randomly selected student from the entire college is both in Culinary Arts and male. However, the condition that the student is in the Culinary Arts program defines the initial group as 177.)
- 0.02 (Did you calculate 94 out of 5802? You have misidentified the condition. The condition that the student is in the Culinary Arts program defines the initial group as 177.)
- 0.53 (Of the 177 Culinary Arts students, 94 are male. 94/177 = 0.53)
A: 0.53

Q: The following table summarizes the full-time enrollment at a community college located in a West Coast city. There are a total of 12,000 full-time students enrolled at the college. The two categorical variables here are gender and program. The programs include academic and vocational programs at the college. Assume that a student can enroll in only one program.

 | Arts-Sci | Bus-Econ | Info Tech | Health Science | Graphics Design | Culinary Arts | Row Totals
Female | 4,660 | 435 | 494 | 421 | 105 | 83 | 6,198
Male | 4,334 | 490 | 564 | 223 | 97 | 94 | 5,802
Column Totals | 8,994 | 925 | 1,058 | 644 | 202 | 177 | 12,000

What is P(Info Tech | Male)?
- 490 of 925 (This is P(Male | Info Tech). The probability you found describes the chance that a randomly selected student from the Info Tech program is a male. We want the conditional probability that a randomly selected male student is in Info Tech. You should first focus on the condition that the student is a male and look at the row for Male.)
- 494 of 6198 (This is P(Info Tech | Female). The problem we wish to find is the probability that a randomly chosen student is in the Info Tech program, given that they are male. You have computed the probability that a random student is in the Info Tech program, given that they are female. Be sure to read both the table and the question carefully.)
- 490 of 12000 (You’ve found the joint probability that a randomly selected student is both male and in Info Tech. We want the conditional probability that a randomly selected male student is in Info Tech. You should first focus on the condition that the student is a male and look at the row for Male.)
- 564 of 5802 (Of the 5802 males, 564 are in Info Tech.)
A: 564 of 5802

Q: The following table summarizes the full-time enrollment at a community college located in a West Coast city. There are a total of 12,000 full-time students enrolled at the college. The two categorical variables here are gender and program. The programs include academic and vocational programs at the college. Assume that a student can enroll in only one program.

 | Arts-Sci | Bus-Econ | Info Tech | Health Science | Graphics Design | Culinary Arts | Row Totals
Female | 4,660 | 435 | 494 | 421 | 105 | 83 | 6,198
Male | 4,334 | 490 | 564 | 223 | 97 | 94 | 5,802
Column Totals | 8,994 | 925 | 1,058 | 644 | 202 | 177 | 12,000

What is the probability that a randomly selected student is in the Culinary Arts program?
- 94 of 5802 (This is the probability that a randomly selected male is in Culinary arts. We were looking for the probability that any student is in Culinary Arts, regardless of his/her gender.)
- 94 of 12000 (This is a joint probability that a randomly selected student is male and in culinary arts. We were looking for a marginal probability that a student is in the culinary arts program, only. Of the 12000 students, 177 are in culinary arts.)
- 83 of 177 (This is the probability that a randomly selected culinary arts student is female. We were looking for the probability that any student is in Culinary Arts, regardless of her/his gender.)
- 177 of 12000 (Of the 12000 total students, 177 are in culinary arts.)
A: 177 of 12000

Q: The following table summarizes the full-time enrollment at a community college located in a West Coast city. There are a total of 12,000 full-time students enrolled at the college. The two categorical variables here are gender and program. The programs include academic and vocational programs at the college. Assume that a student can enroll in only one program.

 | Arts-Sci | Bus-Econ | Info Tech | Health Science | Graphics Design | Culinary Arts | Row Totals
Female | 4,660 | 435 | 494 | 421 | 105 | 83 | 6,198
Male | 4,334 | 490 | 564 | 223 | 97 | 94 | 5,802
Column Totals | 8,994 | 925 | 1,058 | 644 | 202 | 177 | 12,000

What is the probability that a randomly selected graphics design student is female?
- 0.48 (You’ve computed the probability that a graphics design student is male (97 / 202). You are on the right track because you computed a conditional probability. But the problem asked about females, and not males.)
- 0.02 (You’ve reversed the condition in this conditional probability statement. We wished to find P(female | graphics design). Instead, 0.02 represents P(graphics design | female). You should focus on the condition that the student is in graphics design and look at the column margin.)
- 0.009 (You’ve computed the joint probability that a randomly selected student is both female and in graphics design (105 / 12000). However we want the conditional probability that a randomly selected graphics design student is female.)
- 0.52 (Of the 202 graphics design students 105 are female. This is 0.52, or 52%.)
A: 0.52

Q: The following table summarizes the full-time enrollment at a community college located in a West Coast city. There are a total of 12,000 full-time students enrolled at the college. The two categorical variables here are gender and program. The programs include academic and vocational programs at the college. Assume that a student can enroll in only one program.

 | Arts-Sci | Bus-Econ | Info Tech | Health Science | Graphics Design | Culinary Arts | Row Totals
Female | 4,660 | 435 | 494 | 421 | 105 | 83 | 6,198
Male | 4,334 | 490 | 564 | 223 | 97 | 94 | 5,802
Column Totals | 8,994 | 925 | 1,058 | 644 | 202 | 177 | 12,000

What is P(Female and Info Tech)?
- 0.52 (You found the marginal probability that any randomly selected student is female (6198 / 12000). It includes females in all programs. We are only interested in females that are also in the Info Tech program. This is a joint probability.)
- 0.08 (You found the conditional probability that a randomly selected female is in info tech (494 / 6198). We wish to compute a joint probability, and not a conditional probability. In a joint probability we are randomly selecting a student from the entire sample.)
- 0.47 (You found the conditional probability that a randomly selected Info Tech student is female (494 / 1058). We wish to compute a joint probability, and not a conditional probability. In a joint probability we are randomly selecting a student from the entire sample.)
- 0.04 (Of the 12000 total students, 494 are both female and in Info Tech. 494 / 12000 = 0.04.)
A: 0.04

Q:  | No Alcohol Problems | Moderate to Severe Alcohol Problems | Row Totals
Participated in 9/11 Rescue | 793 | 309 | 1,102
Did Not Participate in 9/11 Rescue | 441 | 110 | 551
Column Totals | 1,234 | 419 | 1,653

The table above summarizes the results of a survey of firefighters and other crisis responders (such as police, EMT, etc.) in the New York area.

Do the firefighters and other “first responders” to the 9/11 crisis at the World Trade Center have a higher risk of alcohol-related problems? Which of the following calculations are the most helpful for answering this question?
- 309 / 1,653 and 110 / 1,653 (These are joint probabilities. We need conditional probabilities that compare alcohol problems for those involved in 9/11 rescue and those not involved.)
- 309 / 419 and 110 / 419 (These are the conditional probabilities: P( 9/11 rescuer | alcohol problems) and P( not a 9/11 rescuer | alcohol problems). The condition is the same in both. We need to compare the rate of alcohol problems in the two conditions that are defined by the explanatory variable, Participation in 9/11 Rescue.)
- 309 / 419 and 793 / 1,234 (You are right that we want conditional probabilities. But these probabilities misidentify the condition. We want to compare those who participated in the 9/11 rescue and those who did not. So we will only be working with numbers in the rows in this table.)
- 309 /1,102 and 110 / 551 (These are the conditional probabilities that compare alcohol problems in the two groups defined by the 9/11 Rescue variable: P( alcohol problems | 9/11 rescuer) and P( alcohol problems | not a 9/11 rescuer).)
A: 309 /1,102 and 110 / 551

Q:  | No Alcohol Problems | Moderate to Severe Alcohol Problems | Row Totals
Participated in 9/11 Rescue | 793 | 309 | 1,102
Did Not Participate in 9/11 Rescue | 441 | 110 | 551
Column Totals | 1,234 | 419 | 1,653

The table above summarizes the results of a survey of firefighters and other crisis responders (such as police, EMT, etc.) in the New York area.

We can calculate the risk of alcohol-related problems for those involved in 9/11 response and those not involved.

P( alcohol problems | 9/11 rescuer) = 0.28 and P( alcohol problems | not a 9/11 rescuer) = 0.20

Is the following a valid statement?

Based on this data, those that participated in the 9/11 rescue are more likely to have alcohol-related problems.
- invalid (“More likely” means a higher risk. The risk of alcohol-related problems in the 9/11 rescuers is higher.)
- valid (The risk of alcohol-related problems in the 9/11 rescuers is higher.)
A: valid

Q:  | No Alcohol Problems | Moderate to Severe Alcohol Problems | Row Totals
Participated in 9/11 Rescue | 793 | 309 | 1,102
Did Not Participate in 9/11 Rescue | 441 | 110 | 551
Column Totals | 1,234 | 419 | 1,653

The table above summarizes the results of a survey of firefighters and other crisis responders (such as police, EMT, etc.) in the New York area.

We can calculate the risk of alcohol-related problems for those involved in 9/11 response and those not involved.

P( alcohol problems | 9/11 rescuer) = 0.28 and P( alcohol problems | not a 9/11 rescuer) = 0.20

Is the following a valid statement?

Based on this data, participation in the 9/11 rescue increases the risk of alcohol-related problems by 40%.
- invalid (The correct calculation is (0.28 – 0.20) / 0.20 = 0.40, which is a 40% increase in risk!)
- valid (The correct calculation is (0.28 – 0.20) / 0.20 = 0.40, which is a 40% increase in risk!)
A: valid

Q: Assume that 1% of the population have a very early stage of a bone disease. If this disease is detected early enough, medical treatment can prevent it from becoming a problem. It is therefore desirable to develop a test that can be used for early diagnosis.

The following information is available on one such test.

1. If a person with the disease is tested, the test will come out positive 80% of the time.
2. If a person does not have the disease, it is still possible to get a positive test result 2% of the time.

- Question: How likely is it that a person who gets a positive test result actually has the bone disease?

Your task is to construct a hypothetical two-way table consistent with the given information and use it to answer the question. Base your two-way table on a total count of 10,000. (Here we chose a total of 10,000 so that the counts in the table end up as whole numbers. Don’t worry about determining this ahead of time.)

How likely is it that a person that gets a positive test result actually has the disease?

Which of the following is a correct representation of the probability above:
- P(no disease | positive) (You have correctly identified this as a conditional probability. You have also found the correct given condition - a positive test result. But we want the probability that a person has the disease, given that the test is positive.)
- P(positive and disease) (We want a conditional probability, not a joint probability. We know that the test came back positive, so this is the given condition.)
- P(positive | disease) (You are right that this is a conditional probability. But, you have incorrectly identified the given condition. We know that the test came back positive, so that is the given condition.)
- P(disease | positive) (We know that the test came back positive, so this is the given condition. The outcome we’re looking for is that the person has the disease given that the test is positive. This describes a case where the test is correct!)
A: P(disease | positive)

Q: Assume that 1% of the population have a very early stage of a bone disease. If this disease is detected early enough, medical treatment can prevent it from becoming a problem. It is therefore desirable to develop a test that can be used for early diagnosis.

The following information is available on one such test.

1. If a person with the disease is tested, the test will come out positive 80% of the time.
2. If a person does not have the disease, it is still possible to get a positive test result 2% of the time.

- Question: How likely is it that a person who gets a positive test result actually has the bone disease?

Your task is to construct a hypothetical two-way table consistent with the given information and use it to answer the question. Base your two-way table on a total count of 10,000. (Here we chose a total of 10,000 so that the counts in the table end up as whole numbers. Don’t worry about determining this ahead of time.)

If a person’s test comes back positive, what is the probability that the test is correct and the person does indeed have the disease?
- 0.71 (You’ve computed the probability that a person does not have the disease, given that the test is positive. Instead, we wish to find the probability that the person does have the disease, given that the test is positive. Be sure to read the table and question carefully.)
- 0.80 (You’ve confused the given condition here. The probability you computed is P(positive | disease), or the probability that the test comes back positive, given that the person has the disease. You should reverse the conditions, and concentrate on the row for test positive instead of the column for has disease.)
- 0.008 (You’ve computed the joint probability that a person has the disease, and tested positive. Instead, we wish to compute the probability that a person has the disease given that the test came back with a positive result.)
- 0.29 (We get 0.29 by computing 80 / 278. That is, the number of people who actually have the disease (80) divided by the total amount of people who test positive (278).)
A: 0.29