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Joint, marginal, and conditional relative frequency

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Difficulty: 1 – 5

54718112

A headline says, “\(30\%\) of all survey respondents were satisfied customers who used chat.” Which relative frequency is this: joint, marginal, or conditional? What denominator does the statement require?

Hints

- Identify whether one category or a category pair is being described. - Notice the phrase “of all survey respondents.” - Conditional statements restrict the denominator to a subgroup.

Solution

1. The statement describes the intersection of two categories: satisfied and used chat. 2. It is a joint relative frequency. 3. Its denominator is the grand total of all survey respondents.

Answer

It is a joint relative frequency, using all survey respondents as the denominator.

55619012

The table summarizes whether students in a school club also have a part-time job. <table><tr><th></th><th>Part-time job</th><th>No part-time job</th><th>Total</th></tr><tr><td>Club member</td><td>\(18\)</td><td>\(42\)</td><td>\(60\)</td></tr><tr><td>Not a club member</td><td>\(30\)</td><td>\(30\)</td><td>\(60\)</td></tr><tr><td>Total</td><td>\(48\)</td><td>\(72\)</td><td>\(120\)</td></tr></table> What denominator should be used for the conditional relative frequency of having a part-time job among club members, and what is that conditional relative frequency?

Hints

- The words “among club members” identify the group that forms the denominator. - Use the total for that restricted group rather than the grand total. - The numerator is the count that satisfies both descriptions.

Solution

1. The condition is being a club member, so restrict the denominator to the club-member row total, \(60\). 2. Of those \(60\) students, \(18\) have a part-time job. 3. The conditional relative frequency is \(\frac{18}{60}=0.30\).

Answer

The denominator is \(60\), and the conditional relative frequency is \(0.30\), or \(30\%\).

54716112

The table summarizes science fair projects. <table> <tr><th></th><th>used a physical prototype</th><th>did not use a physical prototype</th><th>Total</th></tr> <tr><td>team project</td><td>\(20\)</td><td>\(15\)</td><td>\(35\)</td></tr> <tr><td>individual project</td><td>\(12\)</td><td>\(18\)</td><td>\(30\)</td></tr> <tr><td>Total</td><td>\(32\)</td><td>\(33\)</td><td>\(65\)</td></tr> </table> Find the joint relative frequency of “team project and used a physical prototype,” the marginal relative frequency of “team project,” and the conditional relative frequency of “used a physical prototype” among those who are “team project.” Give decimals to three places.

Hints

- A joint relative frequency uses the grand total as its denominator. - A marginal relative frequency uses the corresponding row or column total count over the grand total. - A conditional relative frequency uses only the stated subgroup as its denominator.

Solution

1. Joint relative frequency: \(20/65\approx 0.308\). 2. Marginal relative frequency for “team project”: \(35/65\approx 0.538\). 3. Conditional relative frequency: \(20/35\approx 0.571\).

Answer

Joint: \(0.308\). Marginal: \(0.538\). Conditional: \(0.571\).

54716912

A joint relative-frequency table has columns \(C_1\) and \(C_2\). The column margins are \(P(C_1)=0.35\) and \(P(C_2)=0.65\). Within column \(C_1\), \(60\%\) of observations are in row \(R_1\); within column \(C_2\), \(30\%\) are in row \(R_1\). Construct the complete joint relative-frequency table, including row and column margins.

Hints

- A within-column percentage must be multiplied by that column’s marginal share. - Fill the complementary row within each column before calculating row margins. - Check that all four joint entries add to \(1\).

Solution

1. In column \(C_1\), \(P(R_1\cap C_1)=0.35\cdot0.60=0.21\), so \(P(R_2\cap C_1)=0.35-0.21=0.14\). 2. In column \(C_2\), \(P(R_1\cap C_2)=0.65\cdot0.30=0.195\), so \(P(R_2\cap C_2)=0.65-0.195=0.455\). 3. Row margins are \(0.21+0.195=0.405\) and \(0.14+0.455=0.595\).

Answer

<table> <tr><th></th><th>\(C_1\)</th><th>\(C_2\)</th><th>Total</th></tr> <tr><td>\(R_1\)</td><td>\(0.210\)</td><td>\(0.195\)</td><td>\(0.405\)</td></tr> <tr><td>\(R_2\)</td><td>\(0.140\)</td><td>\(0.455\)</td><td>\(0.595\)</td></tr> <tr><td>Total</td><td>\(0.350\)</td><td>\(0.650\)</td><td>\(1.000\)</td></tr> </table>

54717112

A table includes an “unknown” response category. <table> <tr><th></th><th>yes</th><th>no</th><th>unknown</th></tr> <tr><td>Group A</td><td>\(30\)</td><td>\(10\)</td><td>\(10\)</td></tr> <tr><td>Group B</td><td>\(20\)</td><td>\(20\)</td><td>\(10\)</td></tr> </table> Remove the unknown responses and construct the complete-case joint relative-frequency table for yes/no by group. Explain why the original grand total cannot be used as the new denominator.

Hints

- Identify which records remain after the exclusion. - Recompute the denominator before converting counts to relative frequencies. - Check that all retained joint frequencies sum to \(1\).

Solution

1. The complete cases total \(30+10+20+20=80\). 2. Divide the four retained counts by \(80\), giving joint frequencies \(0.375,0.125,0.250,\) and \(0.250\). 3. Row margins are \(0.50,0.50\), and column margins are \(0.625,0.375\). 4. The original total \(100\) includes records excluded from the complete-case table.

Answer

The complete-case joint table is \(\begin{pmatrix}0.375&0.125\\0.250&0.250\end{pmatrix}\), with row margins \(0.50,0.50\) and column margins \(0.625,0.375\). The new denominator is \(80\), not \(100\), because the unknown-response records are excluded.

54717612

Two row groups have the same conditional distribution: <table> <tr><th></th><th>choice X</th><th>choice Y</th><th>choice Z</th></tr> <tr><td>Group 1</td><td>\(0.20\)</td><td>\(0.50\)</td><td>\(0.30\)</td></tr> <tr><td>Group 2</td><td>\(0.20\)</td><td>\(0.50\)</td><td>\(0.30\)</td></tr> </table> The entries are row conditional relative frequencies. What descriptive conclusion can be made about the association between group and choice? Why are row totals unnecessary for this comparison?

Hints

- Compare corresponding percentages across the two rows. - Association appears as a change in the conditional distribution. - Normalization removes the effect of unequal group sizes.

Solution

1. Each choice has the same conditional relative frequency in both groups. 2. Therefore the displayed conditional distributions show no descriptive association between group and choice. 3. Row totals are unnecessary because the comparison is already normalized within each group.

Answer

The two conditional distributions are identical, so the table shows no descriptive association between group and choice. Row totals are not needed because each row is already expressed relative to its own group.

54717712

Convert the following count table to column conditional relative frequencies. <table> <tr><th></th><th>completed survey</th><th>did not complete</th></tr> <tr><td>email invitation</td><td>\(72\)</td><td>\(18\)</td></tr> <tr><td>text invitation</td><td>\(48\)</td><td>\(42\)</td></tr> </table> Each column must sum to \(1\).

Hints

- Find each column total before dividing. - Use a different denominator for each column. - Check each column, not each row, for a sum of \(1\).

Solution

1. The completed-survey column total is \(120\), giving \(72/120=0.60\) and \(48/120=0.40\). 2. The did-not-complete column total is \(60\), giving \(18/60=0.30\) and \(42/60=0.70\). 3. Each column sums to \(1\).

Answer

<table> <tr><th></th><th>completed survey</th><th>did not complete</th></tr> <tr><td>email invitation</td><td>\(0.60\)</td><td>\(0.30\)</td></tr> <tr><td>text invitation</td><td>\(0.40\)</td><td>\(0.70\)</td></tr> </table>

54717812

In a study, \(18\%\) of remote employees and \(12\%\) of office employees miss a weekly deadline. Compute the relative risk of missing a deadline for remote employees compared with office employees. Interpret the result.

Hints

- Compare the two conditional rates multiplicatively. - Put the named comparison group in the numerator. - Interpret the quotient as a rate multiple, not a percentage-point difference.

Solution

1. The relative risk is \(\frac{0.18}{0.12}=1.5\). 2. The remote group’s conditional rate is \(1.5\) times the office group’s conditional rate. 3. This is a descriptive comparison and does not by itself establish causation.

Answer

The relative risk is \(1.5\). Remote employees have \(1.5\) times the observed missed-deadline rate of office employees.

54718212

A \(3\times2\) joint relative-frequency table has cells \( \begin{pmatrix} 0.12&0.18\\ 0.20&0.10\\ 0.08&0.32 \end{pmatrix}. \) Find the relative frequency of the compound event “row 1 or column 1.” Show which joint cells belong to the event and avoid double-counting their overlap.

Hints

- Mark the cells satisfying each part of the “or” statement. - Identify any cell selected twice. - Sum the distinct joint cells exactly once.

Solution

1. Row 1 contains \(0.12+0.18=0.30\). 2. Column 1 contains \(0.12+0.20+0.08=0.40\). 3. The upper-left cell \(0.12\) lies in both descriptions. 4. The union frequency is \(0.30+0.40-0.12=0.58\), equivalently the sum of cells \(0.12+0.18+0.20+0.08\).

Answer

\(0.58\).

54718412

A sample has \(400\) observations and the joint relative-frequency table: <table> <tr><th></th><th>yes</th><th>no</th></tr> <tr><td>Group A</td><td>\(0.18\)</td><td>\(0.27\)</td></tr> <tr><td>Group B</td><td>\(0.32\)</td><td>\(0.23\)</td></tr> </table> Convert every joint relative frequency to a count and add all margins.

Hints

- Every joint frequency uses the same grand-total denominator. - Multiply rather than divide when recovering counts. - Verify that the four counts sum to the sample size.

Solution

1. Multiply each joint relative frequency by \(400\): \(72\), \(108\), \(128\), and \(92\). 2. Row totals are \(180\) and \(220\). 3. Column totals are \(200\) and \(200\), with grand total \(400\).

Answer

<table> <tr><th></th><th>yes</th><th>no</th><th>Total</th></tr> <tr><td>Group A</td><td>\(72\)</td><td>\(108\)</td><td>\(180\)</td></tr> <tr><td>Group B</td><td>\(128\)</td><td>\(92\)</td><td>\(220\)</td></tr> <tr><td>Total</td><td>\(200\)</td><td>\(200\)</td><td>\(400\)</td></tr> </table>

54718512

A table displays these percentages: <table> <tr><th></th><th>uses app</th><th>does not use app</th></tr> <tr><td>under 30</td><td>\(40\%\)</td><td>\(60\%\)</td></tr> <tr><td>30 or older</td><td>\(25\%\)</td><td>\(75\%\)</td></tr> </table> Can these entries be interpreted as joint relative frequencies? What interpretation is supported by the row sums?

Hints

- Check what total the full table has. - Then check each row and each column separately. - The location of the \(100\%\) sum identifies the denominator structure.

Solution

1. The four entries sum to \(200\%\), so they cannot all be joint relative frequencies for one sample. 2. Each row sums to \(100\%\). 3. The supported interpretation is row conditional relative frequencies: app use within each age group.

Answer

No. They are row conditional relative frequencies, because each row—not the whole table—sums to \(100\%\).

54718812

Convert this count table to a complete joint relative-frequency table. <table> <tr><th></th><th>morning</th><th>evening</th></tr> <tr><td>weekday</td><td>\(45\)</td><td>\(30\)</td></tr> <tr><td>weekend</td><td>\(15\)</td><td>\(10\)</td></tr> </table> Include margins and express every value as a decimal.

Hints

- Find the grand total before converting any cell. - Joint relative frequencies all use that one denominator. - Margins can be found by adding relative frequencies or converting count totals.

Solution

1. The grand total is \(100\). 2. Divide every count by \(100\), giving joint cells \(0.45,0.30,0.15,0.10\). 3. Row margins are \(0.75,0.25\); column margins are \(0.60,0.40\).

Answer

<table> <tr><th></th><th>morning</th><th>evening</th><th>Total</th></tr> <tr><td>weekday</td><td>\(0.45\)</td><td>\(0.30\)</td><td>\(0.75\)</td></tr> <tr><td>weekend</td><td>\(0.15\)</td><td>\(0.10\)</td><td>\(0.25\)</td></tr> <tr><td>Total</td><td>\(0.60\)</td><td>\(0.40\)</td><td>\(1.00\)</td></tr> </table>

54718912

A certification program compares pass rates for two preparation formats in two years. <table> <tr><th></th><th>Self-paced</th><th>Instructor-led</th></tr> <tr><td>Year 1</td><td>\(36/60\)</td><td>\(28/40\)</td></tr> <tr><td>Year 2</td><td>\(63/90\)</td><td>\(36/60\)</td></tr> </table> For each year, compute the self-paced pass rate minus the instructor-led pass rate. Then describe how the conditional-rate gap changed from Year 1 to Year 2.

Hints

- Form each rate within its own year-and-format group. - Keep the subtraction order the same in both years. - Compare both the magnitude and the sign of the two gaps.

Solution

1. In Year 1, the rates are \(0.60\) and \(0.70\), so the gap is \(0.60-0.70=-0.10\). 2. In Year 2, the rates are \(0.70\) and \(0.60\), so the gap is \(0.70-0.60=0.10\). 3. The gap increased by \(0.20\), or \(20\) percentage points, and its direction reversed.

Answer

Year 1 gap: \(-10\) percentage points. Year 2 gap: \(+10\) percentage points. The self-paced-minus-instructor-led gap increased by \(20\) percentage points and reversed direction.

54719612

In a table, \(P(\text{online}\mid\text{renewed})=0.80\). A manager restates this as “\(80\%\) of online customers renewed.” Explain why the restatement is not justified and name the relative frequency it actually describes.

Hints

- Read the category after the conditioning bar as the reference group. - Reversing the order changes the denominator. - The same joint cell can yield two different conditional values.

Solution

1. The given value uses renewed customers as the denominator. 2. It says \(80\%\) of renewed customers were online. 3. “Of online customers, renewed” is the reversed conditional frequency and generally has a different denominator and value.

Answer

The statement means \(80\%\) of renewed customers were online. It does not give \(P(\text{renewed}\mid\text{online})\).

54720012

A row-conditional relative-frequency table compares two groups across three outcomes. <table> <tr><th></th><th>Outcome X</th><th>Outcome Y</th><th>Outcome Z</th></tr> <tr><td>Group 1</td><td>\(0.20\)</td><td>\(0.50\)</td><td>\(0.30\)</td></tr> <tr><td>Group 2</td><td>\(0.35\)</td><td>\(0.40\)</td><td>\(0.25\)</td></tr> </table> For each outcome, find the absolute difference between the two conditional frequencies. Which outcome contributes the largest visible difference between the group profiles? Explain why unequal group sizes would not change these three within-row comparisons.

Hints

- Compare corresponding entries rather than raw group counts. - Use absolute differences so direction does not affect the size of a gap. - Recall what denominator is used in a row-conditional table.

Solution

1. The absolute differences are \(|0.20-0.35|=0.15\), \(|0.50-0.40|=0.10\), and \(|0.30-0.25|=0.05\). 2. Outcome X has the largest conditional-frequency difference. 3. Each row is already normalized within its own group, so changing the number of observations in a group would change joint frequencies but not these within-row rates.

Answer

The absolute differences are \(0.15,0.10,\) and \(0.05\). Outcome X differs most. Unequal row sizes do not alter row-conditional comparisons.

55619112

Panel a) shows morning-shift employees and panel b) shows evening-shift employees. In each panel, R means remote and O means on-site. a) Find the conditional relative frequency of remote work within each shift. b) Which shift has the higher remote-work conditional relative frequency, and by how many percentage points?

Figure for problem 556191

Hints

- Treat each panel as one conditioning group. - Within a panel, the denominator is the sum of its R and O counts. - Compare the two resulting proportions with a percentage-point difference.

Solution

1. In panel a), the morning shift has \(30\) remote and \(50\) on-site employees, for \(80\) total. Its remote conditional relative frequency is \(\frac{30}{80}=0.375\). 2. In panel b), the evening shift has \(40\) remote and \(60\) on-site employees, for \(100\) total. Its remote conditional relative frequency is \(\frac{40}{100}=0.40\). 3. The evening-shift rate exceeds the morning-shift rate by \(0.40-0.375=0.025\), or \(2.5\) percentage points.

Answer

a) Morning: \(0.375\); evening: \(0.40\). b) The evening shift is higher by \(2.5\) percentage points.

53083912

A simplified survival table shows how many men from an initial group of \(100{,}000\) live births reached selected ages in two different years. <table> <tr><th>Age in years</th><th>\(1960\)</th><th>\(2020\)</th></tr> <tr><td>\(0\)</td><td>\(100{,}000\)</td><td>\(100{,}000\)</td></tr> <tr><td>\(40\)</td><td>\(92{,}400\)</td><td>\(98{,}800\)</td></tr> <tr><td>\(70\)</td><td>\(52{,}100\)</td><td>\(84{,}500\)</td></tr> <tr><td>\(85\)</td><td>\(12{,}300\)</td><td>\(42{,}600\)</td></tr> </table> a) For each year, estimate the probability that a man who has reached age \(40\) will reach age \(70\). b) For each year, estimate the probability that a man who has reached age \(40\) will reach at least age \(85\). c) Explain why these are empirical probabilities rather than probabilities based on equally likely outcomes.

Hints

- For each year, use the number reaching the later age as the numerator. - Because the person has already reached age \(40\), use the age-\(40\) count as the denominator. - Distinguish probabilities estimated from data from probabilities derived from equally likely theoretical outcomes. - Identify the population represented by each column.

Solution

1. For 1960, the conditional proportion reaching age \(70\) among those reaching age \(40\) is \(\frac{52100}{92400}\approx 0.5639\). For 2020, it is \(\frac{84500}{98800}\approx 0.8553\). 2. For 1960, the conditional proportion reaching age \(85\) among those reaching age \(40\) is \(\frac{12300}{92400}\approx 0.1331\). For 2020, it is \(\frac{42600}{98800}\approx 0.4312\). 3. The values are estimated from observed relative frequencies in population data. They are not derived from a sample space in which all elementary outcomes are assumed to be equally likely.

Answer

a) 1960: \(0.5639\), or about \(56.39\%\). 2020: \(0.8553\), or about \(85.53\%\). b) 1960: \(0.1331\), or about \(13.31\%\). 2020: \(0.4312\), or about \(43.12\%\). c) They are empirical probabilities because they are based on observed relative frequencies.

53084012

An electronics company tracks the reliability of \(50{,}000\) smartphones. The table shows how many phones are still fully functional after selected numbers of months. <table> <tr><th>Months in use</th><th>Functional phones</th></tr> <tr><td>\(0\)</td><td>\(50{,}000\)</td></tr> <tr><td>\(12\)</td><td>\(48{,}500\)</td></tr> <tr><td>\(24\)</td><td>\(42{,}000\)</td></tr> <tr><td>\(36\)</td><td>\(25{,}000\)</td></tr> <tr><td>\(48\)</td><td>\(8000\)</td></tr> </table> a) Estimate the probability that a phone still working after \(12\) months is also working after \(36\) months. b) Estimate the probability that a phone working after \(24\) months fails during the third year, between months \(24\) and \(36\). c) Explain why failure time cannot be modeled here by assuming all possible months of failure are equally likely.

Hints

- In a conditional probability, use only the phones that satisfy the stated starting condition. - Find the number of failures during a time interval by subtracting the later survivor count from the earlier survivor count. - Consider whether wear makes failure equally likely at every age.

Solution

1. For part a, restrict the group to the \(48{,}500\) phones working after \(12\) months. Of those, \(25{,}000\) are working after \(36\) months, so the estimate is \(\frac{25000}{48500}\approx 0.5155\). 2. Between months \(24\) and \(36\), \(42{,}000-25{,}000=17{,}000\) phones fail. Among phones working at month \(24\), the estimated conditional probability is \(\frac{17000}{42000}\approx 0.4048\). 3. Failure rates can change with age and wear. The data do not support treating failure in each month as an equally likely outcome.

Answer

a) \(\frac{25000}{48500}\approx 0.5155\), or about \(51.55\%\). b) \(\frac{17000}{42000}\approx 0.4048\), or about \(40.48\%\). c) The probability of failure changes over time, so the possible failure months are not equally likely.

53095112

A survival table gives the number \(l_x\) of people from an initial population who reach age \(x\). <table> <tr><th>Age \(x\)</th><th>Number \(l_x\) reaching age \(x\)</th></tr> <tr><td>\(25\)</td><td>\(98{,}412\)</td></tr> <tr><td>\(45\)</td><td>\(95{,}120\)</td></tr> <tr><td>\(65\)</td><td>\(80{,}455\)</td></tr> </table> Estimate each probability. 1. A person who is currently \(25\) reaches age \(65\). 2. Two people who are currently \(45\) both reach age \(65\). Assume their survival events are independent. 3. A person who is currently \(25\) dies after reaching age \(45\) but before reaching age \(65\).

Hints

- Estimate a conditional probability by dividing the later-age count by the count for the current age. - For two independent events that must both occur, multiply their probabilities. - To count deaths between two ages, subtract the later survivor count from the earlier survivor count. - Use the population that has already reached the stated current age as the denominator.

Solution

1. Among people reaching age \(25\), the estimated proportion reaching age \(65\) is \(\frac{80455}{98412}\approx 0.8175\). 2. For one person currently age \(45\), the estimated probability of reaching age \(65\) is \(\frac{80455}{95120}\approx 0.8458\). Under the stated independence assumption, the probability that both people reach age \(65\) is \(\left(\frac{80455}{95120}\right)^2\approx 0.7154\). 3. The number reaching age \(45\) but not age \(65\) is \(95{,}120-80{,}455=14{,}665\). Relative to those reaching age \(25\), the estimated probability is \(\frac{14665}{98412}\approx 0.1490\).

Answer

1. \(0.8175\), or about \(81.75\%\). 2. \(0.7154\), or about \(71.54\%\). 3. \(0.1490\), or about \(14.90\%\).

54716212

Two campuses used the same two categorical variables. Their joint relative-frequency tables are shown below. Campus A, \(n=80\): <table> <tr><th></th><th>Completed</th><th>Did not complete</th></tr> <tr><td>Online</td><td>\(0.30\)</td><td>\(0.20\)</td></tr> <tr><td>In person</td><td>\(0.10\)</td><td>\(0.40\)</td></tr> </table> Campus B, \(n=220\): <table> <tr><th></th><th>Completed</th><th>Did not complete</th></tr> <tr><td>Online</td><td>\(0.10\)</td><td>\(0.30\)</td></tr> <tr><td>In person</td><td>\(0.25\)</td><td>\(0.35\)</td></tr> </table> Construct the pooled joint relative-frequency table for all \(300\) participants. Explain why averaging corresponding cells of the two displayed tables would be incorrect.

Hints

- Recover each campus’s cell counts before combining the data. - A pooled relative frequency uses the combined grand total. - Ask what weighting an ordinary average of the two table entries would impose.

Solution

1. Convert Campus A to counts: \(\begin{pmatrix}24&16\\8&32\end{pmatrix}\). 2. Convert Campus B to counts: \(\begin{pmatrix}22&66\\55&77\end{pmatrix}\). 3. Add corresponding counts to obtain \(\begin{pmatrix}46&82\\63&109\end{pmatrix}\) out of \(300\). 4. Divide each count by \(300\): the pooled joint frequencies are approximately \(\begin{pmatrix}0.153&0.273\\0.210&0.363\end{pmatrix}\). 5. A cellwise average would give equal weight to campuses of sizes \(80\) and \(220\), so it would not represent the combined population.

Answer

<table> <tr><th></th><th>Completed</th><th>Did not complete</th></tr> <tr><td>Online</td><td>\(0.153\)</td><td>\(0.273\)</td></tr> <tr><td>In person</td><td>\(0.210\)</td><td>\(0.363\)</td></tr> </table> The entries are rounded. Pool counts first because the campuses have unequal sample sizes.

54716312

For recreation center members, use the table below. <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>\(27\)</td><td>\(63\)</td></tr> <tr><td>weekend visit</td><td>\(32\)</td><td>\(42\)</td><td>\(74\)</td></tr> <tr><td>Total</td><td>\(68\)</td><td>\(69\)</td><td>\(137\)</td></tr> </table> A student reports \(42/69\) as the joint relative frequency of “weekend visit and did not use the pool.” Diagnose the error, then give the correct joint relative frequency and the conditional relative frequency of “weekend visit” given “did not use the pool.”

Hints

- Distinguish the whole sample from a row or column subgroup. - Use counts from the same reference group when comparing percentages. - Separate numerical comparison from contextual interpretation.

Solution

1. A joint relative frequency uses the grand total, so the correct joint value is \(42/137\approx 0.307\). 2. Dividing by the “did not use the pool” column total gives a conditional relative frequency: \(42/69\approx 0.609\).

Answer

Correct joint relative frequency: \(0.307\). The student's value is conditional, and \(P(\text{weekend visit}\mid\text{did not use the pool})\approx 0.609\).

54716412

Two studies have the following joint relative-frequency tables. Both use variables \(X\in\{X_1,X_2\}\) and \(Y\in\{Y_1,Y_2\}\). Study A: <table> <tr><th></th><th>\(Y_1\)</th><th>\(Y_2\)</th><th>Total</th></tr> <tr><td>\(X_1\)</td><td>\(0.25\)</td><td>\(0.25\)</td><td>\(0.50\)</td></tr> <tr><td>\(X_2\)</td><td>\(0.25\)</td><td>\(0.25\)</td><td>\(0.50\)</td></tr> <tr><td>Total</td><td>\(0.50\)</td><td>\(0.50\)</td><td>\(1.00\)</td></tr> </table> Study B: <table> <tr><th></th><th>\(Y_1\)</th><th>\(Y_2\)</th><th>Total</th></tr> <tr><td>\(X_1\)</td><td>\(0.45\)</td><td>\(0.05\)</td><td>\(0.50\)</td></tr> <tr><td>\(X_2\)</td><td>\(0.05\)</td><td>\(0.45\)</td><td>\(0.50\)</td></tr> <tr><td>Total</td><td>\(0.50\)</td><td>\(0.50\)</td><td>\(1.00\)</td></tr> </table> Verify that the two studies have identical marginal relative frequencies. Then compare \(P(Y_1\mid X_1)\) and explain what the comparison shows about what margins can conceal.

Hints

- Compare row totals and column totals before comparing interior cells. - A conditional frequency uses one row as its reference group. - Consider whether margins determine how probability is distributed inside the table.

Solution

1. In both studies, each row margin and each column margin is \(0.50\). 2. Study A has \(P(Y_1\mid X_1)=\frac{0.25}{0.50}=0.50\). 3. Study B has \(P(Y_1\mid X_1)=\frac{0.45}{0.50}=0.90\). 4. Identical margins describe the separate distributions of \(X\) and \(Y\), but they do not determine how the variables are associated within the table.

Answer

Both studies have row and column margins \((0.50,0.50)\). However, \(P(Y_1\mid X_1)=0.50\) in Study A and \(0.90\) in Study B. Margins alone can conceal very different joint relationships.

54716512

A count table initially has rows \(A=(20,30)\) and \(B=(30,20)\). Every record in row A is duplicated once, while row B is unchanged. Compare the row-conditional distribution for A, the row marginal for A, and the first-column marginal before and after duplication.

Hints

- Separate within-row denominators from the grand-total denominator. - Duplicating every record in one row preserves that row’s internal proportions. - Recompute margins using the new grand total.

Solution

1. Initially, A’s conditional distribution is \((20/50,30/50)=(0.40,0.60)\), its row margin is \(0.50\), and the first-column margin is \(50/100=0.50\). 2. After duplication, A becomes \((40,60)\), B remains \((30,20)\), and the new total is \(150\). 3. A’s conditional distribution remains \((0.40,0.60)\). 4. A’s row margin becomes \(100/150=2/3\), and the first-column margin becomes \(70/150=7/15\approx0.467\).

Answer

A’s row-conditional distribution remains \((0.40,0.60)\). Its row margin changes from \(0.50\) to \(\frac23\), and the first-column margin changes from \(0.50\) to \(\frac{7}{15}\approx0.467\).

54716612

A survey of \(200\) residents records location and support for a transit proposal. Some responses are unresolved. <table> <tr><th></th><th>Support</th><th>Oppose</th><th>Unresolved</th><th>Total</th></tr> <tr><td>Urban</td><td>\(72\)</td><td>\(36\)</td><td>\(12\)</td><td>\(120\)</td></tr> <tr><td>Rural</td><td>\(28\)</td><td>\(44\)</td><td>\(8\)</td><td>\(80\)</td></tr> </table> Assume every unresolved response will eventually be classified as either support or oppose. a) Find the possible interval for the overall support relative frequency. b) Find the possible interval for the conditional support relative frequency among urban residents. c) Explain why reporting either unresolved group as all support or all oppose gives a bound rather than an estimate.

Hints

- Keep unresolved observations in the denominator because they remain part of the survey. - To obtain an extreme rate, place every unresolved case on the same side. - Distinguish a range of logically possible values from a prediction.

Solution

1. There are \(100\) confirmed supporters and \(20\) unresolved responses. The final support count can range from \(100\) to \(120\), so the overall relative frequency lies in \([0.50,0.60]\). 2. Among urban residents, the support count can range from \(72\) to \(84\) out of \(120\). The conditional interval is \([0.60,0.70]\). 3. Assigning every unresolved case to one endpoint gives the smallest or largest possible value. It does not describe a typical or most likely allocation.

Answer

a) Overall support relative frequency: \([0.50,0.60]\). b) Urban conditional support relative frequency: \([0.60,0.70]\). c) The endpoint assignments are worst-case bounds, not estimates of how unresolved responses will actually split.

54716712

A training program has two departments. Department A has \(20\) employees and an \(80\%\) completion rate. Department B has \(180\) employees and a \(40\%\) completion rate. a) Find the unweighted mean of the two department completion rates. b) Find the completion relative frequency among all \(200\) employees. c) Explain what population each result describes and why neither calculation is automatically “the correct average.”

Hints

- Decide whether departments or employees are the units receiving equal weight. - Convert each department rate to a count before pooling employees. - Interpret a numerical average by naming its reference population.

Solution

1. The unweighted department mean is \((0.80+0.40)/2=0.60\). 2. Completed counts are \(0.80\cdot20=16\) and \(0.40\cdot180=72\). The employee-level rate is \((16+72)/200=0.44\). 3. The value \(0.60\) treats the two departments as equally weighted units. The value \(0.44\) treats each employee as equally weighted. The appropriate result depends on the stated reference population.

Answer

a) Unweighted department mean: \(0.60\). b) Overall employee completion relative frequency: \(0.44\). c) The first describes an equally weighted “typical department”; the second describes a randomly selected employee.

54716812

A report gives \(P(A)=0.40\), \(P(B)=0.50\), and \(P(A\cap B)=0.18\). It also states that \(P(B\mid A)=0.60\). a) Check whether all four values can describe the same two-way relative-frequency table. b) If the first three values are correct, find the correct value of \(P(B\mid A)\). c) Construct the corresponding \(2\times2\) joint relative-frequency table.

Hints

- A conditional relative frequency, its conditioning margin, and the corresponding joint cell must agree. - Use the given margins to determine the other cells after checking the disputed value. - Verify every row sum, column sum, and the grand total.

Solution

1. The first three values imply \(P(B\mid A)=P(A\cap B)/P(A)=0.18/0.40=0.45\), not \(0.60\). The four claims are inconsistent. 2. The remaining cells are \(P(A\cap B^c)=0.40-0.18=0.22\), \(P(A^c\cap B)=0.50-0.18=0.32\), and \(P(A^c\cap B^c)=1-(0.18+0.22+0.32)=0.28\). 3. These cells have the stated margins and total \(1\).

Answer

a) No; the reported conditional value is inconsistent. b) \(P(B\mid A)=0.45\). c) <table> <tr><th></th><th>\(B\)</th><th>\(B^c\)</th><th>Total</th></tr> <tr><td>\(A\)</td><td>\(0.18\)</td><td>\(0.22\)</td><td>\(0.40\)</td></tr> <tr><td>\(A^c\)</td><td>\(0.32\)</td><td>\(0.28\)</td><td>\(0.60\)</td></tr> <tr><td>Total</td><td>\(0.50\)</td><td>\(0.50\)</td><td>\(1.00\)</td></tr> </table>

54717012

A \(2\times2\) joint relative-frequency table is rounded to the nearest hundredth. The four displayed cells are \(0.24,0.26,0.25,\) and \(0.24\), summing to \(0.99\). Can rounding alone explain the missing \(0.01\)? Give the possible total range of the unrounded cells.

Hints

- Bound the rounding error for one cell. - Accumulate the possible error across all four cells. - Compare the required correction with that total bound.

Solution

1. Under ordinary rounding to the nearest hundredth, the two cells displayed as \(0.24\) can each lie in \([0.235,0.245)\), the cell displayed as \(0.26\) can lie in \([0.255,0.265)\), and the cell displayed as \(0.25\) can lie in \([0.245,0.255)\). 2. Adding the interval endpoints gives a possible unrounded total in \([0.970,1.010)\). 3. Since \(1.000\) lies in this interval, rounding alone can explain the displayed total of \(0.99\).

Answer

Yes. The possible unrounded total is \([0.970,1.010)\), so a true total of \(1.000\) is compatible with the rounded cells.

54717212

Two departments report internship placements. Department A: \(36\) placements among \(60\) applicants. Department B: \(45\) placements among \(90\) applicants. Which department has the larger joint relative frequency in the combined sample? Which has the larger conditional placement rate within its own applicants? Explain why the rankings need not match.

Hints

- Joint frequencies use the combined grand total. - Conditional rates use each department’s own applicant total. - Unequal group sizes can reverse the ranking.

Solution

1. In the combined sample of \(150\), joint placement frequencies are \(\frac{36}{150}=0.24\) for A and \(\frac{45}{150}=0.30\) for B. 2. Conditional placement rates are \(\frac{36}{60}=0.60\) for A and \(\frac{45}{90}=0.50\) for B. 3. B has the larger joint share because it has more applicants; A has the larger within-department rate.

Answer

Department B has the larger joint relative frequency, \(0.30\) versus \(0.24\). Department A has the larger conditional placement rate, \(0.60\) versus \(0.50\).

54717312

Let a cell have positive joint relative frequency \(j=P(A\cap B)\), and let the conditioning margin be \(P(B)=b\). Compare \(P(A\mid B)=j/b\) with the joint relative frequency \(j\). When are they equal, and when is the conditional value strictly larger?

Hints

- Compare the two expressions through their denominators. - A conditioning margin cannot exceed \(1\). - Identify when dividing by the margin changes the value.

Solution

1. A probability margin satisfies \(0<b\le1\). 2. Dividing positive \(j\) by \(b\le1\) gives \(j/b\ge j\). 3. Equality holds when \(b=1\). 4. If \(0<b<1\), then \(j/b>j\). Thus a positive conditional cell frequency is larger than its joint frequency unless the conditioning event is the whole sample space.

Answer

For \(j>0\), \(P(A\mid B)=j/b\ge j\). Equality occurs when \(P(B)=1\); it is strictly larger when \(0<P(B)<1\).

54717412

Row marginal relative frequencies are \(0.30\) for Group A and \(0.70\) for Group B. Their row conditional distributions for outcomes \(X\) and \(Y\) are \((0.80,0.20)\) and \((0.40,0.60)\). Find the column conditional distribution of group membership among observations with outcome \(X\).

Hints

- Convert row-conditional values to joint frequencies first. - Add the relevant joint cells for the column margin. - Renormalize within the column.

Solution

1. Joint \(X\) frequencies are \(0.30\cdot0.80=0.24\) for A and \(0.70\cdot0.40=0.28\) for B. 2. The \(X\) marginal is \(0.24+0.28=0.52\). 3. Within outcome \(X\), the group distribution is \(\frac{0.24}{0.52}=\frac{6}{13}\) for A and \(\frac{0.28}{0.52}=\frac{7}{13}\) for B.

Answer

Among observations with outcome \(X\), Group A has relative frequency \(\frac{6}{13}\) and Group B has relative frequency \(\frac{7}{13}\).

54717512

A company has two shifts. The day shift has an on-time rate of \(0.92\), and the night shift has an on-time rate of \(0.80\). Overall, the on-time rate is \(0.875\). Find the marginal relative frequency of employees on the day shift.

Hints

- Express the overall rate as a weighted combination of the two shift rates. - Let one group’s marginal share be the unknown. - The two group shares must add to \(1\).

Solution

1. Let \(x\) be the day-shift share. Then \(0.92x+0.80(1-x)=0.875\). 2. Solving gives \(0.12x=0.075\), so \(x=0.625\). 3. The night-shift share is \(0.375\).

Answer

The day-shift marginal relative frequency is \(0.625\), or \(62.5\%\).

54717912

A clinic reports vaccination rates of \(0.74\) for Group A and \(0.61\) for Group B. Group A has \(200\) people and Group B has \(300\). Find the percentage-point difference in rates and the difference in vaccinated counts.

Hints

- A rate difference and a count difference answer different questions. - Convert each rate using its own group size. - Do not infer the larger count from the larger percentage.

Solution

1. The rate difference is \(0.74-0.61=0.13\), or \(13\) percentage points. 2. Vaccinated counts are \(0.74\cdot200=148\) and \(0.61\cdot300=183\). 3. Group B has \(35\) more vaccinated people despite its lower rate.

Answer

The rate difference is \(13\) percentage points in favor of Group A. Group B has \(35\) more vaccinated people, \(183\) versus \(148\).

54718012

A joint relative-frequency table is <table> <tr><th></th><th>Outcome 1</th><th>Outcome 2</th></tr> <tr><td>Group A</td><td>\(0.30\)</td><td>\(0.20\)</td></tr> <tr><td>Group B</td><td>\(0.10\)</td><td>\(0.40\)</td></tr> </table> For each group, compute the conditional odds of Outcome 1 rather than Outcome 2. Then compute the ratio of those two odds. Explain why the same odds ratio would result from any count table obtained by multiplying every cell by the same positive total.

Hints

- Conditional odds compare the two cells within one row. - The ratio of row odds can be written as a cross-product ratio. - Check how a common scale factor affects both cross-products.

Solution

1. Within Group A, the conditional odds are \(\frac{0.30}{0.20}=1.5\). 2. Within Group B, the conditional odds are \(\frac{0.10}{0.40}=0.25\). 3. The odds ratio is \(\frac{1.5}{0.25}=6\). Equivalently, it is the cross-product ratio \(\frac{0.30\cdot0.40}{0.20\cdot0.10}=6\). 4. Multiplying every cell by the same total multiplies numerator and denominator cross-products by the same squared factor, which cancels.

Answer

Group A odds: \(1.5\). Group B odds: \(0.25\). The odds ratio is \(6\), and it is unchanged by converting the relative frequencies to proportional counts.

54718612

A dashboard displays the value \(0.27\), rounded to the nearest hundredth using the usual school rule that a halfway value rounds up. The full sample contains \(200\) observations, and the relevant row contains \(120\) observations. a) If \(0.27\) is a joint relative frequency, what integer cell counts are possible? b) If \(0.27\) is instead the conditional relative frequency within that row, what integer cell counts are possible? c) Explain why the two interpretations give different possible counts even though the displayed decimal is the same.

Hints

- Identify the denominator for each interpretation before doing any rounding work. - Translate the displayed decimal into the interval of exact values that would round to it. - A table cell contains a whole-number count, so keep only integers compatible with each denominator.

Solution

1. A value that rounds to \(0.27\) lies in \([0.265,0.275)\). 2. For a joint relative frequency, the denominator is the full sample of \(200\). Thus the cell count \(k\) must satisfy \(53\le k<55\), giving \(k=53\) or \(k=54\). 3. For a row-conditional relative frequency, the denominator is \(120\). Thus \(31.8\le k<33\), so the only integer possibility is \(k=32\). 4. Joint and conditional relative frequencies use different denominators, so the same rounded decimal need not represent the same count.

Answer

a) \(53\) or \(54\) b) \(32\) c) A joint relative frequency uses all \(200\) observations as its denominator, while the row-conditional relative frequency uses only the \(120\) observations in that row.

54718712

Two rows of a conditional relative-frequency table use the same three displayed percentages, each rounded to the nearest whole percent. <table> <tr><th>Group</th><th>Outcome 1</th><th>Outcome 2</th><th>Outcome 3</th><th>Row total</th></tr> <tr><td>A</td><td>\(17\%\)</td><td>\(42\%\)</td><td>\(42\%\)</td><td>\(12\)</td></tr> <tr><td>B</td><td>\(17\%\)</td><td>\(42\%\)</td><td>\(42\%\)</td><td>\(24\)</td></tr> </table> a) Explain why each displayed row may total \(101\%\) without making the table invalid. b) Recover the three integer cell counts in each row. c) There are \(36\) observations altogether. For Outcome 2, compute the joint relative frequency for Group A and for Group B. Explain why the nearly identical conditional percentages do not imply equal joint relative frequencies.

Hints

- Conditional percentages in a row use that row total as their denominator. - Convert each rounded percentage back to a count that is compatible with the stated row total. - For a joint relative frequency, switch to the grand total as the denominator.

Solution

1. Rounding each conditional percentage separately can make the displayed row total differ slightly from \(100\%\). 2. In Group A, the counts are \(2,5,5\), since \(2/12\approx16.67\%\) and \(5/12\approx41.67\%\). 3. In Group B, the counts are \(4,10,10\), since \(4/24\approx16.67\%\) and \(10/24\approx41.67\%\). 4. For Outcome 2, the joint relative frequencies are \(5/36\approx0.139\) for Group A and \(10/36\approx0.278\) for Group B. 5. Conditional percentages use different row denominators, while joint relative frequencies use the common grand total of \(36\).

Answer

a) The \(101\%\) total is a rounding effect. b) Group A: \((2,5,5)\). Group B: \((4,10,10)\). c) Group A and Outcome 2: \(\frac{5}{36}\approx0.139\). Group B and Outcome 2: \(\frac{10}{36}\approx0.278\). The conditional percentages use different row totals, so they do not determine equal joint shares of the full sample.

54719012

The entries below are exact joint relative frequencies. <table> <tr><th></th><th>Uses app</th><th>Does not use app</th></tr> <tr><td>New customers</td><td>\(12.5\%\)</td><td>\(37.5\%\)</td></tr> <tr><td>Returning customers</td><td>\(25\%\)</td><td>\(25\%\)</td></tr> </table> a) Find all row and column marginal relative frequencies. b) Find the conditional relative frequency of app use among new customers and the conditional relative frequency of returning customers among app users. c) Find the smallest positive integer sample size compatible with all four exact joint relative frequencies, and give the corresponding four cell counts.

Hints

- Marginal relative frequencies come from adding the appropriate joint cells. - A conditional relative frequency changes the denominator to the row or column named by the condition. - For the smallest integer table, express the exact joint percentages as reduced fractions.

Solution

1. The row margins are \(50\%\) new and \(50\%\) returning. The column margins are \(37.5\%\) uses app and \(62.5\%\) does not use app. 2. Among new customers, the app-use conditional relative frequency is \(12.5/50=0.25\). 3. Among app users, the returning-customer conditional relative frequency is \(25/37.5=\frac{2}{3}\). 4. The four joint percentages are \(\frac18,\frac38,\frac14,\frac14\). The smallest common sample size that makes all cell counts integers is \(8\). 5. With \(8\) observations, the cell counts are \(1,3,2,2\) in the same order as the table.

Answer

a) Row margins: new \(50\%\), returning \(50\%\). Column margins: uses app \(37.5\%\), does not use app \(62.5\%\). b) App use among new customers: \(0.25\). Returning customers among app users: \(\frac{2}{3}\). c) Smallest sample size: \(8\). Cell counts: \(1,3,2,2\).

54719112

Two groups have conditional success rates \(0.75\) and \(0.35\). The share of the population in the first group is not known exactly, but it lies between \(0.20\) and \(0.50\). Find the complete range of possible marginal success rates. State where the smallest and largest values occur.

Hints

- Express the overall rate as a weighted average using one unknown group share. - Use the fact that the two group shares add to \(1\). - Determine whether the weighted average increases or decreases as the unknown share changes.

Solution

1. Let \(w\) be the first group’s population share. The marginal rate is \(0.75w+0.35(1-w)=0.35+0.40w\). 2. This expression increases as \(w\) increases. 3. At \(w=0.20\), the marginal rate is \(0.35+0.40(0.20)=0.43\). 4. At \(w=0.50\), it is \(0.35+0.40(0.50)=0.55\).

Answer

The possible marginal success rates form the interval \([0.43,0.55]\). The minimum occurs when the first group’s share is \(0.20\), and the maximum occurs when it is \(0.50\).

54719212

A table has three rows with counts: \( (18,12),\quad(30,20),\quad(24,36). \) The first two rows will be combined. Find the conditional distribution of the two columns within the combined row, and compare it with the third row’s conditional distribution.

Hints

- Add counts before computing the combined conditional percentages. - Use the new combined row total as its denominator. - Normalize the comparison row separately.

Solution

1. The combined first-two rows are \((18+30,12+20)=(48,32)\), with total \(80\). 2. Its conditional distribution is \((0.60,0.40)\). 3. The third row totals \(60\), with distribution \((0.40,0.60)\).

Answer

The combined row has conditional distribution \((0.60,0.40)\). The third row has \((0.40,0.60)\).

54719312

Two full tables describe different populations. Population A: <table> <tr><th></th><th>Outcome 1</th><th>Outcome 2</th></tr> <tr><td>Eligible</td><td>\(30\)</td><td>\(20\)</td></tr> <tr><td>Not eligible</td><td>\(10\)</td><td>\(40\)</td></tr> </table> Population B: <table> <tr><th></th><th>Outcome 1</th><th>Outcome 2</th></tr> <tr><td>Eligible</td><td>\(30\)</td><td>\(20\)</td></tr> <tr><td>Not eligible</td><td>\(40\)</td><td>\(10\)</td></tr> </table> A report filters both populations to eligible observations and then converts each eligible row to conditional relative frequencies. a) Find the filtered conditional distribution for each population. b) Find the full-population relative frequency of Outcome 1 in each population. c) Explain what information the filtering hides.

Hints

- After filtering, use only the retained row as the reference population. - For the full-population rate, restore the excluded row and use the grand total. - Compare what remains visible before and after renormalization.

Solution

1. In both populations, the eligible row is \((30,20)\), so the filtered distribution is \((30/50,20/50)=(0.60,0.40)\). 2. Population A has \((30+10)/100=0.40\) in Outcome 1. Population B has \((30+40)/100=0.70\). 3. Filtering discards the not-eligible row. Renormalization then makes the retained rows identical even though the full tables differ substantially.

Answer

a) Both filtered distributions are \((0.60,0.40)\). b) Full-population Outcome 1 rates are \(0.40\) for Population A and \(0.70\) for Population B. c) Filtering hides all differences in the excluded row and can make different full populations look identical.

54719412

A report initially shows these success counts: <table> <tr><th></th><th>Success</th><th>Failure</th><th>Total</th></tr> <tr><td>Group A</td><td>\(20\)</td><td>\(10\)</td><td>\(30\)</td></tr> <tr><td>Group B</td><td>\(19\)</td><td>\(10\)</td><td>\(29\)</td></tr> </table> One successful observation was assigned to Group A but actually belongs to Group B. a) Compare the two conditional success rates before correction. b) Correct the table and compare the rates again. c) Explain how one reclassification reverses the ranking.

Hints

- A reclassification transfers one observation rather than deleting it. - Recompute both the numerator and denominator for each affected group. - Compare the rates before and after the correction, not just the success counts.

Solution

1. Before correction, Group A’s rate is \(20/30\approx0.667\), while Group B’s is \(19/29\approx0.655\), so Group A is higher. 2. Move one success from Group A to Group B. The corrected rows are \((19,10)\) and \((20,10)\). 3. Corrected rates are \(19/29\approx0.655\) for Group A and \(20/30\approx0.667\) for Group B, so Group B is higher. 4. The original rates were close, and the correction changes both the numerator and denominator of each group in opposite directions.

Answer

a) Before correction: Group A \(\approx0.667\), Group B approximately \(0.655\); Group A is higher. b) After correction: Group A approximately \(0.655\), Group B \(\approx0.667\); Group B is higher. c) Moving one successful observation changes both groups and reverses the ranking.

54719712

Two populations have the same row-conditional distributions: Group A \((0.75,0.25)\) and Group B \((0.30,0.70)\). Construct one joint relative-frequency table when the row margins are \((0.50,0.50)\), and another when the row margins are \((0.20,0.80)\). Compare the column margins and explain what the conditional profiles do not determine.

Hints

- Multiply each row’s margin by its conditional profile. - Add down columns after constructing each joint table. - Compare what changes when only the row weights change.

Solution

1. With row margins \((0.50,0.50)\), the joint rows are \((0.375,0.125)\) and \((0.150,0.350)\); column margins are \((0.525,0.475)\). 2. With row margins \((0.20,0.80)\), the joint rows are \((0.150,0.050)\) and \((0.240,0.560)\); column margins are \((0.390,0.610)\). 3. The same within-row profiles can produce different joint and marginal distributions because row weights differ.

Answer

With row margins \((0.50,0.50)\): <table> <tr><th></th><th>Outcome 1</th><th>Outcome 2</th><th>Total</th></tr> <tr><td>Group A</td><td>\(0.375\)</td><td>\(0.125\)</td><td>\(0.500\)</td></tr> <tr><td>Group B</td><td>\(0.150\)</td><td>\(0.350\)</td><td>\(0.500\)</td></tr> <tr><td>Total</td><td>\(0.525\)</td><td>\(0.475\)</td><td>\(1.000\)</td></tr> </table> With row margins \((0.20,0.80)\): <table> <tr><th></th><th>Outcome 1</th><th>Outcome 2</th><th>Total</th></tr> <tr><td>Group A</td><td>\(0.150\)</td><td>\(0.050\)</td><td>\(0.200\)</td></tr> <tr><td>Group B</td><td>\(0.240\)</td><td>\(0.560\)</td><td>\(0.800\)</td></tr> <tr><td>Total</td><td>\(0.390\)</td><td>\(0.610\)</td><td>\(1.000\)</td></tr> </table> Conditional profiles alone do not determine joint or column-marginal frequencies.

54719812

A sample consists of \(25\%\) Group A with success rate \(0.60\), \(50\%\) Group B with success rate \(0.40\), and \(25\%\) Group C with success rate \(0.20\). Find each group’s contribution to the marginal success frequency and identify which group contributes most.

Hints

- A group’s contribution combines its size and within-group rate. - Compute joint success frequencies for all groups. - Do not rank contributions by conditional rate alone.

Solution

1. Contributions are \(0.25\cdot0.60=0.15\), \(0.50\cdot0.40=0.20\), and \(0.25\cdot0.20=0.05\). 2. The marginal success frequency is \(0.40\). 3. Group B contributes most, even though Group A has the highest conditional rate.

Answer

The contributions are \(0.15,0.20,\) and \(0.05\); Group B contributes the most.

54719912

Two groups have conditional success rates \(0.80\) and \(0.50\). In the observed sample their row margins are \(0.30\) and \(0.70\). A target population instead has row margins \(0.60\) and \(0.40\). Find the observed marginal success rate and the standardized marginal rate for the target population. Explain what is held fixed during standardization.

Hints

- Treat a marginal rate as a weighted combination of group rates. - Use the sample weights first, then the target-population weights. - Identify which quantities remain unchanged between the two calculations.

Solution

1. The observed marginal rate is \(0.30\cdot0.80+0.70\cdot0.50=0.59\). 2. Using target weights gives \(0.60\cdot0.80+0.40\cdot0.50=0.68\). 3. Standardization holds the group-specific conditional rates fixed while replacing the row-margin weights.

Answer

Observed marginal rate: \(0.59\). Target-standardized marginal rate: \(0.68\). The conditional rates are held fixed and the group weights are changed.

54720112

A report gives this percentage table: <table> <tr><th></th><th>yes</th><th>no</th></tr> <tr><td>Group A</td><td>\(20\%\)</td><td>\(30\%\)</td></tr> <tr><td>Group B</td><td>\(40\%\)</td><td>\(60\%\)</td></tr> </table> The analyst says the first row uses percentages of the grand total, while the second row uses percentages within Group B. Explain why the table is not a coherent relative-frequency table and what must be added to make it interpretable.

Hints

- Determine the reference group for each displayed percentage. - A table’s margins are meaningful only when the entries share a compatible denominator system. - Mixed normalizations require explicit labels rather than one unlabeled table.

Solution

1. The first row’s denominator is the full sample, while the second row’s denominator is only Group B. 2. Values with different denominator systems cannot be compared or summed as one table. 3. The table must use one consistent normalization, or each entry must be explicitly labeled with its denominator and corresponding margins.

Answer

The table mixes joint and row-conditional percentages, so it is not coherent as one relative-frequency table. Use a single denominator convention or label every denominator explicitly.

54735712

In a sample of \(100\) observations, event \(A\) occurs \(30\) times, event \(B\) occurs \(50\) times, and both occur \(15\) times. a) Do the sample relative frequencies satisfy the product criterion for independence exactly? b) What may be concluded about this sample table? c) Why does this not prove that the population events are independent?

Hints

- Distinguish sample relative frequencies from population probabilities. - Apply the product criterion to the observed table first. - Consider whether one sample can establish an exact population property.

Solution

1. The sample margins are \(0.30\) and \(0.50\), and the sample joint frequency is \(0.15\). 2. Since \(0.15=0.30\cdot0.50\), the empirical table factors exactly and shows no sample association by this criterion. 3. Population independence is a statement about the underlying probability distribution. A finite sample can match the product relationship because of sampling variation, so inference would require a sampling model and an assessment of uncertainty.

Answer

a) Yes; \(0.15=0.30\cdot0.50\). b) The sample table is exactly independent descriptively. c) This does not prove population independence because the table is one finite random sample.

54737712

A \(2 \times 2\) table has counts: <table> <tr><th></th><th>\(B\)</th><th>not \(B\)</th></tr> <tr><td>\(A\)</td><td>\(24\)</td><td>\(16\)</td></tr> <tr><td>not \(A\)</td><td>\(36\)</td><td>\(24\)</td></tr> </table> Use the cross-product criterion to determine whether \(A\) and \(B\) are independent.

Hints

- Compare products of diagonally opposite cells. - This criterion is equivalent to equal conditional distributions in a \(2 \times 2\) table. - Use counts directly; a grand-total conversion is unnecessary.

Solution

1. For a \(2 \times 2\) count table, independence requires the diagonal cross-products to match. 2. \(24\cdot24=576\) and \(16\cdot36=576\). 3. The criterion is satisfied, so the sample table has independent row and column proportions.

Answer

Yes. The cross-products are equal: \(24\cdot24=16\cdot36=576\).

52213012

A company operates two IT support centers, \(\alpha\) and \(\beta\). Each center handles hardware requests \((H)\) and software requests \((S)\). Let \(L\) be the event that a request is resolved. The data from the most recent quarter are shown below. <table> <tr><td>Center</td><td>Request type</td><td>Total requests</td><td>Resolved requests</td></tr> <tr><td>\(\alpha\)</td><td>Hardware \((H)\)</td><td>\(200\)</td><td>\(40\)</td></tr> <tr><td>\(\alpha\)</td><td>Software \((S)\)</td><td>\(800\)</td><td>\(720\)</td></tr> <tr><td>\(\beta\)</td><td>Hardware \((H)\)</td><td>\(800\)</td><td>\(200\)</td></tr> <tr><td>\(\beta\)</td><td>Software \((S)\)</td><td>\(200\)</td><td>\(190\)</td></tr> </table> a) For each center, find \(P(L\mid H)\) and \(P(L\mid S)\). b) A manager claims that center \(\alpha\) is much more effective because it resolved \(76\%\) of all requests, while center \(\beta\) resolved only \(39\%\). Verify these overall rates. c) Evaluate the manager's claim. Which center performs better within each request type? Explain why the type-specific rates and overall rates lead to different conclusions.

Hints

- Interpret \(P(L\mid H)\) as a rate within the hardware-request group. - Add the total and resolved requests separately for each center. - Compare the two centers within hardware, then compare them within software. - Examine the mix of hardware and software requests at each center.

Solution

1. For center \(\alpha\), \(P(L\mid H)=\frac{40}{200}=0.20\) and \(P(L\mid S)=\frac{720}{800}=0.90\). 2. For center \(\beta\), \(P(L\mid H)=\frac{200}{800}=0.25\) and \(P(L\mid S)=\frac{190}{200}=0.95\). 3. Center \(\alpha\) resolved \(40+720=760\) of \(1000\) requests, so its overall rate is \(\frac{760}{1000}=0.76\). Center \(\beta\) resolved \(200+190=390\) of \(1000\) requests, so its overall rate is \(\frac{390}{1000}=0.39\). 4. Center \(\beta\) has the higher resolution rate for both hardware and software. Its lower overall rate occurs because \(80\%\) of its requests are hardware requests, which have much lower resolution rates at both centers. Center \(\alpha\) receives mostly software requests. This reversal is an example of how aggregated data can obscure subgroup comparisons.

Answer

a) Center \(\alpha\): \(P(L\mid H)=20\%\), \(P(L\mid S)=90\%\). Center \(\beta\): \(P(L\mid H)=25\%\), \(P(L\mid S)=95\%\). b) Center \(\alpha\): \(76\%\). Center \(\beta\): \(39\%\). c) Center \(\beta\) performs better for both hardware and software. Its overall rate is lower because it handles a much larger proportion of the harder-to-resolve hardware requests.

54718312

A \(2\times2\) joint relative-frequency table has row margin \(P(A)=0.65\) and column margin \(P(B)=0.55\). Let \(x=P(A\cap B)\). a) Express every cell of the table in terms of \(x\). b) Determine the complete interval of values of \(x\) that makes every cell nonnegative. c) Find the value of \(x\) for which the row and column variables are independent, and verify that it lies in the feasible interval.

Hints

- Use the row and column margins to write the three remaining cells from one unknown intersection. - Every joint relative frequency must be at least zero. - Independence selects one particular interior table from the feasible family.

Solution

1. The four cells are \(x\), \(0.65-x\), \(0.55-x\), and \(1-0.65-0.55+x=x-0.20\). 2. Nonnegativity requires \(x\ge0\), \(x\le0.65\), \(x\le0.55\), and \(x\ge0.20\). Therefore \(0.20\le x\le0.55\). 3. Independence requires \(x=P(A)P(B)=0.65\cdot0.55=0.3575\), which is inside the feasible interval.

Answer

a) <table> <tr><th></th><th>\(B\)</th><th>\(B^c\)</th><th>Total</th></tr> <tr><td>\(A\)</td><td>\(x\)</td><td>\(0.65-x\)</td><td>\(0.65\)</td></tr> <tr><td>\(A^c\)</td><td>\(0.55-x\)</td><td>\(x-0.20\)</td><td>\(0.35\)</td></tr> <tr><td>Total</td><td>\(0.55\)</td><td>\(0.45\)</td><td>\(1\)</td></tr> </table> b) \(x\in[0.20,0.55]\). c) Independence occurs at \(x=0.3575\).

54719512

A tutoring center compares Program A and Program B for students with strong or limited prior preparation. The results are: <table> <tr><th></th><th>Program A</th><th>Program B</th></tr> <tr><td>Strong preparation</td><td>\(9/10\) passed</td><td>\(72/90\) passed</td></tr> <tr><td>Limited preparation</td><td>\(3/10\) passed</td><td>\(2/10\) passed</td></tr> </table> a) Compare the conditional pass rates for the two programs within each preparation group. b) Compare the overall pass rates after pooling the preparation groups. c) Explain why the direction of the comparison reverses and why the pooled comparison alone is misleading.

Hints

- Compute each pass rate using the total for its own preparation-group-and-program cell. - Pool counts within each program only after comparing the two preparation groups separately. - Compare the preparation-group composition of the two programs to explain the reversal.

Solution

1. Among students with strong preparation, Program A has pass rate \(\frac{9}{10}=0.90\), while Program B has \(\frac{72}{90}=0.80\). Program A is higher. 2. Among students with limited preparation, Program A has \(\frac{3}{10}=0.30\), while Program B has \(\frac{2}{10}=0.20\). Program A is again higher. 3. Overall, Program A has \(\frac{12}{20}=0.60\), while Program B has \(\frac{74}{100}=0.74\). The pooled comparison favors Program B. 4. Program B has a much larger share of students with strong preparation, the group with higher pass rates under either program. Unequal group composition creates Simpson’s paradox: the pooled association reverses the within-group associations.

Answer

a) Strong preparation: Program A \(0.90\), Program B \(0.80\). Limited preparation: Program A \(0.30\), Program B \(0.20\). Program A is higher in both groups. b) Overall: Program A \(0.60\), Program B \(0.74\). c) Program B includes a much larger proportion of strongly prepared students. The unequal weights create Simpson’s paradox, so the pooled rates hide the within-group pattern.

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