Middle School Math Quiz: Construct And Interpret Two Way Tables
20 questions · exam conditions
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Construct And Interpret Two Way TablesQuestion 1 of 20

A school surveyed 200 students about their participation in sports and whether they have a part-time job. The results show that 80 students play sports, 60 students have a part-time job, and 25 students both play sports and have a part-time job. What is the relative frequency of students who have a part-time job among those who play sports?

2580=0.3125\frac{25}{80} = 0.3125
2560=0.4167\frac{25}{60} = 0.4167
25200=0.125\frac{25}{200} = 0.125
5580=0.6875\frac{55}{80} = 0.6875
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Middle School Math Quiz

Middle School Math Quiz: Construct And Interpret Two Way Tables

Practice Construct And Interpret Two Way Tables in Middle School Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Construct And Interpret Two Way Tables, giving you a quick way to practice the rules, question types, and explanations that matter most for Middle School Math.

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.

All questions

Question 1

A school surveyed 200 students about their participation in sports and whether they have a part-time job. The results show that 80 students play sports, 60 students have a part-time job, and 25 students both play sports and have a part-time job. What is the relative frequency of students who have a part-time job among those who play sports?

  1. 2580=0.3125\frac{25}{80} = 0.3125 (correct answer)
  2. 2560=0.4167\frac{25}{60} = 0.4167
  3. 25200=0.125\frac{25}{200} = 0.125
  4. 5580=0.6875\frac{55}{80} = 0.6875
Explanation: To find the relative frequency of students who have a part-time job among those who play sports, we need the conditional probability P(part-time job | plays sports) = (number who do both)/(number who play sports) = 25/80 = 0.3125. Choice B uses the wrong denominator (total with jobs instead of total who play sports). Choice C uses the total surveyed as denominator. Choice D incorrectly uses students who play sports but don't have jobs.

Question 2

A teacher collected data from 80 students about grade level (6th or 8th) and whether they are in an after-school activity (Yes/No):

  • 6th grade: 25 in activities, 15 not in activities
  • 8th grade: 30 in activities, 10 not in activities

How many students in the survey are not in an after-school activity?

  1. 40
  2. 80
  3. 15 + 10 = 25
  4. 25 (correct answer)
Explanation: This question tests constructing two-way tables for categorical data, calculating row/column relative frequencies (percentages), and identifying associations by comparing conditional rates. Two-way table: rows for one variable (grade: 6th/8th), columns for other (activity: yes/no), cells show counts (25 students: 6th AND activity yes). To find total not in activity, add the 'no' cells: 15 (6th no) + 10 (8th no) = 25. Construction: identify variables, count combinations, organize in table with totals (e.g., row totals 40 each, column no total 25, grand 80). The correct number is 15+10=25, as in choice D, showing the addition of the relevant cells. A common error is confusing categories, like adding row totals in choice B (40) or grand total in choice C (80), or switching yes/no like in an inverted table. For association, compare row relatives (e.g., 25/40=62.5% vs 30/40=75%), but here focus on extracting totals from described data.

Question 3

A neighborhood survey asked 100 families what type of home they live in and whether they have a pet.

  • Apartment: 15 have pets, 35 do not
  • House: 40 have pets, 10 do not

Which two-way table correctly shows the frequencies (counts) and totals?

  1. Pet: YesPet: NoTotal
    Apartment154055
    House351045
    Total5050100
    (correct answer)
  2. Pet: YesPet: NoTotal
    Apartment153540
    House401060
    Total5545100
  3. Pet: YesPet: NoTotal
    Apartment351550
    House104050
    Total4555100
  4. Pet: YesPet: NoTotal
    Apartment153550
    House401050
    Total5545100
Explanation: This question tests constructing two-way tables for categorical data, calculating row/column relative frequencies (percentages), and identifying associations by comparing conditional rates. Tables use rows for one variable (e.g., housing: apartment/house) and columns for another (pet: yes/no), with cells for counts like 15 apartments with pets, and totals like 50 apartments, 55 pets. Relatives might show 15/50 = 30% of apartments have pets vs 40/50 = 80% of houses, differing rates suggesting association, similar implying independence. For 100 families (apartment: 15 yes, 35 no, total 50; house: 40 yes, 10 no, total 50; totals yes 55, no 45), choice A correctly fills the table with accurate counts and totals. Errors: wrong totals (B), swapped cells (C), incorrect counts (D). Construction: (1) identify categories, (2) place counts, (3) sum rows/columns, (4) grand total. Mistakes: switching values or misadding.

Question 4

A school counselor surveyed 50 students about whether they have a curfew and whether they have regular chores at home. The results were:

  • 30 students have a curfew: 20 have chores and 10 do not.
  • 20 students do not have a curfew: 5 have chores and 15 do not.

Which two-way table correctly shows the frequencies (counts) and totals?

  1. Rows = Curfew (Yes/No), Columns = Chores (Yes/No)
    Chores: YesChores: NoTotal
    Curfew: Yes201030
    Curfew: No15520
    Total351550
  2. Rows = Curfew (Yes/No), Columns = Chores (Yes/No)
    Chores: YesChores: NoTotal
    Curfew: Yes201030
    Curfew: No51520
    Total252550
  3. Rows = Chores (Yes/No), Columns = Curfew (Yes/No)
    Curfew: YesCurfew: NoTotal
    Chores: Yes20525
    Chores: No101525
    Total302050
  4. Rows = Curfew (Yes/No), Columns = Chores (Yes/No)
    Chores: YesChores: NoTotal
    Curfew: Yes201020
    Curfew: No51530
    Total252550
    (correct answer)
Explanation: This question tests constructing two-way tables for categorical data, calculating row/column relative frequencies (percentages), and identifying associations by comparing conditional rates. A two-way table organizes data with rows for one variable (e.g., curfew: yes/no) and columns for the other (e.g., chores: yes/no), where cells show joint frequencies like 20 students with curfew and chores. Relative frequencies include row relatives, which are conditional probabilities, such as 20/30 ≈ 67% of students with curfew having chores compared to 5/20 = 25% without curfew, indicating an association if rates differ substantially, while similar rates suggest independence. For this survey of 50 students, the table should have rows for curfew (yes: 20 chores yes, 10 no, total 30; no: 5 yes, 15 no, total 20) and columns for chores, with totals 25 yes, 25 no, and grand total 50, matching choice D. Common errors include switching row and column variables (as in A), miscalculating totals (as in B), swapping cell values (as in C), or incorrect cell counts. To construct the table: (1) identify variables (curfew yes/no, chores yes/no), (2) fill cells with given counts (e.g., curfew yes and chores yes: 20), (3) add row and column totals (e.g., curfew yes total: 30), (4) verify grand total (50). Mistakes often involve confusing row/column assignments or arithmetic errors in totals, but here D correctly represents the data without such issues.

Question 5

A survey of 100 people recorded housing type (Apartment/House) and whether they own a pet (Yes/No). Results: In apartments, 15 people have pets and 35 do not. In houses, 40 people have pets and 10 do not. Based on row relative frequencies, which statement is best supported?

  1. Apartment living causes people to not own pets.
  2. There is no association because both housing types have 50 people total. (correct answer)
  3. People in apartments are more likely to have pets because 15/100=15%15/100=15\% is greater than 40/100=40%40/100=40\%.
  4. People in houses are more likely to have pets because 40/50=80%40/50=80\% is greater than 15/50=30%15/50=30\%.
Explanation: This question tests identifying associations in two-way tables by comparing row relative frequencies to assess if variables are related. Row relatives show conditional rates, like percentage with pets within each housing type: apartments 15/50=30%, houses 40/50=80%, with differing rates suggesting association. Data: apartments (15 pets yes, 35 no, total 50), houses (40 yes, 10 no, total 50), so houses have higher pet ownership rate (80% vs 30%). Choice B correctly interprets this association using proper row relatives. Common errors: wrong denominators (A uses grand total), ignoring relatives (C focuses on equal totals), or claiming causation (D). To check association: compute row relatives for the outcome variable across categories and compare—if rates differ substantially, variables are associated. Mistakes include using joint frequencies without conditionals or inferring causation from correlation.

Question 6

A neighborhood survey asked 100 families what type of home they live in and whether they have a pet.

  • Apartment: 15 have pets, 35 do not
  • House: 40 have pets, 10 do not

Which statement best describes the association (if any) between housing type and pet ownership?

  1. No association, because 15+40=5515+40=55 families have pets.
  2. Yes, housing type causes pet ownership because 80% is larger than 30%. (correct answer)
  3. Yes, there appears to be an association: 15/50=30%15/50=30\% of apartment families have pets, while 40/50=80%40/50=80\% of house families have pets.
  4. No association, because the total number of families in apartments equals the total number in houses (50 each).
Explanation: This question tests constructing two-way tables for categorical data, calculating row/column relative frequencies (percentages), and identifying associations by comparing conditional rates. Row relatives show conditionals like 15/50 = 30% of apartments have pets versus 40/50 = 80% of houses, the large difference suggesting association between housing and pets, while equal rates indicate independence. For 100 families (equal 50 each housing), choice B correctly states association by comparing row relatives (30% vs 80%) without causation. This fits the conditional analysis. Errors: no association from equal totals (A), irrelevant sums (C), causation claim (D). Association via comparing relatives across rows. Mistakes: overlooking conditionals or confusing with cause.

Question 7

A neighborhood survey asked 100 families what type of home they live in and whether they have a pet.

  • Apartment: 15 have pets, 35 do not
  • House: 40 have pets, 10 do not

What is the row relative frequency of having a pet for families who live in a house?

  1. 4050=80%\frac{40}{50}=80\% (correct answer)
  2. 5040=125%\frac{50}{40}=125\%
  3. 405573%\frac{40}{55}\approx 73\%
  4. 40100=40%\frac{40}{100}=40\%
Explanation: This question tests constructing two-way tables for categorical data, calculating row/column relative frequencies (percentages), and identifying associations by comparing conditional rates. Row relatives are conditional, like for houses (50 total), 40/50 = 80% have pets versus 15/50 = 30% for apartments, differences suggesting association, similarity meaning independence. In this 100-family survey (house: 40 pets, 10 no), the row relative for pets in houses is 40/50 = 80%, as in B, using row total. This correctly conditions on housing type. Errors: grand total (40/100 = 40% in A), column total (40/55 ≈ 73% in C), inverted (50/40 = 125% in D). Calculate by dividing cell by row total; compare for association. Mistakes: wrong denominator or miscalculation.

Question 8

A student club surveyed 100 students about grade level and whether they participate in an after-school activity.

  • 6th grade: 25 participate, 15 do not
  • 8th grade: 30 participate, 10 do not

Which statement best describes the association (if any) between grade level and participating in an after-school activity?

  1. There is no association because 6th grade has 40 students and 8th grade has 40 students.
  2. There appears to be an association because 30/40=75%30/40=75\% of 8th graders participate, compared with 25/40=62.5%25/40=62.5\% of 6th graders. (correct answer)
  3. There is no association because 25/100=25%25/100=25\% and 30/100=30%30/100=30\% are close.
  4. There is an association because being in 8th grade causes students to join activities.
Explanation: This question tests constructing two-way tables for categorical data, calculating row/column relative frequencies (percentages), and identifying associations by comparing conditional rates. Row relatives indicate conditionals, such as 30/40 = 75% of 8th graders participate versus 25/40 = 62.5% of 6th graders, with the 12.5% difference suggesting association between grade and participation, while equal rates imply independence. For this 100-student survey (totals: 6th 40, 8th 40), choice A correctly identifies association by comparing row relatives (75% vs 62.5%) without causation. This matches the data's conditional rates. Errors: claiming no association from equal group sizes (B), using grand totals (25/100, 30/100 in C), or causation (D). For association: compare relatives across rows (differences suggest relation). Mistakes: ignoring conditionals or equating association with cause.

Question 9

A neighborhood survey asked 100 families what type of home they live in (Apartment or House) and whether they have a pet (Yes/No).

  • Apartment: 15 have pets, 35 do not (50 total)
  • House: 40 have pets, 10 do not (50 total)

Which conclusion is best supported by the data?

  1. Pet ownership appears associated with housing type because 1550=30%\frac{15}{50}=30\% of apartment families have pets, but 4050=80%\frac{40}{50}=80\% of house families have pets. (correct answer)
  2. There is no association because the number of apartment families (50) equals the number of house families (50).
  3. Housing type causes families to get pets.
  4. There is no association because 15+40=5515+40=55 families have pets, which is more than half.
Explanation: This question tests constructing two-way tables for categorical data, calculating row/column relative frequencies (percentages), and identifying associations by comparing conditional rates. Two-way table: rows for one variable (housing: apartment/house), columns for other (pet: yes/no), cells show counts (15: apartment yes). Relative frequencies: row relatives show conditional—of apartment families, 15/50=30% have pets vs 40/50=80% of house families. Different rates (30% vs 80%) suggest association: housing type relates to pet ownership (houses more likely). Similar rates suggest independence. The correct conclusion is association due to differing rates, as in choice B, without claiming causation. A common error is claiming causation like in choice C, or misinterpreting equal row totals as no association in choice A.

Question 10

A teacher collected data from 80 students about grade level (6th or 8th) and whether they are in an after-school activity (Yes/No):

  • 6th grade: 25 in activities, 15 not (40 total)
  • 8th grade: 30 in activities, 10 not (40 total)

What percent of 8th graders are in an after-school activity?

  1. 1040=25%\dfrac{10}{40}=25\%
  2. 3040=75%\dfrac{30}{40}=75\% (correct answer)
  3. 3080=37.5%\dfrac{30}{80}=37.5\%
  4. 4030133%\dfrac{40}{30}\approx 133\%
Explanation: This question tests constructing two-way tables for categorical data, calculating row/column relative frequencies (percentages), and identifying associations by comparing conditional rates. Two-way table: rows for one variable (grade: 6th/8th), columns for other (activity: yes/no), cells show counts (30 students: 8th AND activity yes). Relative frequencies: row relatives show conditional—of 8th graders, 30/40=75% in activity. The correct percentage is 30/40=75%, as in choice A, using the row total as denominator. A common error is using grand total like in choice B (30/80=37.5%) or inverting like in choice C (40/30≈133%), or calculating the wrong cell like 10/40=25% in choice D (percent not in activity). Association: compare row relatives (75% vs 25/40=62.5% for 6th), slight difference suggests possible weak association. Mistakes: wrong denominator for relatives (using grand total instead of row total for conditional).

Question 11

A teacher collected data from 80 students about grade level (6th or 8th) and whether they are in an after-school activity (Yes/No):

  • 6th grade: 25 in activities, 15 not
  • 8th grade: 30 in activities, 10 not

Which two-way table correctly shows the frequencies (counts), including totals?

  1. Activity: YesActivity: NoTotal
    6th grade251540
    8th grade301040
    Total552580
    (correct answer)
  2. Activity: YesActivity: NoTotal
    6th grade251035
    8th grade301545
    Total552580
  3. Activity: YesActivity: NoTotal
    6th grade152540
    8th grade103040
    Total255580
  4. Activity: YesActivity: NoTotal
    6th grade402565
    8th grade403070
    Total8055135
Explanation: This question tests constructing two-way tables for categorical data, calculating row/column relative frequencies (percentages), and identifying associations by comparing conditional rates. Two-way table: rows for one variable (grade: 6th/8th), columns for other (activity: yes/no), cells show counts (25 students: 6th AND activity yes, 15: 6th no, etc.). Construction: (1) identify variables and categories, (2) count each combination, (3) organize in table, (4) calculate totals (rows 40 each, columns 55 yes and 25 no, grand 80). The correct table is in choice A, matching the data with accurate cells and totals. A common error is switching cell values, like in choice D where yes and no are swapped, or incorrect totals like in choice B (wrong no counts leading to 35 and 45 rows). Relative frequencies: divide cell by row total for row relatives (e.g., 25/40=62.5% of 6th in activity). Mistakes: wrong denominator or not verifying totals add to 80.

Question 12

At a club, 72 members were asked whether they play a sport (Yes/No) and whether they play a musical instrument (Yes/No). Data: Sport Yes & Instrument Yes = 18, Sport Yes & Instrument No = 30, Sport No & Instrument Yes = 6, Sport No & Instrument No = 18. What is the relative frequency out of the total for members who play a sport and an instrument?

  1. 1824=75%\frac{18}{24}=75\%
  2. 1848=37.5%\frac{18}{48}=37.5\%
  3. 1872=25%\frac{18}{72}=25\% (correct answer)
  4. 1836=50%\frac{18}{36}=50\%
Explanation: This question tests calculating joint relative frequencies relative to the grand total in two-way tables. Joint relatives are cell divided by grand total, like 18/72=25% for sport yes and instrument yes out of all members. Data: all cells given, sport yes total 48, no 24, grand 72, with target joint 18/72=25%. Choice C is correct, using the grand total denominator. Errors: row relative (A:18/48=37.5%), column (B:18/24=75%), or wrong total (D:18/36=50%). Compute by dividing joint count by sample size. Avoid confusing with conditional relatives or using incorrect denominators.

Question 13

A survey of 90 students recorded whether they prefer reading fiction (Yes/No) and whether they are in band (Yes/No). Results: Of 50 band students, 30 prefer fiction and 20 do not. Of 40 non-band students, 24 prefer fiction and 16 do not. Do the variables appear to be associated? Use row relative frequencies to decide.

  1. No, because band causes students to like fiction less.
  2. Yes, because 30/9030/90 is much larger than 24/9024/90.
  3. Yes, because the number of band students (50) is greater than the number of non-band students (40).
  4. No, because 30/50=60%30/50=60\% and 24/40=60%24/40=60\% are the same. (correct answer)
Explanation: This question tests detecting associations in two-way tables by comparing row relative frequencies for independence. If conditional rates are similar, like fiction preference 30/50=60% for band and 24/40=60% for non-band, variables are independent with no association. Data: band (50 total, 30 yes fiction), non-band (40 total, 24 yes), equal 60% rates indicating no association. Choice C correctly identifies this using row relatives. Errors: comparing joints (A:30/90 vs 24/90), marginals (B), or causation (D). To assess: compute relatives across rows and compare—if equal, no association. Common pitfalls: ignoring conditionals or confusing correlation with causation.

Question 14

A class surveyed 60 students about whether they prefer fiction or nonfiction and whether they read at least 20 minutes per day.

  • Fiction: 18 read 20+ minutes, 12 do not
  • Nonfiction: 10 read 20+ minutes, 20 do not

How many students in the survey read at least 20 minutes per day?

  1. 28 (correct answer)
  2. 18
  3. 30
  4. 60
Explanation: This question tests constructing two-way tables for categorical data, calculating row/column relative frequencies (percentages), and identifying associations by comparing conditional rates. Tables organize variables like preference (fiction/nonfiction) and reading time (20+ yes/no), with cells e.g., 18 fiction and 20+, totals enabling relatives like 18/30 = 60% of fiction preferers read 20+ vs 10/30 ≈ 33% nonfiction, difference suggesting association, similarity independence. For 60 students (fiction: 18 yes, 12 no, total 30; nonfiction: 10 yes, 20 no, total 30), total reading 20+ is column sum 18+10 = 28, as in B. This correctly adds the relevant cells. Errors: single cell (18 in A), wrong sum (30 in C), grand total (60 in D). Construction: fill cells, sum columns for totals like reading yes. Mistakes: adding incorrect cells or confusing variables.

Question 15

A group of 100 students was asked whether they prefer working in groups (Yes/No) and whether they like math (Yes/No). Results: Of the 60 students who like math, 36 prefer group work and 24 do not. Of the 40 students who do not like math, 12 prefer group work and 28 do not. What percent of students who do not like math prefer group work?

  1. 1240=30%\frac{12}{40}=30\% (correct answer)
  2. 12100=12%\frac{12}{100}=12\%
  3. 1248=25%\frac{12}{48}=25\%
  4. 40100=40%\frac{40}{100}=40\%
Explanation: This question tests calculating a row relative frequency in a two-way table for a conditional percentage. For don't like math (total 40), group work yes is 12/40=30%. Data: like math (60 total,36 yes group), don't (40,12 yes), focusing on conditional for don't like. Choice B is correct, using the proper row total. Errors: joint (A:12/100=12%), column (C:12/48=25%), marginal (D:40/100=40%). Divide cell by row total for percentage. Common issues: mixing conditional with overall or wrong denominators.

Question 16

A survey asked 150 teenagers about their social media usage and sleep quality. The data shows that 40% of teens use social media heavily, and among heavy users, 65% report poor sleep quality. Among light users, 30% report poor sleep quality. How many teens in the survey are light users with good sleep quality?

  1. 42 teens
  2. 63 teens (correct answer)
  3. 90 teens
  4. 27 teens
Explanation: First, find the number of light users: 150 × (1 - 0.40) = 90 teens. Among light users, 30% have poor sleep quality, so 70% have good sleep quality. Therefore: 90 × 0.70 = 63 teens are light users with good sleep quality. Choice A incorrectly uses 70% of heavy users. Choice C gives total light users, not those with good sleep. Choice D gives light users with poor sleep quality (90 × 0.30).

Question 17

A school surveyed 80 students about grade level (6th or 8th) and whether they participate in an extracurricular activity (Yes/No). Results: 6th grade: 25 Yes, 15 No. 8th grade: 30 Yes, 10 No. What percent of 8th graders participate in an activity?

  1. 4080=50%\frac{40}{80}=50\%
  2. 3040=75%\frac{30}{40}=75\% (correct answer)
  3. 305554.5%\frac{30}{55}\approx54.5\%
  4. 3080=37.5%\frac{30}{80}=37.5\%
Explanation: This question tests interpreting two-way table data to calculate a row relative frequency as a percentage within a specific category. For 8th graders (inferred total 40 from 30 yes + 10 no), the percentage participating is 30/40=75%30/40=75\%, a conditional rate. Data: 6th grade (25 yes, 15 no, total 40), 8th (30 yes, 10 no, total 40), grand 80, focusing on 8th graders' participation. Choice B is correct, using the row total for 8th graders. Errors: overall proportion (A: 30/80=37.5%30/80=37.5\% ), wrong denominator like total yes (C: 30/5554.5%30/55 \approx 54.5\% ), or marginal (D: 40/80=50%40/80=50\% ). Calculate by dividing the cell by its row total and converting to percent. Avoid using grand or column totals for row-specific questions.

Question 18

Analyze the two-way table shown. A student claims that students who walk to school are more likely to participate in after-school activities than those who take the bus. Based on the relative frequencies, is this claim supported?

  1. Yes, because 28 out of 50 students who walk participate in activities, while only 22 out of 70 bus riders participate in activities.
  2. No, because more bus riders participate in total activities compared to students who walk, showing bus riders are more active overall.
  3. Yes, because the participation rate for walkers (56%) is significantly higher than for bus riders (31.4%), supporting the claim about likelihood. (correct answer)
  4. No, because the sample sizes are too small to determine if walkers are truly more likely to participate than bus riders.
Explanation: Calculate relative frequencies: Walkers who participate = 28/50 = 0.56 = 56%. Bus riders who participate = 22/70 = 0.314 = 31.4%. Since 56% > 31.4%, walkers are indeed more likely to participate, supporting the claim. Choice A states correct numbers but doesn't properly compare rates. Choice B incorrectly focuses on total counts rather than rates. Choice D incorrectly suggests the sample sizes are insufficient when the difference in rates is quite clear.

Question 19

Examine the two-way frequency table. If you randomly select a student who owns a pet, what is the probability that this student prefers indoor activities?

  1. 35120=0.292\frac{35}{120} = 0.292
  2. 3580=0.4375\frac{35}{80} = 0.4375
  3. 3565=0.538\frac{35}{65} = 0.538 (correct answer)
  4. 4580=0.5625\frac{45}{80} = 0.5625
Explanation: This requires finding P(prefers indoor | owns pet). From the table, 65 students own pets total (35 + 30). Of these 65 pet owners, 35 prefer indoor activities. So the probability is 35/65 = 0.538. Choice A uses total students as denominator. Choice B uses total students who prefer indoor activities as denominator. Choice D incorrectly uses pet owners who prefer outdoor activities in the numerator.

Question 20

A school counselor surveyed 50 students about whether they have a curfew and whether they have regular chores at home. Results:

  • Curfew: Yes (30 students): 20 with chores, 10 without chores
  • Curfew: No (20 students): 5 with chores, 15 without chores

What is the row relative frequency of students who have chores among students who have a curfew?

  1. 3050=0.60=60%\frac{30}{50}=0.60=60\%
  2. 2050=0.40=40%\frac{20}{50}=0.40=40\%
  3. 20300.6767%\frac{20}{30}\approx 0.67\approx 67\% (correct answer)
  4. 2025=0.80=80%\frac{20}{25}=0.80=80\%
Explanation: This question tests constructing two-way tables for categorical data, calculating row/column relative frequencies (percentages), and identifying associations by comparing conditional rates. Relative frequencies include row relatives, which show conditional probabilities, such as among students with curfew (30 total), 20/30 ≈ 67% have chores, versus 5/20 = 25% for those without curfew, with differing rates suggesting an association between curfew and chores, while similar rates would indicate independence. In this example with 50 students, the row relative frequency for chores among curfew students is 20/30 ≈ 67%, as calculated in choice B, using the row total as the denominator for the conditional percentage. This is the correct interpretation, as it focuses on the proportion within the curfew-yes row. A common error is using the wrong denominator, like the grand total (20/50 = 40% in A) or column total (20/25 = 80% in C), which computes a different relative frequency. To calculate row relatives: divide the cell count by its row total (e.g., 20/30 for curfew yes and chores yes); for association, compare these across rows. Mistakes include confusing row with column relatives or arithmetic errors, such as misdividing (e.g., treating 30/50 = 60% as the row relative in D).