AP Statistics Quiz: Expected Counts In Two Way Tables

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Expected Counts In Two Way TablesQuestion 1 of 20

A study examined the relationship between pet ownership (dog, cat, no pet) and allergy status (yes, no) in a random sample of 600 people. In the sample, 250 people owned a dog, 150 owned a cat, and 200 had no pet. A total of 180 people reported having allergies.

Which of the following expressions correctly calculates the expected number of people who own a cat and have allergies, under the null hypothesis of no association?

\frac{(150)(180)}{600}

\frac{(150)(250)}{600}

\frac{(180)(200)}{600}

\frac{(600)}{(150)(180)}

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AP Statistics Quiz

AP Statistics Quiz: Expected Counts In Two Way Tables

Practice Expected Counts In Two Way Tables in AP Statistics 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 Expected Counts In Two Way Tables, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Statistics.

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

Which of the following expressions correctly calculates the expected number of people who own a cat and have allergies, under the null hypothesis of no association?

$\frac{(150)(180)}{600}$ (correct answer)
$\frac{(150)(250)}{600}$
$\frac{(180)(200)}{600}$
$\frac{(600)}{(150)(180)}$

Explanation: The formula for the expected count in a cell of a two-way table is (row total × column total) / grand total. Here, the row total for 'owning a cat' is 150, the column total for 'having allergies' is 180, and the grand total sample size is 600. The correct expression is $\frac{(150)(180)}{600}$ .

Question 2

Data was collected to see if there is an association between the day of the week and the number of customers at a restaurant. A random sample of 200 days was observed. The totals are: 80 weekdays and 120 weekend days. The customer traffic was categorized as Low, Medium, or High. There were 50 days with Low traffic, 90 with Medium, and 60 with High. On weekdays, there were 30 days with Low traffic.

What is the expected number of weekdays with Low customer traffic if traffic level is independent of the type of day?

$20$ (correct answer)
$30$
$40$
$50$

Explanation: The expected count is (row total × column total) / grand total. The row total for weekdays is 80. The column total for Low traffic is 50. The grand total is 200. The expected count is $(80 \times 50) / 200 = 4000 / 200 = 20$ . The value of 30 is the observed count, which is a common distractor.

Question 3

A research study produced the following two-way table of counts for two categorical variables, A and B. A total of 200 subjects were studied. For variable A, the counts are 80 for level 1 and 120 for level 2. For variable B, the counts are 100 for level X and 100 for level Y.

For a chi-square test of independence, which of the following comparisons between expected counts is correct?

The expected count for (Level 1, Level X) is equal to the expected count for (Level 2, Level X).
The expected count for (Level 1, Level X) is less than the expected count for (Level 1, Level Y).
The expected count for (Level 2, Level X) is greater than the expected count for (Level 1, Level X). (correct answer)
The expected count for (Level 1, Level Y) is greater than the expected count for (Level 2, Level Y).

Explanation: First, calculate the four expected counts using the formula (row total × column total) / grand total. The expected count for (Level 1, Level X) is $(80 \times 100) / 200 = 40$ . The expected count for (Level 1, Level Y) is $(80 \times 100) / 200 = 40$ . The expected count for (Level 2, Level X) is $(120 \times 100) / 200 = 60$ . The expected count for (Level 2, Level Y) is $(120 \times 100) / 200 = 60$ . Comparing the values as per the choices, only C is correct because the expected count for (Level 2, Level X), which is 60, is greater than the expected count for (Level 1, Level X), which is 40.

Question 4

A study was conducted to investigate whether the genre of a movie seen (Action, Comedy, Drama) is independent of the age group of the moviegoer (Child, Teen, Adult). Data was collected from a random sample of 300 moviegoers.

Under the null hypothesis that movie genre preference is independent of age group, how is the expected number of teens who prefer action movies calculated?

By multiplying the total number of teens by the total number of people who prefer action movies, then dividing by the total number of moviegoers. (correct answer)
By dividing the number of teens who were observed to prefer action movies by the total number of teens.
By multiplying the proportion of all moviegoers who are teens by the total number of action movies available.
By averaging the number of moviegoers across all combinations of age group and genre.

Explanation: The expected count for a cell under the null hypothesis of independence is calculated by the formula: (row total × column total) / grand total. In this context, this corresponds to (total number of teens × total number of people who prefer action movies) / total number of moviegoers.

Question 5

A gym tracked 500 members by whether they attend group classes and whether they renewed their membership. Assuming independence, which expression calculates the expected count for the Group Classes & Renewed cell?

$\dfrac{(200)(350)}{500}$ (correct answer)
$\dfrac{200}{500}$
$140$
$\dfrac{350}{500}$
$(200)(350)$

Explanation: To find expected counts assuming independence, we apply (row total × column total) ÷ grand total. For Group Classes & Renewed, we multiply members attending group classes (200) by members who renewed (350), then divide by all 500 members. Choice A shows this correctly: (200)(350)/500. Choice C gives 140, which is the calculated result but not the expression itself. Choices B and D show individual proportions, while E shows the product without division. The expected count formula helps us test whether the observed counts differ significantly from what independence would predict.

Question 6

A researcher classified 120 plants by whether they received fertilizer and whether they bloomed. Under the assumption of independence, which expression calculates the expected count for the No Fertilizer & Bloomed cell?

$\dfrac{(50)(70)}{120}$ (correct answer)
$\dfrac{50}{120}$
$30$
$\dfrac{70}{120}$
$(50)(70)$

Explanation: Expected counts in two-way tables use the formula (row total × column total) ÷ grand total. For No Fertilizer & Bloomed, we multiply plants without fertilizer (50) by plants that bloomed (70), then divide by all 120 plants. Choice A correctly shows (50)(70)/120. Choice C shows 30, which might be an observed count but isn't the expression. Choices B and D show marginal proportions that don't calculate expected counts, while E multiplies totals without dividing, yielding 3,500 instead of the reasonable expected count of about 29.

Question 7

A clinic categorized 180 patients by whether they received a flu shot and whether they later reported flu symptoms. Under the assumption of independence, which expression calculates the expected count for the Shot & Symptoms cell?

$\dfrac{(120)(45)}{180}$ (correct answer)
$\dfrac{45}{180}$
$\dfrac{120}{180}$
$20$
$(120)(45)$

Explanation: This problem requires calculating the expected count for a cell in a two-way table assuming independence between variables. The expected count formula is (row total × column total) ÷ grand total. For the Shot & Symptoms cell, we multiply the total who got shots (120) by the total with symptoms (45), then divide by all 180 patients. Choice A correctly represents this: (120)(45)/180. Choice D shows 20, which might be an observed count, while B and C show individual proportions. Choice E multiplies the totals but forgets the crucial step of dividing by the grand total.

Question 8

A study categorized 90 commuters by whether they bike to work and whether their commute is under 5 miles or 5 miles and over. Assuming independence, which expression calculates the expected count for the Bike & Under 5 miles cell?

$\dfrac{50}{90}$
$\dfrac{(36)(50)}{90}$ (correct answer)
$\dfrac{36}{90}$
$18$
$(36)(50)$

Explanation: Expected counts under independence use the formula $(row\ total \times column\ total) \div grand\ total$ . For Bike & Under 5 miles, we multiply commuters who bike (36) by those with commutes under 5 miles (50), then divide by all 90 commuters. Choice B shows this correctly: $(36)(50)/90$ . Choice D gives 18, which might be an observed count rather than the expression. Choices A and C show individual proportions, while E shows only the product. Understanding this formula is essential for testing whether categorical variables are associated or independent.

Question 9

A school surveyed 200 students about whether they participate in a sport and whether they prefer morning or afternoon classes. Under the assumption that sport participation and class-time preference are independent, which expression calculates the expected count for the Sport & Morning cell?

$\dfrac{(120)(110)}{200}$ (correct answer)
$\dfrac{120}{200}$
$\dfrac{110}{200}$
$70$
$(120)(110)$

Explanation: This question tests your ability to calculate expected counts in a two-way table under the assumption of independence. The formula for expected count is (row total × column total) ÷ grand total. Since we need the expected count for Sport & Morning, we multiply the total number of students who play sports (120) by the total number who prefer morning classes (110), then divide by the grand total of 200 students. Choice A correctly shows this formula: (120)(110)/200. Choice D gives 70, which might be the actual count but not the expression for calculating it. The other choices show only partial calculations or incorrect formulas.

Question 10

A university surveyed 300 students about whether they live on campus and whether they own a car. Assuming these variables are independent, which expression calculates the expected count for the On Campus & Owns Car cell?

$\dfrac{(180)(120)}{300}$ (correct answer)
$\dfrac{180}{300}$
$\dfrac{120}{300}$
$80$
$(180)(120)$

Explanation: Expected counts in two-way tables are calculated using (row total × column total) ÷ grand total when assuming independence. For the On Campus & Owns Car cell, we multiply students living on campus (180) by students owning cars (120), then divide by the total 300 students. Choice A shows this correctly: (180)(120)/300. Choice D gives 80, which could be the actual count but isn't the expression. Choices B and C show individual proportions, while E shows the product without dividing by the grand total, a common error.

Question 11

A company recorded whether 150 employees work remotely and whether they are in the Sales or Engineering department. Assuming the two variables are independent, which expression calculates the expected count for the Remote & Sales cell?

$\dfrac{60}{150}$
$\dfrac{(90)(60)}{150}$ (correct answer)
$30$
$\dfrac{90}{150}$
$(90)(60)$

Explanation: To find expected counts in two-way tables, we use the formula (row total × column total) ÷ grand total. For the Remote & Sales cell, we need to multiply the total number of remote workers (90) by the total number in Sales (60), then divide by the grand total of 150 employees. Choice B correctly shows this: (90)(60)/150. Choice C shows 30, which is likely the result of the calculation but not the expression itself. Choices A and D show individual proportions rather than the complete formula, while E forgets to divide by the grand total.

Question 12

A bookstore recorded 160 purchases by whether the customer used a coupon and whether the purchase was fiction or nonfiction. If coupon use and genre are independent, which expression calculates the expected count for the Coupon & Fiction cell?

$\dfrac{(40)(100)}{160}$ (correct answer)
$\dfrac{40}{160}$
$\dfrac{100}{160}$
$20$
$(40)(100)$

Explanation: This problem requires the expected count formula: (row total × column total) ÷ grand total. For Coupon & Fiction, we multiply customers using coupons (40) by fiction purchases (100), then divide by all 160 purchases. Choice A correctly represents this: (40)(100)/160. Choice D shows 20, which might be an actual count but not the expression. Choices B and C show marginal proportions that don't give expected counts, while E forgets to divide by the grand total. Expected counts are crucial for chi-square tests of independence.

Question 13

A poll of 400 adults recorded whether they support a policy and whether they are under 30 or 30 and over. Assuming independence, which expression calculates the expected count for the Under 30 & Support cell?

$\dfrac{160}{400}$
$\dfrac{(150)(240)}{400}$ (correct answer)
$90$
$\dfrac{240}{400}$
$(150)(240)$

Explanation: This question tests the expected count formula: (row total × column total) ÷ grand total. For Under 30 & Support, we multiply those under 30 (150) by those who support (240), then divide by the total 400 adults. Choice B shows this correctly: (150)(240)/400. Choice C gives 90, which is the calculated result (36,000/400 = 90) but not the expression. Choices A and D show individual proportions, while E shows the product without division. Remember that expected counts tell us what we'd expect if the variables were truly independent.

Question 14

A movie theater recorded 250 customers by whether they bought popcorn and whether they attended a matinee or evening show. If popcorn purchase and show time are independent, which expression calculates the expected count for the Popcorn & Evening cell?

$\dfrac{150}{250}$
$\dfrac{(100)(150)}{250}$ (correct answer)
$60$
$\dfrac{100}{250}$
$(100)(150)$

Explanation: To calculate expected counts under independence, we use (row total × column total) ÷ grand total. For Popcorn & Evening, we need the total who bought popcorn (100) times the total at evening shows (150), divided by all 250 customers. Choice B correctly shows (100)(150)/250. Choice C shows 60, likely the calculated result rather than the expression. Choices A and D show individual proportions that don't give expected counts, while E forgets to divide by the grand total, which would give an impossibly large expected count.

Question 15

For a chi-square test, data on two categorical variables, X and Y, were collected from a sample of 250 individuals. The data are summarized in a two-way table with 3 rows and 4 columns.

What is the sum of all the expected counts for the 12 cells in the table?

It depends on the marginal distributions of variables X and Y.
It is equal to 12, the number of cells in the table.
It is equal to 250, the grand total of the table. (correct answer)
It is equal to $(3-1)(4-1) = 6$ , the degrees of freedom.

Explanation: A fundamental property of expected counts is that their sum over all cells in the table must be equal to the grand total of the observed counts. In this case, the grand total is the sample size, 250. Therefore, the sum of all expected counts is 250.

Question 16

The expected count for a cell in a two-way table for a chi-square test of independence can be interpreted as the expected number of observations in that cell if the null hypothesis is true. This is equivalent to which of the following?

The sample size multiplied by the sum of the marginal probabilities for that cell's row and column.
The sample size multiplied by the product of the marginal probabilities for that cell's row and column. (correct answer)
The product of the row total and column total for that cell.
The average of the observed counts in that cell's row and column.

Explanation: The marginal probability for a row is (row total / grand total), and for a column is (column total / grand total). If two variables are independent, the joint probability is the product of their marginal probabilities. The expected count is the sample size (grand total) times this joint probability: $n \times P(\text{row}) \times P(\text{col}) = n \times \frac{\text{row total}}{n} \times \frac{\text{col total}}{n} = \frac{(\text{row total})(\text{col total})}{n}$ .

Question 17

A survey of 400 randomly selected office workers was conducted to investigate a possible association between primary commuting method and reported job satisfaction. The results are summarized as follows: 250 workers reported high job satisfaction, while 150 reported low job satisfaction. Among those with high satisfaction, 100 commute by car and 150 use public transit. Overall, 180 workers commute by car.

For a chi-square test for independence, what is the expected count of workers who commute by car and report high job satisfaction, assuming commuting method and job satisfaction are independent?

$112.5$ (correct answer)
$100.0$
$156.25$
$180.0$

Explanation: The expected count for a cell in a two-way table is calculated as (row total × column total) / grand total. The row total for high job satisfaction is 250. The column total for commuting by car is 180. The grand total is 400. Therefore, the expected count is $(250 \times 180) / 400 = 45000 / 400 = 112.5$ .

Question 18

A technology company surveyed 1,000 smartphone users to determine if there is a relationship between the phone's operating system (OS) and the primary type of app used. The results showed 600 users had OS A and 400 had OS B. Overall, 700 users primarily used social media apps, and 300 users primarily used gaming apps.

To perform a chi-square test for independence, the company needs to calculate expected counts. What is the expected number of users with OS A who primarily use gaming apps, assuming OS and primary app type are independent?

$120$
$180$ (correct answer)
$240$
$420$

Explanation: The expected count is calculated as (row total × column total) / grand total. The total number of users with OS A (row total) is 600. The total number of users who prefer gaming apps (column total) is 300. The grand total is 1,000. The expected count is $(600 \times 300) / 1000 = 180000 / 1000 = 180$ .

Question 19

A sociologist surveyed 500 adults, classified by their highest education level (High School, Bachelor's, Graduate) and their primary source of news (Online, Television). The data collected are: 200 have a High School education, 180 have a Bachelor's, and 120 have a Graduate degree. A total of 350 adults use Online sources as their primary news source.

For a chi-square test of association between education level and primary news source, what is the expected count of adults with a Bachelor's degree whose primary news source is Television?

$54$ (correct answer)
$126$
$150$
$180$

Explanation: The expected count is (row total × column total) / grand total. The row total for Bachelor's degree is 180. The column total for Television is the total number of adults minus those who use Online sources: $500 - 350 = 150$ . The grand total is 500. So, the expected count is $(180 \times 150) / 500 = 27000 / 500 = 54$ .

Question 20

A marketing firm is analyzing survey data from 800 consumers to see if there is an association between age group (Under 30, 30-50, Over 50) and preferred type of vacation (Beach, City, Adventure). There are 300 consumers in the 'Under 30' age group.

For a chi-square test, what is the sum of the expected counts for the 'Under 30' age group across all three vacation types?

It is equal to the average number of consumers per age group, which is $800/3$ .
It is equal to the row total for the 'Under 30' age group, which is 300. (correct answer)
It is equal to the grand total of the table, which is 800.
It cannot be determined without knowing the column totals for each vacation type.

Explanation: A property of expected counts in a two-way table is that the sum of the expected counts in any given row is equal to the total for that row. Similarly, the sum of expected counts in any column equals the column total. Therefore, the sum of expected counts for the 'Under 30' group is its row total, 300.