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Expected counts in two-way tables

Learning goals

  • Computes expected counts under the null model.
  • Checks expected-count conditions for chi-square procedures.
  • Compares observed and expected counts conceptually.
  • Identifies which cells contribute most to discrepancy when relevant.

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Expected counts in two-way tables 1
A school surveys \(100\) students about their grade level and whether they bring lunch. Of these students, \(50\) are juniors and \(30\) bring lunch. If grade level and bringing lunch are independent, how many juniors would be expected to bring lunch?

Hints

- Which total describes juniors, and which describes students who bring lunch? - Think about what fraction of all surveyed students are juniors. - Under independence, that fraction should be the same among students who bring lunch.

Solution

1. Under independence, use the junior total, the bring-lunch total, and the grand total. 2. The expected number is \(E=\frac{50\cdot30}{100}=15\). 3. This predicts \(15\) juniors bringing lunch if the two variables are independent; it is not necessarily the observed number.

Answer

\(15\) juniors.
Expected counts in two-way tables 2
In a two-way table, Neema says, “To find an expected count under independence, I need the observed count from that same cell.” Is Neema correct? State the information that actually determines an expected cell count.

Hints

- Imagine covering the inside cells of the table. What information is still visible? - What would an independence model use to predict a hidden count? - Would the prediction change if the inside counts changed but the totals stayed fixed?

Solution

1. The statement is incorrect. 2. Under independence, an expected cell count is determined by the cell's row total, the cell's column total, and the grand total. 3. The observed count in that individual cell is not used to calculate its expected count.

Answer

No. The expected count is determined by the relevant row total, column total, and grand total.
Expected counts in two-way tables 3
A table contains \(120\) students. One row contains \(\frac{1}{4}\) of all students, and one column contains \(\frac{1}{3}\) of all students. If the row and column variables are independent, what is the expected count where that row and column meet?

Hints

- Convert the row and column information into proportions of the whole table. - Independence connects the two marginal proportions multiplicatively. - After finding the expected cell proportion, convert it back to a count.

Solution

1. Under independence, the expected proportion in the cell is the product of the row and column proportions. 2. The expected count is \(120\cdot\frac{1}{4}\cdot\frac{1}{3}=10\).

Answer

\(10\)
Expected counts in two-way tables 4
A school surveys \(60\) seniors and \(40\) juniors. In all, \(30\) of the \(100\) students walk to school. If walking and grade level are independent, how many walkers would be expected in each grade? Explain why you use the same walking percentage for both grades.

Hints

- Find the fraction of all students who walk, then ask how independence affects that fraction within each grade. - Each grade has a different number of students. What happens when the same percentage is applied to both?

Solution

1. Overall, \(\frac{30}{100}=30\%\) walk. Independence predicts that this same percentage applies within each grade. 2. The expected counts are \(0.30\cdot60=18\) seniors and \(0.30\cdot40=12\) juniors. They add to the \(30\) walkers overall.

Answer

\(18\) seniors and \(12\) juniors. Independence predicts the same walking rate in each grade, not the same number of walkers.
Expected counts in two-way tables 5
A movie theater counted \(14\) afternoon-show customers who bought popcorn. If show time and buying popcorn were independent, the expected count for that group would be \(8\). A manager says, “So \(8\) afternoon-show customers bought popcorn.” Explain the mistake. What does the expected count tell us?

Hints

- Could a model still be useful when its prediction differs from what happened? - What can the theater learn by counting customers that a model alone cannot tell it?

Solution

1. The theater counted \(14\) afternoon-show customers who bought popcorn, not \(8\). 2. The expected count \(8\) is a prediction under independence, not a count of what actually happened.

Answer

The manager confused a model prediction with an observed count. The theater counted \(14\); the expected count \(8\) is what independence predicts, not what the theater observed.
Expected counts in two-way tables 6
A school randomly samples students in ninth and tenth grades. In the samples, \(25\%\) of ninth graders and \(45\%\) of tenth graders take the bus. If grade and bus use are independent in the school's student population, what must be true of the two population bus-use percentages? Explain why the two sample percentages alone cannot settle whether the variables are associated.

Hints

- Separate the school's population from the students chosen for the survey. - Think about how results might change if the school selected a different random sample.

Solution

1. Under independence, the population bus-use percentage would be the same for ninth and tenth graders. 2. The observed sample rates are different, but random samples can differ by chance even when the population rates are equal. 3. The sample difference suggests an association; by itself it does not establish that the population variables are associated.

Answer

The two population bus-use percentages would be equal under independence. The sample percentages can differ by chance, so the observed difference alone does not establish a population association.
Expected counts in two-way tables 7
A school surveys juniors and seniors who travel to school either by bus or on foot. The table shows the totals. <table><tr><th></th><th>Bus</th><th>Walk</th><th>Total</th></tr><tr><th>Juniors</th><td>?</td><td>?</td><td>\(40\)</td></tr><tr><th>Seniors</th><td>?</td><td>?</td><td>\(60\)</td></tr><tr><th>Total</th><td>\(50\)</td><td>\(50\)</td><td>\(100\)</td></tr></table> Complete the four expected counts if grade level and travel method are independent.

Hints

- The last row gives travel-method totals, while the last column gives grade totals. - Each expected count combines one grade total with one travel-method total. - Check that the four expected counts give the same totals as the supplied margins.

Solution

1. For each cell, use \(E=\frac{(\text{row total})(\text{column total})}{\text{grand total}}\). 2. The expected counts for juniors are \(\frac{40\cdot50}{100}=20\) for bus and \(20\) for walking. 3. The expected counts for seniors are \(\frac{60\cdot50}{100}=30\) for bus and \(30\) for walking.

Answer

<table><tr><th></th><th>Bus</th><th>Walk</th></tr><tr><th>Juniors</th><td>\(20\)</td><td>\(20\)</td></tr><tr><th>Seniors</th><td>\(30\)</td><td>\(30\)</td></tr></table>
Expected counts in two-way tables 8
A \(2 \times 3\) table has row totals \(30\) and \(70\), column totals \(20\), \(50\), and \(30\), and grand total \(100\). Find all six expected counts under independence.

Hints

- Keep the same row total while moving across a row, and change only the column total. - The grand total is the denominator for every expected count. - Use the margins to check your completed expected table.

Solution

1. Use \(E=\frac{(\text{row total})(\text{column total})}{100}\) for each cell. 2. The first row expectations are \(6\), \(15\), and \(9\). 3. The second row expectations are \(14\), \(35\), and \(21\). 4. These expected counts reproduce the given row and column totals.

Answer

First row: \((6, 15, 9)\). Second row: \((14, 35, 21)\).
Expected counts in two-way tables 9
A \(2 \times 2\) table has row totals \(45\) and \(55\), column totals \(30\) and \(70\), and grand total \(100\). Find the expected count in every cell under independence.

Hints

- Expected counts do not have to be whole numbers. - Use one row total and one column total for each cell. - Check that each row of expected counts sums to its row total.

Solution

1. Apply \(E=\frac{(\text{row total})(\text{column total})}{100}\) to each cell. 2. The first row expectations are \(\frac{45\cdot 30}{100}=13.5\) and \(\frac{45\cdot 70}{100}=31.5\). 3. The second row expectations are \(\frac{55\cdot 30}{100}=16.5\) and \(\frac{55\cdot 70}{100}=38.5\).

Answer

First row: \((13.5, 31.5)\). Second row: \((16.5, 38.5)\).
Expected counts in two-way tables 10
A \(3 \times 2\) table has row totals \(20\), \(30\), and \(50\), column totals \(40\) and \(60\), and grand total \(100\). Complete the expected-count table under independence.

Hints

- The same two column proportions apply to every row under independence. - Compute one row at a time from its row total. - Verify the column sums after completing the table.

Solution

1. Multiply each row total by each column total and divide by \(100\). 2. The first row expectations are \((8, 12)\). 3. The second row expectations are \((12, 18)\). 4. The third row expectations are \((20, 30)\).

Answer

Expected rows: \((8, 12)\), \((12, 18)\), and \((20, 30)\).
Expected counts in two-way tables 11
A community survey asks \(180\) people whether they use the public library and whether they are ages \(18\)–\(24\). Of those surveyed, \(84\) use the library and \(45\) are ages \(18\)–\(24\). If library use and age group are independent, how many library users ages \(18\)–\(24\) would be expected?

Hints

- Identify the totals for library use and for the stated age group. - Under independence, the proportion of library users should be the same within that age group as in the whole sample. - Check that your expected count is smaller than either of the two category totals.

Solution

1. The library-user total is \(84\), the age-group total is \(45\), and the grand total is \(180\). 2. Under independence, the expected count is \(E=\frac{84\cdot45}{180}=21\).

Answer

\(21\) people.
Expected counts in two-way tables 12
A \(2 \times 3\) table has row totals \(80\) and \(120\), column totals \(50\), \(70\), and \(80\), and grand total \(200\). Find the three expected counts in the first row under independence, then identify which column has the largest expected first-row count.

Hints

- You only need the first-row total, each column total, and the grand total. - The expected counts in one row are proportional to the column totals. - Compare the three completed expectations after calculating them.

Solution

1. Use the first-row total \(80\) with each column total. 2. The expected counts are \(\frac{80\cdot 50}{200}=20\), \(\frac{80\cdot 70}{200}=28\), and \(\frac{80\cdot 80}{200}=32\). 3. The largest expected count is \(32\), in the third column.

Answer

Expected first row: \((20, 28, 32)\). The third column has the largest expected count.
Expected counts in two-way tables 13
A \(2 \times 2\) table has row totals \(12\) and \(88\), column totals \(20\) and \(80\), and grand total \(100\). a) Find all four expected counts under independence. b) Which expected counts are below \(5\)? c) Does the usual high-school chi-square condition that every expected count be at least \(5\) hold?

Hints

- Compute the entire expected table before checking the condition. - Expected-count conditions concern the expected values, not the observed cell counts. - Compare each expectation with the stated threshold separately.

Solution

1. The expected counts are \(\frac{12\cdot 20}{100}=2.4\), \(\frac{12\cdot 80}{100}=9.6\), \(\frac{88\cdot 20}{100}=17.6\), and \(\frac{88\cdot 80}{100}=70.4\). 2. Only the cell with expected count \(2.4\) is below \(5\). 3. Therefore, the stated expected-count condition does not hold.

Answer

a) Expected rows: \((2.4, 9.6)\) and \((17.6, 70.4)\). b) Only \(2.4\) is below \(5\). c) No.
Expected counts in two-way tables 14
In a two-way table with grand total \(150\), one column total is \(45\). Under independence, the expected count in a cell in that column is \(18\). What is the total for the row containing that cell?

Hints

- Write the expected-count relationship with the row total as the unknown. - The known expected count, column total, and grand total determine the missing margin. - Rearrange the equation before substituting or simplifying.

Solution

1. Let the unknown row total be \(r\). 2. The expected-count equation is \(18=\frac{45r}{150}\). 3. Solving gives \(r=\frac{18\cdot 150}{45}=60\).

Answer

\(60\)
Expected counts in two-way tables 15
A two-way table has grand total \(100\). One row total is \(35\), and the expected count in one cell of that row is \(14\). Under independence, find the column total for that cell.

Hints

- Treat the expected-count formula as an equation rather than a one-way calculation. - The row total and grand total are already known. - Isolate the missing column total algebraically.

Solution

1. Let the unknown column total be \(c\). 2. Use \(14=\frac{35c}{100}\). 3. Solving gives \(c=\frac{14\cdot 100}{35}=40\).

Answer

\(40\)
Expected counts in two-way tables 16
A cell belongs to a row with total \(30\) and a column with total \(40\) in a table with grand total \(100\). Camille calculates its expected count as \(\frac{30\cdot 40}{30}=40\). Identify the error and give the correct expected count under independence.

Hints

- Check each part of the expected-count formula against the table margins. - Ask what quantity should normalize the product of a row total and a column total. - Test whether Camille's result is plausible relative to the row and column totals.

Solution

1. The denominator in the expected-count formula must be the grand total, not the row total. 2. The correct calculation is \(\frac{30\cdot 40}{100}=12\). 3. Camille's value \(40\) would incorrectly assign the entire column total to this one row.

Answer

Camille divided by the row total instead of the grand total. The correct expected count is \(12\).
Expected counts in two-way tables 17
A school records where \(60\) students prefer to study. <table><tr><th></th><th>Library</th><th>Home</th></tr><tr><th>Juniors</th><td>\(24\)</td><td>\(12\)</td></tr><tr><th>Seniors</th><td>\(9\)</td><td>\(15\)</td></tr></table> Find the four expected counts under independence. Explain what the expected junior–library count means.

Hints

- First add each row and each column of the observed table. - Each expected cell uses its row total, column total, and the overall number surveyed. - Use the category names when explaining the expected count.

Solution

1. The row totals are \(36\) and \(24\), and the column totals are \(33\) and \(27\). 2. The junior expected counts are \(\frac{36\cdot33}{60}=19.8\) and \(\frac{36\cdot27}{60}=16.2\). 3. The senior expected counts are \(\frac{24\cdot33}{60}=13.2\) and \(\frac{24\cdot27}{60}=10.8\). 4. The predicted junior–library count is \(19.8\) under independence, whereas the observed count is \(24\).

Answer

Expected rows: Juniors \((19.8,16.2)\), Seniors \((13.2,10.8)\). The expected junior–library count predicts about \(19.8\) juniors who prefer the library if grade and study location are independent.
Expected counts in two-way tables 18
A garden center surveys shoppers. Of \(30\) weekday shoppers, \(12\) prefer native plants; of \(40\) weekend shoppers, \(24\) prefer native plants. Find the percentage who prefer native plants in each group. Would equal percentages be expected if shopping day and plant preference were independent? Explain using your results.

Hints

- Which total belongs with the weekday native-plant count? What about weekends? - Would comparing just the counts be fair when the group sizes differ? - Ask whether plant preference seems to change when you know which day the customer visited.

Solution

1. On weekdays, \(\frac{12}{30}=40\%\) prefer native plants. 2. On weekends, \(\frac{24}{40}=60\%\) prefer native plants. 3. Under independence, the two percentages would be equal. The observed percentages differ by \(20\) percentage points.

Answer

Weekday shoppers: \(40\%\). Weekend shoppers: \(60\%\). Independence would predict equal percentages, but these differ by \(20\) percentage points.
Expected counts in two-way tables 19
A school survey includes \(30\) juniors and \(170\) seniors. Of all the students, \(20\) bike to school and \(180\) use another way to travel. If grade level and travel method are independent, find the smallest expected count. Does the usual condition that every expected count be at least \(5\) hold?

Hints

- Find the survey's total before using the expected-count formula. - Consider which grade and travel method might have the smallest independence prediction, then check it. - Compare the predicted count with the stated lower bound.

Solution

1. The overall total is \(30+170=200\), and the smallest row and column totals are \(30\) and \(20\). 2. The expected junior–bike count is \(\frac{30\cdot20}{200}=3\), the smallest expected count. The other expectations are \(27\), \(17\), and \(153\). 3. Because \(3<5\), the stated expected-count condition fails.

Answer

The smallest expected count is \(3\) for juniors who bike. No, the expected-count condition is not met.
Expected counts in two-way tables 20
A school surveys \(40\) juniors and \(60\) seniors. The chart shows how many students in all choose each study location. If grade level and study location are independent, how many juniors would be expected to choose the library, and how many seniors would be expected to choose the cafe?
Figure for problem 563189

Hints

- Read the overall totals for the two named locations from the horizontal bars. - Combine each group size with the matching overall category total. - Independence preserves the overall location proportions within each grade.

Solution

1. The chart gives column totals \(50\) for Library, \(30\) for Home, and \(20\) for Cafe, totaling \(100\). 2. The expected junior–library count is \(\frac{40\cdot50}{100}=20\). 3. The expected senior–cafe count is \(\frac{60\cdot20}{100}=12\).

Answer

\(20\) juniors at the library and \(12\) seniors at the cafe.
Expected counts in two-way tables 21
A \(2 \times 3\) table has row totals \(40\) and \(60\), column totals \(30\), \(50\), and \(20\), and grand total \(100\). The observed first row is \((18, 14, 8)\). a) Find all expected cell counts under independence. b) Compute observed minus expected for every cell. c) Identify which column shows the largest absolute departure in the first row.

Hints

- Under independence, expected counts are determined entirely by the margins. - Recover the unlisted observed row from the column totals. - Residuals compare observed counts with independence expectations cell by cell.

Solution

1. Expected counts are row total times column total divided by the grand total. The first row expectations are \((12, 20, 8)\), and the second row expectations are \((18, 30, 12)\). 2. The observed second row is obtained from column totals: \((12, 36, 12)\). 3. Residuals are first row \((6, -6, 0)\) and second row \((-6, 6, 0)\). 4. The first two columns tie for the largest absolute first-row residual, each with magnitude \(6\).

Answer

a) Expected table: first row \((12, 20, 8)\), second row \((18, 30, 12)\). b) Residuals: first row \((6, -6, 0)\), second row \((-6, 6, 0)\). c) Columns 1 and 2 tie, with absolute residual \(6\).
Expected counts in two-way tables 22
Under independence, a cell has expected count \(24\). Its row total is \(60\), and its column total is \(80\). Find the grand total of the two-way table.

Hints

- This time the denominator of the expected-count formula is the unknown. - Write an equation that uses the two known margins and the expected count. - Rearrange carefully so the grand total is isolated.

Solution

1. Let the grand total be \(N\). 2. The expected-count equation is \(24=\frac{60\cdot 80}{N}\). 3. Solving gives \(N=\frac{60\cdot 80}{24}=200\).

Answer

\(200\)
Expected counts in two-way tables 23
A school surveys \(200\) students about grade level and where they study. Of these students, \(80\) are juniors and \(50\) prefer the library. Under independence, find the expected number of juniors who prefer the library in two ways: a) Use the expected-count formula. b) Use the idea that juniors should make up the same share of library-preferring students as of all surveyed students. Explain why the two methods agree.

Hints

- First find the share of surveyed students who are juniors. - Think about what independence means for that share within students preferring the library. - Compare the two ways of writing the same multiplication.

Solution

1. Formula: \(E=\frac{80\cdot50}{200}=20\). 2. Juniors make up \(\frac{80}{200}=0.40\) of the survey. Under independence they should also make up \(40\%\) of the \(50\) students preferring the library, so \(0.40\cdot50=20\). 3. Both calculations multiply the junior share of the full sample by the library-preferring total.

Answer

a) \(20\) juniors. b) \(20\) juniors. The formula and proportion method both use the same independence prediction.
Expected counts in two-way tables 24
In one row of a two-way table, the expected counts under independence are \((10, 20, 30)\). The observed counts in the first two cells are \(14\) and \(17\), and the observed row total is \(60\). a) Find the third observed count. b) Find the three residuals \(O-E\). c) Explain why the residuals in this row add to \(0\).

Hints

- Use the observed row total before calculating residuals. - Compare each observed cell with its matching expected cell. - Think about what happens when you subtract two rows that have the same total.

Solution

1. The third observed count is \(60-14-17=29\). 2. The residuals are \(14-10=4\), \(17-20=-3\), and \(29-30=-1\). 3. Their sum is \(4-3-1=0\). 4. Expected counts preserve the row total, so the observed and expected row sums are both \(60\); therefore the row's observed-minus-expected residuals must sum to \(0\).

Answer

a) \(29\) b) \((4, -3, -1)\) c) They sum to \(0\) because the observed and expected counts have the same row total.
Expected counts in two-way tables 25
A \(2 \times 2\) observed table has row totals \(50\) and \(50\), column totals \(30\) and \(70\), and grand total \(100\). Aoife proposes this expected table under independence: <table><tr><th></th><th>Column 1</th><th>Column 2</th></tr><tr><th>Row 1</th><td>\(20\)</td><td>\(30\)</td></tr><tr><th>Row 2</th><td>\(10\)</td><td>\(40\)</td></tr></table> The proposed table has the correct row and column totals. Is it nevertheless a valid expected table under independence? Explain and give the correct expected counts.

Hints

- Preserving margins is one check, but independence imposes an additional proportional structure. - Compare each row's share of the grand total with its share of each column. - Recalculate at least one proposed cell from the expected-count formula.

Solution

1. Matching the margins is necessary but not sufficient for an independence expected table. 2. Because both row totals are \(50\), each row is half of the grand total, so each row should receive half of each column total. 3. The correct expectations are \(15\) and \(35\) in each row. 4. The proposed table is invalid because its interior counts do not follow the independence formula even though its margins match.

Answer

No. The correct expected table has rows \((15, 35)\) and \((15, 35)\).
Expected counts in two-way tables 26
Students in two grades report how they travel to school. Panel a) shows juniors; panel b) shows seniors. a) Read the six counts and find the row totals, column totals, and grand total. b) Find all expected counts under independence. c) Which travel method has the largest absolute residual \(|O-E|\), and in which grade?
Figure for problem 550413

Hints

- The junior and senior totals are not the same. Does the taller bar necessarily mean a greater-than-expected count? - Think about how the total number using each travel method is divided between the two grades. - Could the largest absolute departure occur in more than one cell? Check both grades.

Solution

1. The observed counts for Bus, Walk, and Bike are juniors \((28,21,21)\) and seniors \((35,42,28)\). The row totals are \(70\) and \(105\), the column totals are \(63,63,49\), and the grand total is \(175\). 2. The expected counts are \((25.2,25.2,19.6)\) for juniors and \((37.8,37.8,29.4)\) for seniors. 3. The residuals are \((2.8,-4.2,1.4)\) and \((-2.8,4.2,-1.4)\). 4. Walking has the largest absolute residual, \(4.2\), in both grades.

Answer

a) Observed rows: juniors \((28,21,21)\), seniors \((35,42,28)\); row totals \((70,105)\), column totals \((63,63,49)\), grand total \(175\). b) Expected rows: juniors \((25.2,25.2,19.6)\), seniors \((37.8,37.8,29.4)\). c) Walking in both grades, with \(|O-E|=4.2\).
Expected counts in two-way tables 27
In a \(2 \times 3\) table with grand total \(200\), the first row total is \(60\). Under independence, the expected counts in the first row are \((12, 18, 30)\). a) Recover the three column totals. b) Find the expected counts in the second row.

Hints

- Work backward from each first-row expected count to its column margin. - All three reverse equations share the same row total and grand total. - After finding the column totals, determine the second row total from the grand total.

Solution

1. For each column total \(c\), the first-row expectation satisfies \(E=\frac{60c}{200}\), so \(c=\frac{200E}{60}\). 2. The column totals are \(40\), \(60\), and \(100\). 3. The second row total is \(200-60=140\). 4. Its expected counts are \(\frac{140\cdot 40}{200}=28\), \(\frac{140\cdot 60}{200}=42\), and \(\frac{140\cdot 100}{200}=70\).

Answer

a) Column totals: \((40, 60, 100)\). b) Second-row expected counts: \((28, 42, 70)\).

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