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Two-way tables for two categorical variables

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Difficulty: 1 – 5

54714812

A survey records response channel and satisfaction. Eight respondents did not report satisfaction, but their response channels are known: \(5\) used chat and \(3\) used phone. Explain how to retain all respondents in a two-way table. What new column should be added, and what counts enter it?

Hints

- Missing values can be represented as an explicit category when retaining records is important. - Keep the known variable classification for each incomplete record. - Ensure the added category is distinct from every substantive response.

Solution

1. Add a satisfaction category labeled “not reported.” 2. Enter \(5\) in the chat/not-reported cell and \(3\) in the phone/not-reported cell. 3. This preserves the channel information and keeps each respondent in exactly one cell.

Answer

Add a “satisfaction not reported” column. Put \(5\) in the chat row and \(3\) in the phone row of that column.

54715312

A two-way table has column categories “low,” “medium,” and “high” and row categories “amber,” “blue,” and “green.” A report says, “The row variable increases from amber to green.” Is that interpretation justified by the row categories? Give a defensible way to order the rows in the display and explain why that ordering does not create an ordinal meaning.

Hints

- Ask whether the row labels themselves encode magnitude or rank. - Separate the order used to display categories from the measurement scale of the variable. - A reproducible display convention does not have to imply “less than” or “greater than.”

Solution

1. The row labels amber, blue, and green are names of categories; they do not encode numerical size or rank. 2. Therefore, describing the row variable as increasing from amber to green is not justified. 3. A neutral reproducible convention, such as alphabetical order, is defensible because it organizes the display without claiming that one category is larger than another.

Answer

No. The row categories are nominal, so “increases from amber to green” invents an unsupported ordering. A neutral order such as alphabetical order is defensible because it changes only the display, not the meaning of the categories.

54716012

A published table shows a blank interior cell, while every other cell and all margins are numerical. Give two different interpretations of the blank that would lead to different analyses, and state what metadata is needed before treating it as \(0\).

Hints

- A visual blank is not automatically a numerical zero. - Consider both data absence and observation absence. - Look for documentation that defines display conventions.

Solution

1. The blank might mean an observed count of zero, or it might mean missing/suppressed information. 2. Treating a missing or suppressed count as zero would distort margins and later calculations. 3. A legend, footnote, or data dictionary must define the blank symbol before it is assigned a numerical value.

Answer

The blank could mean zero observations or unavailable/suppressed data. A legend, footnote, or data dictionary is needed before interpreting it as \(0\).

54711512

A report at Northbridge University crosses campus location with course format, but each row of the source file represents one course section, not one student. Daria labels the grand total “students surveyed.” Explain the observational-unit error. What may the table legitimately count, and what additional data would be needed to create a student-level table?

Hints

- Identify what one row of the source file represents. - Ask whether one person can contribute more than one source row. - Match the table’s grand-total label to the observational unit.

Solution

1. Each source row represents a course section, so each interior frequency counts sections classified by campus location and format. 2. The grand total therefore counts course sections, not students. 3. A student-level table would require one record per student, or a rule for assigning students who take more than one section to a single observational unit without double-counting.

Answer

The table may legitimately count course sections. It cannot be labeled as a count of students without student-level records and a rule that prevents students enrolled in multiple sections from being counted more than once.

54711712

A report gives this two-way frequency table for school newspaper readers. <table> <tr><th></th><th>read the sports section</th><th>did not read the sports section</th><th>Total</th></tr> <tr><td>digital edition</td><td>\(39\)</td><td>\(26\)</td><td>\(65\)</td></tr> <tr><td>print edition</td><td>\(41\)</td><td>\(16\)</td><td>\(57\)</td></tr> <tr><td>Total</td><td>\(80\)</td><td>\(42\)</td><td>\(125\)</td></tr> </table> Is the table internally consistent? Identify the incorrect entry and give the value it must have.

Hints

- Add the row totals and compare that sum with the displayed grand total. - Add the column totals as an independent check. - When both margin sums agree, the inconsistent grand total is the entry that must change.

Solution

1. The row totals sum to \(65+57=122\). 2. The column totals also sum to \(80+42=122\). 3. Therefore the displayed grand total \(125\) is inconsistent and must be \(122\).

Answer

No. The grand total is incorrect; it must be \(122\), not \(125\).

54711912

A data clerk is building a two-way table for science fair projects. The four observed category pairs occurred \(22\), \(36\), \(22\), and \(38\) times for \((\text{team project, used a physical prototype})\), \((\text{team project, did not use a physical prototype})\), \((\text{individual project, used a physical prototype})\), and \((\text{individual project, did not use a physical prototype})\), respectively. Find every row and column total and verify the grand total in two different ways.

Hints

- Organize the four category combinations before computing totals. - A column total accounts for every cell in that column. - Check that the row totals and column totals lead to the same grand total.

Solution

1. Row totals are \(22+36=58\) and \(22+38=60\). 2. Column totals are \(22+22=44\) and \(36+38=74\). 3. Both \(58+60\) and \(44+74\) equal \(118\).

Answer

Row totals: \(58\), \(60\). Column totals: \(44\), \(74\). Grand total: \(118\).

54712012

A proposed \(2\times2\) table has row totals \(40\) and \(60\), column totals \(70\) and \(30\), and upper-left cell \(45\). Can such a table exist? Identify the violated constraint without completing all four cells.

Hints

- A joint count is contained within both its row and its column. - Compare the cell with the smaller containing margin. - An impossibility can be detected before reconstructing the table.

Solution

1. The upper-left cell lies in the first row, whose total is only \(40\). 2. A cell count cannot exceed either margin containing it. 3. Since \(45>40\), the proposed information is impossible.

Answer

No. The upper-left cell cannot be \(45\) because its row total is \(40\).

54712112

The completed two-way frequency table below summarizes recreation center members. <table> <tr><th></th><th>used the pool</th><th>did not use the pool</th><th>Total</th></tr> <tr><td>weekday visit</td><td>\(36\)</td><td>\(17\)</td><td>\(53\)</td></tr> <tr><td>weekend visit</td><td>\(40\)</td><td>\(27\)</td><td>\(67\)</td></tr> <tr><td>Total</td><td>\(76\)</td><td>\(44\)</td><td>\(120\)</td></tr> </table> Amina says the entry \(40\) belongs in the row “weekday visit” because it is in the “used the pool” column. Explain precisely what that cell represents and why Amina's placement is wrong.

Hints

- Read a two-way-table cell by naming both its row category and its column category. - Locate the row label aligned with the entry \(40\). - Check how moving the entry would affect the row totals.

Solution

1. A cell belongs simultaneously to one row category and one column category. 2. The entry \(40\) is at the intersection of “weekend visit” and “used the pool.” 3. Moving it to the “weekday visit” row would change the classification of those \(40\) observations and would also break the displayed margins.

Answer

The \(40\) observations are those that are both “weekend visit” and “used the pool.” The column alone does not determine the row, so the entry must stay at that intersection.

54712212

A campus café recorded these 12 orders in the order \((\text{service type},\text{cup type})\): \( \begin{aligned} &(\text{counter},\text{reusable}),(\text{mobile},\text{disposable}),(\text{counter},\text{disposable}),\\ &(\text{mobile},\text{reusable}),(\text{mobile},\text{disposable}),(\text{counter},\text{reusable}),\\ &(\text{counter},\text{disposable}),(\text{mobile},\text{disposable}),(\text{counter},\text{reusable}),\\ &(\text{mobile},\text{reusable}),(\text{counter},\text{disposable}),(\text{mobile},\text{disposable}). \end{aligned} \) Create a two-way frequency table with service type as the row variable and cup type as the column variable. Include all margins.

Hints

- Treat each ordered pair as one observation in exactly one cell. - Tally the four possible category combinations before finding any totals. - Check that both sets of margins give the same grand total.

Solution

1. Tally each ordered pair: counter/reusable \(=3\), counter/disposable \(=3\), mobile/reusable \(=2\), and mobile/disposable \(=4\). 2. The row totals are \(6\) counter orders and \(6\) mobile orders. 3. The column totals are \(5\) reusable cups and \(7\) disposable cups, for a grand total of \(12\).

Answer

<table> <tr><th></th><th>reusable</th><th>disposable</th><th>Total</th></tr> <tr><td>counter</td><td>\(3\)</td><td>\(3\)</td><td>\(6\)</td></tr> <tr><td>mobile</td><td>\(2\)</td><td>\(4\)</td><td>\(6\)</td></tr> <tr><td>Total</td><td>\(5\)</td><td>\(7\)</td><td>\(12\)</td></tr> </table>

54712312

A survey at Cedar Grove High School originally used rows “walk,” “bike,” and “motorized” and columns “arrived before 8 a.m.” and “arrived at or after 8 a.m.” Tomás proposes combining “walk” and “bike” into a row called “active travel.” The original interior counts are: <table> <tr><th></th><th>before 8 a.m.</th><th>at or after 8 a.m.</th></tr> <tr><td>walk</td><td>\(18\)</td><td>\(7\)</td></tr> <tr><td>bike</td><td>\(12\)</td><td>\(9\)</td></tr> <tr><td>motorized</td><td>\(31\)</td><td>\(23\)</td></tr> </table> Create the recoded table and explain which dimension changes and which dimension stays unchanged.

Hints

- Combine categories only within the same column. - Ask which variable is being recoded. - Verify that recoding does not change the grand total.

Solution

1. Add the walk and bike counts within each arrival-time column: \(18+12=30\) and \(7+9=16\). 2. The motorized row remains \((31,23)\). 3. The row variable changes from three categories to two; the two arrival-time columns remain unchanged.

Answer

<table> <tr><th></th><th>before 8 a.m.</th><th>at or after 8 a.m.</th><th>Total</th></tr> <tr><td>active travel</td><td>\(30\)</td><td>\(16\)</td><td>\(46\)</td></tr> <tr><td>motorized</td><td>\(31\)</td><td>\(23\)</td><td>\(54\)</td></tr> <tr><td>Total</td><td>\(61\)</td><td>\(39\)</td><td>\(100\)</td></tr> </table> The row dimension is reduced from three categories to two; the column dimension is unchanged.

54712412

A draft table uses rows “all subscriptions,” “monthly subscriptions,” and “annual subscriptions,” crossed with renewal status. Monthly and annual subscriptions are the only subscription types. Explain why “all subscriptions” must not appear as a third row category. How should the information be displayed instead?

Hints

- Check whether one proposed row contains observations from other rows. - Interior categories should classify each observation exactly once. - Distinguish a category from a subtotal.

Solution

1. “All subscriptions” contains both monthly and annual subscriptions, so it overlaps both proposed row categories. 2. Treating it as a third interior row would count every subscription once in its type row and again in the all-subscriptions row. 3. Monthly and annual should be the two row categories; “all subscriptions” should appear only as the row-total margin or as a separate summary outside the interior table.

Answer

Use monthly and annual as the mutually exclusive interior rows. Display “all subscriptions” only as a total, not as another category row, because it is a subtotal that contains the other rows.

54712612

A completed table originally showed \(26\) evening customers who paid cash and \(34\) evening customers who paid by card. A later audit found that \(5\) of the evening cash payments had been miscoded; they were actually evening card payments. The original table was: <table> <tr><th></th><th>cash</th><th>card</th><th>Total</th></tr> <tr><td>morning</td><td>\(31\)</td><td>\(29\)</td><td>\(60\)</td></tr> <tr><td>evening</td><td>\(26\)</td><td>\(34\)</td><td>\(60\)</td></tr> <tr><td>Total</td><td>\(57\)</td><td>\(63\)</td><td>\(120\)</td></tr> </table> Correct the table and identify which margins change.

Hints

- A recoding within one row transfers observations between columns. - Track the decrease and increase as equal amounts. - Decide which totals are unaffected by moving rather than adding observations.

Solution

1. Move \(5\) observations within the evening row: evening/cash becomes \(26-5=21\), and evening/card becomes \(34+5=39\). 2. The evening row total and grand total do not change. 3. The cash column total becomes \(52\), and the card column total becomes \(68\).

Answer

<table> <tr><th></th><th>cash</th><th>card</th><th>Total</th></tr> <tr><td>morning</td><td>\(31\)</td><td>\(29\)</td><td>\(60\)</td></tr> <tr><td>evening</td><td>\(21\)</td><td>\(39\)</td><td>\(60\)</td></tr> <tr><td>Total</td><td>\(52\)</td><td>\(68\)</td><td>\(120\)</td></tr> </table> Only the two column totals change.

54712712

A poll asks each respondent to select every streaming service they use. The report proposes a two-way table with rows “uses Service A” and “uses Service B” and columns “uses Service C” and “does not use Service C.” Explain why the proposed row categories do not define a standard two-way frequency table. Give one valid way to revise the row variable while preserving all respondents.

Hints

- Check whether one respondent can satisfy both proposed row labels. - A table cell should represent one unique category pair for each observation. - Consider creating exhaustive, nonoverlapping combinations of A and B use.

Solution

1. “Uses Service A” and “uses Service B” are not mutually exclusive; one respondent can belong to both rows. 2. Standard two-way table categories for one variable must place each observation in exactly one row and one column. 3. One valid revision is a four-category row variable: “A only,” “B only,” “both A and B,” and “neither A nor B.”

Answer

The proposed rows overlap, so respondents could be counted twice. A valid row variable is “A only,” “B only,” “both,” and “neither,” crossed with Service C use.

54713112

Two analysts display the same counts in different orientations. <table> <tr><th>Table A</th><th>approved</th><th>not approved</th></tr> <tr><td>new applicant</td><td>\(32\)</td><td>\(18\)</td></tr> <tr><td>returning applicant</td><td>\(41\)</td><td>\(9\)</td></tr> </table> <table> <tr><th>Table B</th><th>new applicant</th><th>returning applicant</th></tr> <tr><td>not approved</td><td>\(18\)</td><td>\(9\)</td></tr> <tr><td>approved</td><td>\(32\)</td><td>\(41\)</td></tr> </table> Determine whether the tables represent the same joint frequency information. Describe the row and column operations that connect them.

Hints

- Compare category pairs rather than physical cell positions. - Ask what transposing a table does to its variables. - Reordering categories changes presentation, not the underlying pairs.

Solution

1. Match counts by category pair: every new/approved, new/not-approved, returning/approved, and returning/not-approved count agrees. 2. Transposing Table A makes approval status the row variable and applicant type the column variable. 3. Reordering the transposed rows to not approved, approved produces Table B. Therefore the tables contain the same joint information.

Answer

Yes. Transpose Table A, then place the “not approved” row before the “approved” row. The category-pair counts are unchanged.

54713712

A company records work location and desk type for \(100\) employees. <table> <tr><th></th><th>Standing desk</th><th>Standard desk</th><th>Total</th></tr> <tr><td>Remote</td><td>\(26\)</td><td>\(34\)</td><td>\(60\)</td></tr> <tr><td>On-site</td><td>\(18\)</td><td>\(22\)</td><td>\(40\)</td></tr> <tr><td>Total</td><td>\(44\)</td><td>\(56\)</td><td>\(100\)</td></tr> </table> One employee was entered as “remote with a standing desk” but should have been “on-site with a standard desk.” Correct the entire table. State which margins change and explain why the grand total does not change.

Hints

- A correction can move one observation from one joint category to another rather than remove it. - Track the row and column lost by the original cell and gained by the corrected cell. - Check whether the number of employees in the data set changes.

Solution

1. Move one observation out of the remote/standing cell: \(26\to25\). 2. Move it into the on-site/standard cell: \(22\to23\). The other two interior cells do not change. 3. The remote total becomes \(59\), and the on-site total becomes \(41\). 4. The standing-desk total becomes \(43\), and the standard-desk total becomes \(57\). 5. The same employee remains in the data set, so the grand total stays \(100\).

Answer

<table> <tr><th></th><th>Standing desk</th><th>Standard desk</th><th>Total</th></tr> <tr><td>Remote</td><td>\(25\)</td><td>\(34\)</td><td>\(59\)</td></tr> <tr><td>On-site</td><td>\(18\)</td><td>\(23\)</td><td>\(41\)</td></tr> <tr><td>Total</td><td>\(43\)</td><td>\(57\)</td><td>\(100\)</td></tr> </table> Both row margins and both column margins change; the grand total does not.

54713812

A study has variables “housing” with categories apartment and house, and “internet service” with categories fiber and nonfiber. The four joint counts are \(46\), \(34\), \(29\), and \(51\), listed in this order: \( (\text{house}, \text{fiber}),\;(\text{apartment}, \text{nonfiber}),\;(\text{apartment}, \text{fiber}),\;(\text{house}, \text{nonfiber}). \) Create a table with internet service as the row variable and housing as the column variable. Include margins.

Hints

- Read the requested row and column variables before placing counts. - Attach each value to its labeled category pair. - Use margins to check that every observation was placed once.

Solution

1. Place each count by its labels, not by the order in which row and column categories might normally be listed. 2. The fiber row is apartment \(29\), house \(46\); the nonfiber row is apartment \(34\), house \(51\). 3. Row totals are \(75\) and \(85\); column totals are \(63\) and \(97\); the grand total is \(160\).

Answer

<table> <tr><th></th><th>apartment</th><th>house</th><th>Total</th></tr> <tr><td>fiber</td><td>\(29\)</td><td>\(46\)</td><td>\(75\)</td></tr> <tr><td>nonfiber</td><td>\(34\)</td><td>\(51\)</td><td>\(85\)</td></tr> <tr><td>Total</td><td>\(63\)</td><td>\(97\)</td><td>\(160\)</td></tr> </table>

54714412

A survey’s response variable has three defined categories: yes, no, and not reported. In one month, no respondent falls in “not reported.” An analyst deletes that zero-count column before publishing the table. Explain one advantage and one disadvantage of keeping the zero-count column. Does deleting it change the observed grand total or the defined variable?

Hints

- Separate a category’s definition from its observed frequency. - Ask what a zero count contributes to the grand total. - Consider comparability across different reporting periods.

Solution

1. Keeping the column documents the complete category system and makes month-to-month tables structurally comparable. 2. Deleting it creates a more compact display but can make readers think the variable was defined with only two categories. 3. Because the deleted column has count \(0\), the observed grand total is unchanged. 4. The variable’s defined domain still includes “not reported” even when that category is absent from this sample.

Answer

Keeping the zero column preserves the declared category structure; deleting it simplifies the display. The grand total is unchanged, but the variable still has three defined categories.

54714712

Two regional offices use the same row and column definitions. Their tables are: <table> <tr><th>Office A</th><th>approved</th><th>not approved</th></tr> <tr><td>new client</td><td>\(24\)</td><td>\(16\)</td></tr> <tr><td>returning client</td><td>\(31\)</td><td>\(9\)</td></tr> </table> <table> <tr><th>Office B</th><th>approved</th><th>not approved</th></tr> <tr><td>new client</td><td>\(18\)</td><td>\(22\)</td></tr> <tr><td>returning client</td><td>\(27\)</td><td>\(13\)</td></tr> </table> Combine the offices into one two-way table. State the condition that makes cell-by-cell addition legitimate.

Hints

- Add only cells representing the same category pair. - Confirm that no observation appears in both office datasets. - Recompute margins after combining the interior counts.

Solution

1. Add corresponding category-pair counts: \(24+18=42\), \(16+22=38\), \(31+27=58\), and \(9+13=22\). 2. Row totals are \(80\) and \(80\); column totals are \(100\) and \(60\); the grand total is \(160\). 3. Cell-by-cell addition is legitimate because both tables use identical category definitions and represent disjoint sets of observations.

Answer

<table> <tr><th></th><th>approved</th><th>not approved</th><th>Total</th></tr> <tr><td>new client</td><td>\(42\)</td><td>\(38\)</td><td>\(80\)</td></tr> <tr><td>returning client</td><td>\(58\)</td><td>\(22\)</td><td>\(80\)</td></tr> <tr><td>Total</td><td>\(100\)</td><td>\(60\)</td><td>\(160\)</td></tr> </table> The definitions must match, and the observation sets must not overlap.

54715012

A report lists the four interior counts of a \(2\times2\) table as the unlabeled multiset \(\{12,18,27,43\}\). The row and column category labels are known, but no count is attached to a category pair. Can the labeled table be reconstructed uniquely? Explain and give two different labeled tables consistent with the information.

Hints

- Counts without positions do not identify category pairs. - Try swapping two counts that are not in the same labeled cell. - One counterexample pair is enough to disprove uniqueness.

Solution

1. The multiset gives values but not their positions. 2. For example, \(\begin{pmatrix}12&18\\27&43\end{pmatrix}\) and \(\begin{pmatrix}12&27\\18&43\end{pmatrix}\) use the same four counts but assign them to different category pairs. 3. Because these labeled tables differ, reconstruction is not unique.

Answer

No. For example, both \(\begin{pmatrix}12&18\\27&43\end{pmatrix}\) and \(\begin{pmatrix}12&27\\18&43\end{pmatrix}\) use the same multiset but represent different labeled tables.

54715112

A hospital table separates discharges into “home,” “rehabilitation,” and “long-term care,” crossed with “readmitted within 30 days” and “not readmitted.” An administrator wants to merge rehabilitation and long-term care into “facility discharge.” Give one decision question for which the merged table is adequate and one for which it is inadequate.

Hints

- Identify exactly which distinction aggregation removes. - A question using only the combined category remains answerable. - A question comparing the merged components does not.

Solution

1. The merged table is adequate for comparing home discharges with all facility discharges. 2. It is inadequate for comparing rehabilitation with long-term care, because those categories are no longer separated. 3. Aggregation is acceptable only when the decision does not require the distinction being removed.

Answer

Adequate: “Is readmission more common for home or facility discharges?” Inadequate: “Is readmission more common after rehabilitation or long-term care?”

54715812

A district combines annual tables for the same students in 2025 and 2026. Each table classifies students by lunch type and bus use. The analyst adds the tables cell by cell and labels the grand total “number of students.” Explain why that label may be wrong even when the arithmetic is correct. Give a more accurate label.

Hints

- Identify the observational unit in each annual table. - Combining periods can create repeated records for the same person. - Choose a label that reflects what is actually counted.

Solution

1. The same student can appear once in each year, so cell-by-cell addition counts student-year records rather than necessarily distinct students. 2. Arithmetic addition is valid for totals of records if category definitions match. 3. A more accurate grand-total label is “student-year observations.”

Answer

The sum may count the same person twice, once per year. Label the total “student-year observations,” not necessarily “students.”

54715912

A medical table leaves one interior cell blank because that combination is impossible: rows are “pregnant” and “not pregnant,” and columns are “received prenatal care” and “did not receive prenatal care.” The study defines prenatal care as care received during the recorded pregnancy. Should the pregnant/no-care cell, the not-pregnant/prenatal-care cell, or both be structural zeros? Explain the distinction between a structural zero and missing data.

Hints

- Use the operational definition of prenatal care. - Ask whether each category pair can exist, not whether it happened to be observed. - Do not confuse an impossible combination with an unreported value.

Solution

1. The not-pregnant/prenatal-care combination is impossible under the study definition, so that cell is a structural zero. 2. The pregnant/no-care combination is possible and must not be forced to zero. 3. A structural zero represents an impossible category pair; missing data represents an unknown value for a possible observation.

Answer

Only the not-pregnant/prenatal-care cell is a structural zero. A structural zero is logically impossible; missing data is unknown information about a possible case.

55618912

The bar chart shows counts for two commuting methods and whether students arrived on time. Each bar label gives a pair of categories. Use the chart to write the complete two-way frequency table for commuting method by arrival status, including row totals, column totals, and the grand total.

Figure for problem 556189

Hints

- Treat each bar as one interior cell of a two-way table. - Put commuting method on one dimension and arrival status on the other. - After entering the four interior counts, compute each margin from the cells it contains.

Solution

1. Read the four joint counts from the chart: bus/on time \(25\), bus/late \(10\), bike/on time \(15\), and bike/late \(5\). 2. The bus row total is \(25+10=35\), and the bike row total is \(15+5=20\). 3. The on-time column total is \(25+15=40\), and the late column total is \(10+5=15\). 4. The grand total is \(35+20=55\).

Answer

<table><tr><th>Commute</th><th>On time</th><th>Late</th><th>Total</th></tr><tr><td>Bus</td><td>\(25\)</td><td>\(10\)</td><td>\(35\)</td></tr><tr><td>Bike</td><td>\(15\)</td><td>\(5\)</td><td>\(20\)</td></tr><tr><td>Total</td><td>\(40\)</td><td>\(15\)</td><td>\(55\)</td></tr></table>

54711412

An image classifier and a human reviewer each label the same \(200\) photos as “damaged” or “not damaged.” Their classifications are summarized below. <table> <tr><th></th><th>Human: damaged</th><th>Human: not damaged</th><th>Total</th></tr> <tr><td>Classifier: damaged</td><td>\(68\)</td><td>\(24\)</td><td>\(92\)</td></tr> <tr><td>Classifier: not damaged</td><td>\(12\)</td><td>\(96\)</td><td>\(108\)</td></tr> <tr><td>Total</td><td>\(80\)</td><td>\(120\)</td><td>\(200\)</td></tr> </table> Using the human review as the reference classification: a) Identify the false-positive and false-negative cells. b) Find the total number of disagreements. c) Explain why the classifier labels \(12\) more photos as damaged than the human reviewer does.

Hints

- Decide what agreement and disagreement look like in a classification table. - The direction of an error depends on which classification is treated as the reference. - Compare the two damaged margins with the difference between the off-diagonal cells.

Solution

1. False positives are photos labeled damaged only by the classifier, so there are \(24\). False negatives are photos labeled damaged only by the human reviewer, so there are \(12\). 2. Disagreements occupy the two off-diagonal cells: \(24+12=36\). 3. The classifier’s damaged total exceeds the human damaged total by \(92-80=12\). Equivalently, false positives exceed false negatives by \(24-12=12\).

Answer

a) False positives: \(24\); false negatives: \(12\). b) There are \(36\) disagreements. c) The classifier has \(12\) more damaged labels because it has \(12\) more false positives than false negatives.

54711612

A three-way dataset records residence (on campus/off campus), meal plan (yes/no), and class year (first-year/upper-year). The two class-year tables are: <table> <tr><th>First-year</th><th>meal plan</th><th>no meal plan</th></tr> <tr><td>on campus</td><td>\(48\)</td><td>\(12\)</td></tr> <tr><td>off campus</td><td>\(9\)</td><td>\(21\)</td></tr> </table> <table> <tr><th>Upper-year</th><th>meal plan</th><th>no meal plan</th></tr> <tr><td>on campus</td><td>\(26\)</td><td>\(34\)</td></tr> <tr><td>off campus</td><td>\(18\)</td><td>\(52\)</td></tr> </table> Marginalize over class year to create one residence-by-meal-plan table.

Hints

- Treat class year as the dimension being removed. - Add only identical residence–meal-plan combinations. - Recompute margins after combining the layers.

Solution

1. Add corresponding cells across the two class-year layers. 2. The four cells are \(48+26=74\), \(12+34=46\), \(9+18=27\), and \(21+52=73\). 3. Row totals are \(120\) and \(100\); column totals are \(101\) and \(119\); the grand total is \(220\).

Answer

<table> <tr><th></th><th>meal plan</th><th>no meal plan</th><th>Total</th></tr> <tr><td>on campus</td><td>\(74\)</td><td>\(46\)</td><td>\(120\)</td></tr> <tr><td>off campus</td><td>\(27\)</td><td>\(73\)</td><td>\(100\)</td></tr> <tr><td>Total</td><td>\(101\)</td><td>\(119\)</td><td>\(220\)</td></tr> </table>

54712512

A \(2\times2\) table has row totals \(40\) and \(35\), and column totals \(32\) and \(43\). Let \(x\) be the upper-left interior count. Write all four interior counts in terms of \(x\). Then find every integer value of \(x\) for which all four counts are nonnegative.

Hints

- Use each margin to express an adjacent cell from the upper-left cell. - The last cell should agree with both its row and column totals. - Translate “count” into nonnegativity constraints.

Solution

1. The upper-right count is \(40-x\), and the lower-left count is \(32-x\). 2. The lower-right count is \(35-(32-x)=x+3\). 3. Nonnegativity requires \(x\ge0\), \(x\le40\), \(x\le32\), and \(x\ge-3\). Therefore \(x\) can be any integer from \(0\) through \(32\).

Answer

The four cells are \(x\), \(40-x\), \(32-x\), and \(x+3\). The feasible integer values are \(0\le x\le32\).

54712912

A transition table uses the same categories A and B for its rows and columns. A symmetric transition table would have the A-to-B count equal to the B-to-A count. The row margins are \(70,50\), while the column margins are \(65,55\). Can any table with these margins be symmetric? Explain without solving for all four cells.

Hints

- Consider what symmetry does to corresponding row and column entries. - Compare a row sum with the matching column sum. - A contradiction in the margins is enough; interior reconstruction is unnecessary.

Solution

1. In a symmetric table, each row sum equals the corresponding column sum because entries mirror across the diagonal. 2. A symmetric table would therefore require first row margin \(=65\) and second row margin \(=55\) when the column margins are \(65,55\). 3. The given row margins \(70,50\) do not match the corresponding column margins. 4. Therefore no table with these margins can be symmetric.

Answer

No. Symmetry would force corresponding row and column margins to match, but \(70\ne65\) and \(50\ne55\).

54713012

A registration report combined “mobile” and “desktop” into one row labeled “online.” <table> <tr><th></th><th>Attended</th><th>Did not attend</th><th>Total</th></tr> <tr><td>Online</td><td>\(54\)</td><td>\(36\)</td><td>\(90\)</td></tr> <tr><td>In person</td><td>\(28\)</td><td>\(22\)</td><td>\(50\)</td></tr> <tr><td>Total</td><td>\(82\)</td><td>\(58\)</td><td>\(140\)</td></tr> </table> An archived note says that \(40\) online registrants used mobile devices and that \(30\) of those mobile registrants attended. a) Reconstruct a table with separate mobile, desktop, and in-person rows. b) Explain why the collapsed table alone could not determine the mobile and desktop rows.

Hints

- Treat the archived mobile information as one complete row of the expanded table. - The desktop counts are the portions of the online counts not assigned to mobile. - Consider how many different pairs of rows could have the same combined row.

Solution

1. The mobile row is \(30\) attended and \(40-30=10\) did not attend. 2. Subtract the mobile row from the online row. The desktop row is \(54-30=24\) attended and \(36-10=26\) did not attend, for a total of \(50\). 3. The in-person row remains \((28,22)\). 4. The collapsed online row stores only column sums across mobile and desktop. Many different mobile-desktop splits could produce those same sums, so the archived information is necessary.

Answer

a) <table> <tr><th></th><th>Attended</th><th>Did not attend</th><th>Total</th></tr> <tr><td>Mobile</td><td>\(30\)</td><td>\(10\)</td><td>\(40\)</td></tr> <tr><td>Desktop</td><td>\(24\)</td><td>\(26\)</td><td>\(50\)</td></tr> <tr><td>In person</td><td>\(28\)</td><td>\(22\)</td><td>\(50\)</td></tr> <tr><td>Total</td><td>\(82\)</td><td>\(58\)</td><td>\(140\)</td></tr> </table> b) Collapsing mobile and desktop discards their separate cell counts, so the split is not recoverable without additional information.

54713212

Four sampled records have category pairs and survey weights: \( (A, \text{yes}, 2.0),\quad(A, \text{no}, 1.0),\quad (B, \text{yes}, 0.5),\quad(B, \text{no}, 1.5). \) Create both the unweighted \(2 \times 2\) frequency table and the weighted table. Include all margins, and explain why the weighted entries should not be described as numbers of sampled records.

Hints

- Tally records once for the unweighted table. - Sum record weights, rather than record indicators, for the weighted table. - Distinguish sample size from weighted population contribution.

Solution

1. Each category pair appears once, so every unweighted interior count is \(1\); row and column totals are \(2\), with grand total \(4\). 2. Summing weights by cell gives weighted entries \(2.0,1.0,0.5,1.5\). 3. Weighted row totals are \(3.0,2.0\); weighted column totals are \(2.5,2.5\); the weighted grand total is \(5.0\). 4. Weights represent estimated population contribution, so weighted entries are not literal counts of the four sampled records.

Answer

Unweighted table: <table> <tr><th></th><th>yes</th><th>no</th><th>Total</th></tr> <tr><td>A</td><td>\(1\)</td><td>\(1\)</td><td>\(2\)</td></tr> <tr><td>B</td><td>\(1\)</td><td>\(1\)</td><td>\(2\)</td></tr> <tr><td>Total</td><td>\(2\)</td><td>\(2\)</td><td>\(4\)</td></tr> </table> Weighted table: <table> <tr><th></th><th>yes</th><th>no</th><th>Total</th></tr> <tr><td>A</td><td>\(2.0\)</td><td>\(1.0\)</td><td>\(3.0\)</td></tr> <tr><td>B</td><td>\(0.5\)</td><td>\(1.5\)</td><td>\(2.0\)</td></tr> <tr><td>Total</td><td>\(2.5\)</td><td>\(2.5\)</td><td>\(5.0\)</td></tr> </table> The weighted values are estimated population contributions, not counts of sampled records.

54713312

A report says: “Of \(90\) surveyed households, \(52\) have a pet, \(41\) have a garden, and \(70\) have a pet or a garden. The count \(70\) belongs in the ‘pet and garden’ cell.” Explain the error, find the correct “pet and garden” count, and complete the \(2\times2\) table.

Hints

- Distinguish the union from the intersection. - Use the overlap needed to avoid double-counting the two group totals. - The neither cell comes from the complement of the union.

Solution

1. The value \(70\) is a union count, not an intersection count. 2. Inclusion-exclusion gives the both count \(52+41-70=23\). 3. Pet only is \(29\), garden only is \(18\), and neither is \(90-70=20\).

Answer

<table> <tr><th></th><th>garden</th><th>no garden</th><th>Total</th></tr> <tr><td>pet</td><td>\(23\)</td><td>\(29\)</td><td>\(52\)</td></tr> <tr><td>no pet</td><td>\(18\)</td><td>\(20\)</td><td>\(38\)</td></tr> <tr><td>Total</td><td>\(41\)</td><td>\(49\)</td><td>\(90\)</td></tr> </table> The error is treating “or” as “and.”

54713412

A \(2\times3\) table records \(150\) deliveries by vehicle type and time window. The van row total is \(84\). The morning column total is \(51\), the afternoon column total is \(63\), and the evening column total is \(36\). Van deliveries include \(32\) morning and \(34\) afternoon deliveries. Complete all six interior cells and all margins.

Hints

- Complete the known row before using the column totals. - Each lower cell is the corresponding column remainder. - Check that the two row totals reproduce the grand total.

Solution

1. Van evening deliveries are \(84-32-34=18\). 2. Bicycle deliveries are \(51-32=19\) in the morning, \(63-34=29\) in the afternoon, and \(36-18=18\) in the evening. 3. The bicycle row total is \(19+29+18=66\), and \(84+66=150\).

Answer

<table> <tr><th></th><th>morning</th><th>afternoon</th><th>evening</th><th>Total</th></tr> <tr><td>van</td><td>\(32\)</td><td>\(34\)</td><td>\(18\)</td><td>\(84\)</td></tr> <tr><td>bicycle</td><td>\(19\)</td><td>\(29\)</td><td>\(18\)</td><td>\(66\)</td></tr> <tr><td>Total</td><td>\(51\)</td><td>\(63\)</td><td>\(36\)</td><td>\(150\)</td></tr> </table>

54713612

A \(2\times2\) table has row totals \(60\) and \(40\), column totals \(55\) and \(45\), and diagonal-cell total \(65\). Find all four interior counts and show that the extra diagonal condition makes the table unique.

Hints

- Represent one interior count with a variable. - Express the remaining cells using the margins. - Use the diagonal information as the final independent constraint.

Solution

1. Let the upper-left count be \(x\). The other cells are \(60-x\), \(55-x\), and \(x-15\). 2. The diagonal condition gives \(x+(x-15)=65\). 3. Solving yields \(x=40\), so the table is \(\begin{pmatrix}40&20\\15&25\end{pmatrix}\). 4. The single free parameter is fixed by the diagonal total, so the table is unique.

Answer

\(\begin{pmatrix}40&20\\15&25\end{pmatrix}\). The fixed margins leave one free parameter, and the diagonal-total condition determines it.

54713912

A \(3\times3\) table compares a commuter’s usual mode last year with this year. The table is symmetric: the count for “last year bike, this year bus” equals the count for “last year bus, this year bike,” and similarly for every off-diagonal pair. The diagonal counts for car, bus, and bike are \(40\), \(31\), and \(22\). Above the diagonal, the counts are car-to-bus \(=9\), car-to-bike \(=6\), and bus-to-bike \(=8\). Complete the table and find the grand total.

Hints

- Reflect each given off-diagonal count across the main diagonal. - The diagonal entries are already fixed. - Sum complete rows only after applying the symmetry condition.

Solution

1. Symmetry gives bus-to-car \(=9\), bike-to-car \(=6\), and bike-to-bus \(=8\). 2. Row totals are \(55\), \(48\), and \(36\). 3. The grand total is \(55+48+36=139\).

Answer

<table> <tr><th>Last year / this year</th><th>car</th><th>bus</th><th>bike</th><th>Total</th></tr> <tr><td>car</td><td>\(40\)</td><td>\(9\)</td><td>\(6\)</td><td>\(55\)</td></tr> <tr><td>bus</td><td>\(9\)</td><td>\(31\)</td><td>\(8\)</td><td>\(48\)</td></tr> <tr><td>bike</td><td>\(6\)</td><td>\(8\)</td><td>\(22\)</td><td>\(36\)</td></tr> <tr><td>Total</td><td>\(55\)</td><td>\(48\)</td><td>\(36\)</td><td>\(139\)</td></tr> </table>

54714012

A survey began with \(240\) people. Twelve are missing the row variable, \(17\) are missing the column variable, and \(5\) are missing both variables. How many people can be included in a complete-case two-way table? Explain why simply subtracting \(12+17\) is incorrect.

Hints

- Think of the two missingness groups as overlapping sets. - Count the overlap only once. - The table includes only records with both categorical values observed.

Solution

1. The number missing at least one variable is \(12+17-5=24\), because the \(5\) missing both were counted twice. 2. The complete-case count is \(240-24=216\). 3. Subtracting \(12+17\) would remove the five doubly missing records twice.

Answer

The complete-case table contains \(216\) people. Subtracting \(12+17\) is wrong because the \(5\) people missing both variables would be subtracted twice.

54714112

A \(2\times3\) table has row totals \(50\) and \(70\), and column totals \(30\), \(40\), and \(50\). No interior cells are given. Show that the margins do not determine a unique table by constructing two different valid tables with these same margins.

Hints

- Choose a nonnegative first row that has the required row total. - The column totals then determine the second row. - A second valid first row is enough to disprove uniqueness.

Solution

1. One valid first row is \((10,15,25)\), giving second row \((20,25,25)\). 2. Another valid first row is \((20,10,20)\), giving second row \((10,30,30)\). 3. Both tables have row totals \(50,70\) and column totals \(30,40,50\), so the margins alone are insufficient.

Answer

Two valid tables are: <table> <tr><th></th><th>C1</th><th>C2</th><th>C3</th><th>Total</th></tr> <tr><td>R1</td><td>\(10\)</td><td>\(15\)</td><td>\(25\)</td><td>\(50\)</td></tr> <tr><td>R2</td><td>\(20\)</td><td>\(25\)</td><td>\(25\)</td><td>\(70\)</td></tr> <tr><td>Total</td><td>\(30\)</td><td>\(40\)</td><td>\(50\)</td><td>\(120\)</td></tr> </table> and <table> <tr><th></th><th>C1</th><th>C2</th><th>C3</th><th>Total</th></tr> <tr><td>R1</td><td>\(20\)</td><td>\(10\)</td><td>\(20\)</td><td>\(50\)</td></tr> <tr><td>R2</td><td>\(10\)</td><td>\(30\)</td><td>\(30\)</td><td>\(70\)</td></tr> <tr><td>Total</td><td>\(30\)</td><td>\(40\)</td><td>\(50\)</td><td>\(120\)</td></tr> </table>

54714212

A privacy report replaces one small count with “suppressed” but prints every other interior count and all margins. <table> <tr><th></th><th>requested aid</th><th>did not request aid</th><th>Total</th></tr> <tr><td>Program A</td><td>suppressed</td><td>\(28\)</td><td>\(31\)</td></tr> <tr><td>Program B</td><td>\(19\)</td><td>\(42\)</td><td>\(61\)</td></tr> <tr><td>Total</td><td>\(22\)</td><td>\(70\)</td><td>\(92\)</td></tr> </table> Show why the suppression does not protect the count. State one kind of additional suppression that would be needed to prevent direct reconstruction.

Hints

- Try reconstructing the hidden value from its row. - Check whether its column provides the same reconstruction. - Privacy suppression must remove every direct arithmetic path to the value.

Solution

1. The Program A row reveals the hidden count as \(31-28=3\). 2. The requested-aid column independently reveals it as \(22-19=3\). 3. Suppressing only the small cell is ineffective while a containing row or column total and the companion cell remain visible. 4. At least one additional linked value, such as the Program A row total or another cell needed to derive that total, must also be suppressed under a valid complementary-suppression plan.

Answer

The hidden count is \(3\). The printed margins disclose it, so linked totals or companion cells must also be suppressed according to a complementary-suppression rule.

54714912

A population has \(60\) people currently in category A and \(40\) in category B. One year later, \(50\) are in A and \(50\) are in B. A transition table would cross current category with later category. Let \(x\) be the number who moved from A to B and \(y\) the number who moved from B to A. What relationship between \(x\) and \(y\) is forced by the margins? Do the two sets of margins determine the full transition table?

Hints

- Express the later A total as its starting count minus departures plus arrivals. - Distinguish net change from the two directional flows. - Test whether more than one nonnegative pair fits the same margins.

Solution

1. Category A loses \(x\) people and gains \(y\), so its later total is \(60-x+y=50\). 2. Therefore \(x-y=10\). 3. Many nonnegative integer pairs satisfy this relationship, such as \((10, 0)\) and \((15, 5)\). 4. The before-and-after margins determine the net flow but not the two opposing flows, so the full transition table is not unique.

Answer

The margins force \(x-y=10\). They do not determine \(x\) and \(y\) separately, so the transition table is not unique.

54715212

Two completed \(2\times2\) tables have the same row totals and column totals: \( T_1=\begin{pmatrix}40&10\\10&40\end{pmatrix}, \qquad T_2=\begin{pmatrix}25&25\\25&25\end{pmatrix}. \) Explain what the shared margins do and do not determine. Which table shows a stronger association between the variables?

Hints

- Compare interior patterns after confirming the margins match. - Margins summarize each variable separately. - Association is visible in how cell proportions vary across rows.

Solution

1. Both tables have row totals \(50,50\) and column totals \(50,50\). 2. Margins determine one-variable distributions but not how categories pair within the table. 3. \(T_1\) concentrates observations on the diagonal, while \(T_2\) has the same distribution in both rows; \(T_1\) shows stronger association.

Answer

The margins are identical but the joint distributions differ. \(T_1\) shows stronger association; \(T_2\) shows no row-to-row change in the column distribution.

54715412

A data team must change the table below so the “yes” column total becomes \(52\), while the grand total and both row totals remain unchanged. <table> <tr><th></th><th>yes</th><th>no</th><th>Total</th></tr> <tr><td>Group 1</td><td>\(21\)</td><td>\(29\)</td><td>\(50\)</td></tr> <tr><td>Group 2</td><td>\(26\)</td><td>\(24\)</td><td>\(50\)</td></tr> <tr><td>Total</td><td>\(47\)</td><td>\(53\)</td><td>\(100\)</td></tr> </table> What is the minimum number of observations that must be reclassified, and what kind of move achieves it?

Hints

- Preserve row totals by moving observations within rows. - Each reclassification changes the target column total by one. - A lower bound comes from the required total change.

Solution

1. The yes total must increase by \(52-47=5\). 2. Reclassifying one observation from no to yes changes the yes total by \(1\) while preserving its row total and the grand total. 3. At least \(5\) observations must move, and moving any \(5\) no observations to yes within their existing rows achieves the target.

Answer

The minimum is \(5\) reclassifications from “no” to “yes,” with each observation staying in its original row.

54715512

A table counts registrations by workshop and payment status. One record was duplicated in “ceramics and paid,” while another record was omitted from “painting and unpaid.” Describe the net changes to the grand total, the four affected margins, and the two affected interior cells when both errors are corrected.

Hints

- Treat the duplicate correction and omission correction separately. - Combine their effects only after tracking each cell. - Equal removal and addition leave the total number of observations unchanged.

Solution

1. Remove one from ceramics/paid and add one to painting/unpaid. 2. The grand total is unchanged because one record is removed and one is added. 3. Ceramics row and paid column each decrease by \(1\); painting row and unpaid column each increase by \(1\).

Answer

Ceramics/paid decreases by \(1\), and painting/unpaid increases by \(1\). The grand total is unchanged. The ceramics row and paid column decrease by \(1\); the painting row and unpaid column increase by \(1\).

54715612

A two-way table of \(200\) records has row margins \(120,80\) and column margins \(90,110\). The raw file is lost. Can the original order of the \(200\) records be reconstructed from the table? Can the number of records in each category pair be reconstructed from the margins alone? Explain the two different information losses.

Hints

- Ask what information a frequency table keeps about individual records. - Then ask what margins keep about row-column pairings. - Identify whether one or several tables can share the same margins.

Solution

1. A frequency table discards record order, so the original sequence cannot be reconstructed even if every interior count were known. 2. The displayed margins also omit how row and column categories pair. 3. Many interior tables can share these margins, so the four joint counts are not determined either. 4. Record-order loss and joint-association loss are separate: the first occurs when microdata are aggregated, and the second occurs when only margins are retained.

Answer

Neither the original record order nor the interior joint counts can be reconstructed. Aggregation removes order, and retaining only margins also removes the pairing information.

54715712

Office A classifies contact method as phone, email, or web form. Office B classifies it only as phone or digital, where digital combines email and web form. Both offices cross contact method with resolved/not resolved. Can their published tables be added cell by cell without further information? Explain what harmonization is possible and what information is missing.

Hints

- Compare the category definitions before comparing the numbers. - Determine whether a detailed classification can be collapsed. - Ask whether a combined category can be disaggregated from totals alone.

Solution

1. The phone categories can be aligned directly. 2. Office A can be collapsed by adding email and web-form counts to create a digital category. 3. Office B cannot be expanded into separate email and web-form categories because its published table does not contain that split. 4. The tables can be combined only at the coarser phone/digital level, assuming all other definitions and observational units match.

Answer

They cannot be added in their published forms. Collapse Office A to phone/digital, then combine at that coarser level; Office B does not contain enough information to recover separate email and web-form counts.

54711812

For an \(r\times c\) two-way table, all row and column totals are fixed. How many interior cell counts can be chosen freely before the remaining cells are forced? Apply the result to a \(3\times4\) table.

Hints

- Once enough cells are chosen, margins determine the rest. - Avoid counting the final row and final column as independently free. - Test the formula on a small \(2\times2\) table.

Solution

1. Choose the cells in the first \(r-1\) rows and first \(c-1\) columns freely. 2. The remaining cell in each of those rows is forced by its row total; the last row is then forced by column totals. 3. The number of free cells is \((r-1)(c-1)\). For \(3\times4\), this is \(2\cdot3=6\).

Answer

A table with fixed margins has \((r-1)(c-1)\) free interior counts. A \(3\times4\) table has \(6\).

54712812

A report claims that exactly two interior entries in the table below were swapped and that every displayed row and column total is correct. <table> <tr><th></th><th>approved</th><th>not approved</th><th>Total</th></tr> <tr><td>online application</td><td>\(44\)</td><td>\(18\)</td><td>\(68\)</td></tr> <tr><td>paper application</td><td>\(24\)</td><td>\(14\)</td><td>\(32\)</td></tr> <tr><td>Total</td><td>\(58\)</td><td>\(42\)</td><td>\(100\)</td></tr> </table> Determine whether the claim is possible. Support your conclusion by testing the swap that fixes the row totals.

Hints

- Test candidate swaps against both row totals and column totals. - Fixing one set of margins is not enough. - A contradiction may mean the stated error description is incomplete.

Solution

1. The online row needs counts summing to \(68\). Swapping \(18\) and \(24\) gives \(44+24=68\). 2. The paper row then gives \(18+14=32\). 3. The corrected column totals are \(44+18=62\) and \(24+14=38\), not \(58\) and \(42\). Therefore at least one additional entry or margin is wrong, so the claim that all displayed margins are correct is impossible.

Answer

No swap of only two interior entries can make the displayed row and column totals all correct. Swapping \(18\) and \(24\) fixes the row totals but produces column totals \(62\) and \(38\), not \(58\) and \(42\). The report is internally contradictory.

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