Conditional Frequencies & Association - Math 2

1. A marketing team analyzes customer data and finds that P(Purchase|Email Campaign) = 0.28 and P(Purchase|No Email Campaign) = 0.15. They conclude there's positive association between email campaigns and purchases. However, they later discover that P(Email Campaign|Purchase) = 0.35 and P(Email Campaign|No Purchase) = 0.42. How should this additional information affect their conclusion?
   Answer: It indicates a calculation error since positive association should yield P(Email Campaign|Purchase) > P(Email Campaign|No Purchase) If there's positive association between email campaigns and purchases (as shown by P(Purchase|Email) > P(Purchase|No Email)), then we must also have P(Email|Purchase) > P(Email|No Purchase). The given values P(Email|Purchase) = 0.35 < 0.42 = P(Email|No Purchase) contradict positive association. This indicates either a calculation error or data inconsistency. Choices A and B incorrectly accept the contradictory information. Choice D misses the logical contradiction.
2. A survey of 500 employees examines the relationship between job satisfaction and work location. The results show that 180 of 300 office workers are satisfied, while 140 of 200 remote workers are satisfied. A manager argues there's no meaningful association because "the difference is only 10 employees." What is the most appropriate response?
   Answer: The manager is incorrect because conditional probabilities show P(Satisfied|Office) = 0.60 while P(Satisfied|Remote) = 0.70 Association should be assessed using conditional probabilities, not absolute differences. P(Satisfied|Office) = 180/300 = 0.60 and P(Satisfied|Remote) = 140/200 = 0.70. The 10 percentage point difference (0.70 - 0.60 = 0.10) shows a meaningful association favoring remote work satisfaction. Choice A miscalculates the difference and uses inappropriate reasoning. Choice C correctly identifies the manager's error but focuses on absolute numbers rather than rates. Choice D incorrectly invokes arbitrary significance thresholds for absolute differences.
3. A quality control analyst examines defect rates across two production shifts. Shift A produces 60% of total output with a 3% defect rate, while Shift B produces 40% of total output with a 7% defect rate. When examining defective items, what proportion comes from Shift A?
   Answer: 39.1%, found using P(Shift A|Defective) = P(Defective|Shift A) × P(Shift A) ÷ P(Defective) When you encounter a problem asking for the probability that an item with a certain characteristic came from a specific source, you're dealing with conditional probability and Bayes' theorem. This reverses the typical "given cause, find effect" to "given effect, find cause." To find what proportion of defective items comes from Shift A, you need to calculate P(Shift A|Defective). Using Bayes' theorem: P(Shift A|Defective) = P(Defective|Shift A) × P(Shift A) ÷ P(Defective). First, find the total defect rate: P(Defective) = (0.60 × 0.03) + (0.40 × 0.07) = 0.018 + 0.028 = 0.046. Then: P(Shift A|Defective) = (0.03 × 0.60) ÷ 0.046 = 0.018 ÷ 0.046 = 0.391 or 39.1%. Choice A incorrectly assumes the proportion of defective items mirrors total production, ignoring that Shift B has a higher defect rate. Choice B calculates only the proportion of total output that is defective from Shift A, not the proportion of defective items that came from Shift A. Choice C attempts some adjustment but lacks the mathematical rigor of Bayes' theorem and arrives at an incorrect estimate. The correct answer is D: 39.1%. Study tip: Conditional probability problems often involve reversing cause and effect. When you see "given that X occurred, what's the probability it came from source Y," immediately think Bayes' theorem. Always calculate the total probability of the condition occurring across all sources first.
4. A researcher studying the relationship between exercise frequency and sleep quality surveyed 400 adults. The data shows that 60% of regular exercisers report good sleep quality, while 35% of non-regular exercisers report good sleep quality. If 40% of all surveyed adults are regular exercisers, what is the conditional probability that a randomly selected adult who reports good sleep quality is a regular exerciser?
   Answer: 0.52 First, calculate the total probability of good sleep: P(good sleep) = P(good sleep|regular) × P(regular) + P(good sleep|non-regular) × P(non-regular) = 0.60 × 0.40 + 0.35 × 0.60 = 0.24 + 0.21 = 0.45. Then use Bayes' theorem: P(regular|good sleep) = P(good sleep|regular) × P(regular) / P(good sleep) = (0.60 × 0.40) / 0.45 = 0.24 / 0.45 ≈ 0.52. Choice B (0.60) is the conditional probability in the wrong direction. Choice C (0.48) would result from incorrectly calculating 1 - 0.52. Choice D (0.40) is simply the marginal probability of being a regular exerciser.
5. Two researchers analyze the same dataset on income level and shopping frequency. Researcher A claims strong positive association because $P(\text{High Income}|\text{Frequent Shopper}) = 0.68$ while $P(\text{High Income}|\text{Infrequent Shopper}) = 0.32$. Researcher B claims weak association because $P(\text{Frequent Shopper}|\text{High Income}) = 0.51$ while $P(\text{Frequent Shopper}|\text{Low Income}) = 0.48$. Who is correct?
   Answer: Both are correct since they're measuring different aspects of the same association using valid conditional probability comparisons Both researchers are examining the same association from different perspectives. Researcher A examines how income distribution varies between shopping groups, while Researcher B examines how shopping frequency varies between income groups. Both approaches are valid for assessing association - they're measuring the same underlying relationship. The apparent difference in strength (0.36 vs 0.03) reflects different base rates, not different association strengths. Choice A misinterprets the meaning of the differences. Choice B incorrectly suggests only one direction is valid. Choice D is wrong because conditional probability comparisons are the standard method for assessing association.
6. A study of 500 patients examined the relationship between treatment type and recovery outcome. Treatment A had a 75% recovery rate, Treatment B had a 60% recovery rate, and Treatment C had a 45% recovery rate. If treatments were assigned in a 2:3:5 ratio respectively, what is the conditional probability that a patient who did not recover received Treatment C?
   Answer: 0.61 First, determine patient distribution from the 2:3:5 ratio: Treatment A = 100 patients, Treatment B = 150 patients, Treatment C = 250 patients. Calculate non-recovered patients: Treatment A: 100 × 0.25 = 25, Treatment B: 150 × 0.40 = 60, Treatment C: 250 × 0.55 = 137.5 ≈ 138. Total non-recovered = 25 + 60 + 138 = 223. Therefore, P(Treatment C | did not recover) = 138/223 ≈ 0.61. Choice A (0.55) is the non-recovery rate for Treatment C. Choice C (0.50) incorrectly assumes equal treatment distribution. Choice D (0.45) is the recovery rate for Treatment C.
7. A study examined the relationship between exercise frequency and sleep quality among 400 adults. The data shows that 60% of regular exercisers report good sleep quality, while 35% of non-regular exercisers report good sleep quality. If 45% of all participants are regular exercisers, what can be concluded about the association between exercise frequency and sleep quality?
   Answer: There is a positive association since $P(\text{Good Sleep}|\text{Regular Exercise}) > P(\text{Good Sleep}|\text{Non-Regular Exercise})$ To assess association, we compare conditional probabilities. Since P(Good Sleep|Regular Exercise) = 0.60 > 0.35 = P(Good Sleep|Non-Regular Exercise), there is a positive association. Choice B incorrectly tries to use the reverse conditional probabilities for independence. Choice C misinterprets the comparison needed. Choice D is incorrect because we have sufficient information from the given conditional probabilities.
8. A study reports that among smartphone users, 64% prefer iOS, while among tablet users, 58% prefer iOS. However, 45% of all participants are smartphone users and 55% are tablet users. If we want to determine whether device type and OS preference are associated, which comparison is most relevant?
   Answer: Compare 0.64 and 0.58 directly, since these conditional probabilities show the association strength When determining if two categorical variables are associated, you need to examine whether the distribution of one variable changes depending on the value of the other variable. This is fundamentally about comparing conditional probabilities. The correct approach is to compare the conditional probabilities directly: P(iOS|Smartphone) = 0.64 versus P(iOS|Tablet) = 0.58. If device type and OS preference were independent, these conditional probabilities would be equal. Since 0.64 ≠ 0.58, there is an association between device type and OS preference. The magnitude of this difference (0.06) indicates the strength of association. This makes B correct. A is flawed because comparing individual conditional probabilities to an overall rate doesn't directly test for association. The overall rate is a weighted average of the conditional rates, so deviations are expected even without association. C reverses the conditioning. While P(Smartphone|Prefers iOS) versus P(Smartphone|Prefers Other OS) could also demonstrate association, this approach is unnecessarily complex given the information provided and doesn't directly use the stated conditional probabilities. D confuses marginal probabilities with conditional probabilities. The device distribution (45% smartphone, 55% tablet) affects the overall preference rates but doesn't directly test whether the variables are associated. Association is about how preferences change across device types, not about the relative sizes of the groups. Study tip: For association questions, always look for comparisons between conditional probabilities of the same type. If P(Y|X₁) ≠ P(Y|X₂), then X and Y are associated.
9. A marketing researcher claims that gender and product preference are independent based on survey data. However, the conditional relative frequencies show $P(\text{Prefer A}|\text{Male}) = 0.42$ and $P(\text{Prefer A}|\text{Female}) = 0.38$. If males comprise 55% of the sample, what additional information most directly contradicts the independence claim?
   Answer: The joint probabilities $P(\text{Male and Prefer A}) \neq P(\text{Male}) \times P(\text{Prefer A})$ violate independence Independence requires P(Male and Prefer A) = P(Male) × P(Prefer A). We can calculate: P(Prefer A) = 0.55(0.42) + 0.45(0.38) = 0.402, and P(Male and Prefer A) = 0.55(0.42) = 0.231. Since P(Male) × P(Prefer A) = 0.55 × 0.402 = 0.221 ≠ 0.231, independence is violated. Choice A is incorrect because marginal probability not equaling conditional probabilities doesn't necessarily contradict independence. Choice B focuses on difference magnitude without statistical context. Choice C is a valid test but less direct than D.
10. A health study reports that among patients with high blood pressure, 72% take medication regularly, while among patients with normal blood pressure, 28% take medication regularly. If these conditional frequencies suggest an association, what can be inferred about the reverse conditional probabilities?
    Answer: $P(\text{High BP}|\text{Takes Medication})$ must be greater than $P(\text{High BP}|\text{No Medication})$ When P(Takes Medication|High BP) > P(Takes Medication|Normal BP), there is a positive association. This necessarily implies P(High BP|Takes Medication) > P(High BP|No Medication) due to the mathematical relationship between conditional probabilities in associated variables. Choice B incorrectly assumes symmetry that doesn't exist. Choice C is wrong because positive association in one direction guarantees positive association in the reverse direction. Choice D is incorrect because the relationship holds regardless of specific sample sizes.
11. A medical study tracked 800 patients across two treatment groups. In Group X, 45 out of 300 patients experienced side effects. In Group Y, 84 out of 500 patients experienced side effects. A researcher claims there is no meaningful association between treatment group and side effects because the absolute difference in the number of patients with side effects is relatively small. What is the most appropriate critique of this reasoning?
    Answer: The reasoning is flawed because comparing raw counts ignores group sizes: the conditional frequencies are 15% vs 16.8%, suggesting minimal association The researcher incorrectly compares absolute counts (45 vs 84) without considering that the groups have different sizes (300 vs 500). The conditional frequencies are 45/300 = 15% and 84/500 = 16.8%, showing only a 1.8 percentage point difference, which supports minimal association. Choice A incorrectly states the percentages are equivalent when they're not exactly equal. Choice B contains a calculation error. Choice C suggests the small difference is meaningful, which contradicts the actual minimal difference observed.

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