Expected Counts in Two-Way Tables - AP Statistics

1. 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?
   Answer: The sample size multiplied by the product of the marginal probabilities for that cell's row and column. 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: $n \times P(\text{row}) \times P(\text{col}) = n \times \frac{\text{row total}}{n} \times \frac{\text{col total}}{n} = \frac{(\text{row total})(\text{col total})}{n}$.
2. For a chi-square test of independence, which of the following comparisons between expected counts is correct?
   Answer: The expected count for (Level 2, Level X) is greater than the expected count for (Level 1, Level X). 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.
3. For a chi-square test for homogeneity, what is the expected number of plants of Species A that have red flowers, assuming the color distribution is the same for both species?
   Answer: $80$ First, find the column total for red flowers. 40% of 500 plants is $0.40 \times 500 = 200$ red flowers. The expected count is (row total × column total) / grand total. The row total for Species A is 200. The column total for red flowers is 200. The grand total is 500. The expected count is $(200 \times 200) / 500 = 40000 / 500 = 80$.
4. 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?
   Answer: $\dfrac{(150)(240)}{400}$ 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.
5. 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?
   Answer: $\dfrac{(120)(45)}{180}$ 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.
6. 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?
   Answer: 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. 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.
7. 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?
   Answer: $\frac{(150)(180)}{600}$ 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 $\frac{(150)(180)}{600}$.
8. What is the sum of all the expected counts for the 12 cells in the table?
   Answer: It is equal to 250, the grand total of the table. 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.
9. 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?
   Answer: $180$ 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 $(600 \times 300) / 1000 = 180000 / 1000 = 180$.
10. 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?
    Answer: $\dfrac{(36)(50)}{90}$ 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.
11. Of 300 adults, 45% are urban and 40% prefer buses. Under independence, expected urban bus riders?
    Answer: $(135)(120)/300=54$ Forty-five percent of 300 is 135 urban adults, and forty percent is 120 bus riders. Under independence, expected overlap is 300(0.45)(0.40)=54, or equivalently (135)(120)/300=54. The tempting wrong answer is 120, the total bus riders, because it ignores the 45% urban condition.
12. A cell has expected count 30, row total 120, table total 400. Its column total is?
    Answer: $(30)(400)/120=100$ Expected count equals row total times column total divided by table total, so 30 = (120)(column total)/400. Multiply both sides by 400 and divide by 120: column total = (30)(400)/120 = 100. A common error is (30)(120)/400 = 9, which swaps the row total and table total in the formula.
13. 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?
    Answer: $54$ 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 $(180 \times 150) / 500 = 27000 / 500 = 54$.
14. 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?
    Answer: $\dfrac{(180)(120)}{300}$ 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.
15. A cell has observed count 50, row total 80, column total 60, table total 200. Expected count?
    Answer: $(80)(60)/200=24$ Expected count uses the formula row total times column total divided by table total, so 80 times 60 is 4,800, then divided by 200 gives 24. The tempting wrong answer is (50)(60)/80=37.5, which incorrectly uses the observed count in place of the row total.
16. A cell has observed count 3 and expected count 8. Which is the validity concern?
    Answer: Neither value is too small The validity condition for chi-square procedures is about expected counts, not observed counts. Here the expected count is 8, which meets the usual rule of at least 5, so there is no validity concern from this cell. The tempting mistake is to worry that observed 3 is too small, but observed counts are not the standard that determines whether the test is valid.
17. 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?
    Answer: $\dfrac{(120)(110)}{200}$ 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.
18. Given the information provided, what was the overall total number of students surveyed who participate in sports?
    Answer: $200$ The formula for expected count is: Expected = (Row Total × Column Total) / Grand Total. We are given: Expected = 80, Row Total = 200, Grand Total = 500. We can solve for the Column Total (students who participate in sports): $80 = (200 \times \text{Column Total}) / 500$. Rearranging the equation gives: Column Total = $(80 \times 500) / 200 = 40000 / 200 = 200$.
19. For a chi-square test of homogeneity, what would be the expected frequency of morning customers who prefer Espresso, assuming coffee preference is the same for both shifts?
    Answer: $60$ The expected frequency (or count) is (row total × column total) / grand total. The row total for morning customers is 300. The column total for Espresso preference is 100. The grand total is 500. The expected frequency is $(300 \times 100) / 500 = 30000 / 500 = 60$.
20. For a chi-square test, what is the sum of the expected counts for the 'Under 30' age group across all three vacation types?
    Answer: It is equal to the row total for the 'Under 30' age group, which is 300. 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.

Start this quiz