Math 2 Quiz: Conditional Frequencies And Association

What this quiz covers

This quiz focuses on Conditional Frequencies And Association, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. There is a positive association since $P(\text{Good Sleep}|\text{Regular Exercise}) > P(\text{Good Sleep}|\text{Non-Regular Exercise})$ (correct answer)
2. There is no association since $P(\text{Regular Exercise}|\text{Good Sleep}) = P(\text{Regular Exercise}|\text{Poor Sleep})$
3. There is a negative association since $P(\text{Good Sleep}) < P(\text{Good Sleep}|\text{Regular Exercise})$
4. The association cannot be determined without knowing $P(\text{Regular Exercise}|\text{Good Sleep})$

Question 2

1. 60%, since Shift A produces the majority of total output regardless of defect considerations
2. 30%, calculated as the product of Shift A's output proportion and defect rate (0.60 × 0.03 = 0.018)
3. 43%, estimated as the weighted average of output proportions adjusted for relative defect rates
4. 39.1%, found using P(Shift A|Defective) = P(Defective|Shift A) × P(Shift A) ÷ P(Defective) (correct answer)

Question 3

1. It confirms positive association since both sets of conditional probabilities show consistent directional relationships between the variables
2. It contradicts their conclusion since P(Email Campaign|Purchase) < P(Email Campaign|No Purchase) suggests negative association
3. It indicates a calculation error since positive association should yield P(Email Campaign|Purchase) > P(Email Campaign|No Purchase) (correct answer)
4. It suggests the association is weak since the reverse conditional probabilities show smaller differences than the original ones

Question 4

1. The manager is correct because 180 - 140 = 40 employees, and 40/500 = 0.08 represents only 8% of the total sample
2. The manager is incorrect because conditional probabilities show P(Satisfied|Office) = 0.60 while P(Satisfied|Remote) = 0.70 (correct answer)
3. The manager is incorrect because the absolute difference of 40 employees represents a substantial portion of the smaller remote group
4. The manager is correct because statistical significance requires differences of at least 50 employees in samples of this size

Question 5

1. $P(\text{High BP}|\text{Takes Medication})$ must be greater than $P(\text{High BP}|\text{No Medication})$ (correct answer)
2. $P(\text{High BP}|\text{Takes Medication})$ must equal 0.72 since the probabilities are symmetric
3. $P(\text{High BP}|\text{Takes Medication})$ could be less than $P(\text{High BP}|\text{No Medication})$ depending on base rates
4. $P(\text{High BP}|\text{Takes Medication})$ cannot be determined without additional information about sample sizes

Question 6

1. Researcher A, because the difference of 0.36 in conditional probabilities shows stronger association than B's difference of 0.03
2. Researcher B, because conditioning on the explanatory variable (income) gives the proper measure of association strength
3. Both are correct since they're measuring different aspects of the same association using valid conditional probability comparisons (correct answer)
4. Neither is correct since association strength requires comparing joint probabilities, not conditional probabilities

Question 7

1. The overall probability $P(\text{Prefer A}) = 0.401$ is not equal to either conditional probability
2. The difference $|P(\text{Prefer A}|\text{Male}) - P(\text{Prefer A}|\text{Female})| = 0.04$ exceeds the margin of error
3. The conditional probability $P(\text{Male}|\text{Prefer A}) \neq P(\text{Male}|\text{Prefer B})$ when calculated from the data
4. The joint probabilities $P(\text{Male and Prefer A}) \neq P(\text{Male}) \times P(\text{Prefer A})$ violate independence (correct answer)

Question 8

1. Compare the overall iOS preference rate of 0.606 with both 0.64 and 0.58 to see which deviates more
2. Compare 0.64 and 0.58 directly, since these conditional probabilities show the association strength (correct answer)
3. Compare P(Smartphone|Prefers iOS) with P(Smartphone|Prefers Other OS) using the given information
4. Compare 0.45 and 0.55 with the conditional probabilities to determine if device distribution affects preferences

Question 9

1. 0.52 (correct answer)
2. 0.60
3. 0.48
4. 0.40

Question 10

1. 0.55
2. 0.61 (correct answer)
3. 0.50
4. 0.45

Question 11

1. The reasoning is flawed because it ignores the conditional relative frequencies: 15% vs 16.8%, showing the groups are essentially equivalent
2. The reasoning is flawed because it compares raw counts instead of conditional relative frequencies: 15% vs 16.8%, which actually supports the claim
3. The reasoning is flawed because it compares raw counts instead of conditional relative frequencies: 15% vs 16.8%, but this difference is still meaningful
4. The reasoning is flawed because comparing raw counts ignores group sizes: the conditional frequencies are 15% vs 16.8%, suggesting minimal association (correct answer)