Aimathic – Correlation and causation

Correlation and causation

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Difficulty: 1 – 5

5513889

A report says, “Students who use a weekly planner tend to have higher course grades.” A headline rewrites this as, “Using a weekly planner raises students' course grades.” Which sentence states an association, and which sentence makes a causal claim?

Hints

- Look for wording that only says two variables occur together. - Look for wording that says one variable produces a change in the other. - Focus on the difference between “tend to have” and “raises.”

Solution

1. “Students who use a weekly planner tend to have higher course grades” describes two variables occurring together, so it states an association. 2. “Using a weekly planner raises students' course grades” says that one variable produces a change in the other, so it makes a causal claim.

Answer

The report states an association. The headline makes a causal claim.

5513899

Two studies examine whether a new review routine improves quiz scores. Study A asks students whether they already use the routine and compares their quiz scores with scores of students who do not use it. Study B randomly assigns students to use either the new routine or the usual routine for the same amount of study time, then gives everyone the same quiz. Which study provides stronger evidence that the review routine causes a change in quiz scores? Give the key reason.

Hints

- Compare how students enter the groups in the two studies. - Identify which study assigns the review routine instead of merely observing it. - Causal evidence is stronger when the treatment is deliberately randomized.

Solution

1. Study A observes choices students have already made, so other differences between the groups could help explain a score difference. 2. Study B randomly assigns the review routine, so the treatment is deliberately varied rather than chosen by the students. 3. Therefore, Study B provides stronger evidence for a causal effect.

Answer

Study B. Random assignment provides stronger evidence that the review routine, rather than preexisting differences between the groups, caused a change in quiz scores.

5440789

A school compared project completion with attendance at an optional review session. <table><tr><th>Review session</th><th>Completed project</th><th>Did not complete</th><th>Total</th></tr><tr><td>Attended</td><td>\(42\)</td><td>\(18\)</td><td>\(60\)</td></tr><tr><td>Did not attend</td><td>\(30\)</td><td>\(30\)</td><td>\(60\)</td></tr><tr><td>Total</td><td>\(72\)</td><td>\(48\)</td><td>\(120\)</td></tr></table> a) Find the conditional relative frequency of completing the project for each attendance group. b) Describe the association shown by the table. c) Does the table prove that attending the review session caused students to complete the project? Explain.

Hints

- Use each row total as the denominator for a within-group completion rate. - Compare the two percentages using percentage points. - Separate evidence of association from evidence produced by a randomized experiment.

Solution

1. Among students who attended, the completion rate is \(\frac{42}{60}=70\%\). 2. Among students who did not attend, the completion rate is \(\frac{30}{60}=50\%\). 3. The completion rate is \(20\) percentage points higher for students who attended, so the variables show a positive association in this group. 4. The table is observational. Other differences between the groups, such as prior preparation or motivation, could help explain the result, so causation is not established.

Answer

a) Attended: \(70\%\). Did not attend: \(50\%\). b) Project completion is associated with attending the review session; the completion rate is \(20\) percentage points higher among attendees. c) No. The observational table does not rule out confounding variables or self-selection.

5440879

A school recorded arrival status by commute type. <table><tr><th>Commute type</th><th>On time</th><th>Late</th><th>Total</th></tr><tr><td>School bus</td><td>\(54\)</td><td>\(6\)</td><td>\(60\)</td></tr><tr><td>Other commute</td><td>\(72\)</td><td>\(18\)</td><td>\(90\)</td></tr><tr><td>Total</td><td>\(126\)</td><td>\(24\)</td><td>\(150\)</td></tr></table> a) Find the conditional relative-frequency distribution of arrival status for each commute group. b) Compare the on-time rates. c) Find the overall on-time rate and explain why it does not replace the two conditional rates.

Hints

- Divide each row by its own row total. - Compare the same outcome across the two commute groups. - Use the grand total only for the overall rate, not for a conditional rate.

Solution

1. For school-bus riders, the conditional distribution is \(\frac{54}{60}=90\%\) on time and \(\frac{6}{60}=10\%\) late. 2. For other commuters, it is \(\frac{72}{90}=80\%\) on time and \(\frac{18}{90}=20\%\) late. 3. The on-time rate is \(10\) percentage points higher for school-bus riders. 4. Overall, \(\frac{126}{150}=84\%\) of students arrived on time. This pooled percentage combines groups of different sizes and does not show the within-group difference.

Answer

a) School bus: \(90\%\) on time, \(10\%\) late. Other commute: \(80\%\) on time, \(20\%\) late. b) The school-bus on-time rate is \(10\) percentage points higher. c) Overall on-time rate: \(84\%\). The overall rate combines the groups and hides their conditional difference.

5441169

A school surveyed textbook format preference by grade level. <table><tr><th>Grade</th><th>Digital</th><th>Print</th><th>Total</th></tr><tr><td>9</td><td>\(36\)</td><td>\(24\)</td><td>\(60\)</td></tr><tr><td>10</td><td>\(30\)</td><td>\(30\)</td><td>\(60\)</td></tr><tr><td>11</td><td>\(18\)</td><td>\(42\)</td><td>\(60\)</td></tr><tr><td>Total</td><td>\(84\)</td><td>\(96\)</td><td>\(180\)</td></tr></table> a) Find the conditional relative-frequency distribution of textbook preference within each grade. b) Describe the association between grade level and textbook preference. c) Which grade shows the largest preference gap, and what is that gap in percentage points?

Hints

- Divide each grade row by its row total. - Compare the digital percentages in grade order. - Find a preference gap by subtracting the two percentages within one grade.

Solution

1. Grade 9: \(\frac{36}{60}=60\%\) digital and \(40\%\) print. 2. Grade 10: \(50\%\) digital and \(50\%\) print. 3. Grade 11: \(\frac{18}{60}=30\%\) digital and \(70\%\) print. 4. Digital preference decreases as grade level increases in this survey, while print preference increases, so the variables are associated. 5. Grade 11 has the largest gap: \(70\%-30\%=40\) percentage points in favor of print.

Answer

a) Grade 9: \(60\%\) digital, \(40\%\) print. Grade 10: \(50\%\) digital, \(50\%\) print. Grade 11: \(30\%\) digital, \(70\%\) print. b) Higher grade levels show lower digital preference and higher print preference in this survey. c) Grade 11, with a \(40\)-percentage-point gap in favor of print.

5441749

Neighborhood fires attended by more firefighters tend to have greater property damage. Explain the likely causal structure behind this association.

Hints

- Ask what determines how many firefighters are sent. - Identify a variable connected to both response size and damage. - Consider the sequence of events.

Solution

1. More severe fires generally cause more property damage. 2. Those severe fires also lead emergency services to send more firefighters. 3. Fire severity is a common cause, and the likely causal direction is from severity to both staffing and damage, not from firefighters to damage.

Answer

Larger, more severe fires both require more firefighters and cause more damage. The correlation does not mean firefighters cause the damage.

5441759

Two regions report the following annual data. <table><tr><th>Region</th><th>Population</th><th>Hospitals</th><th>Reported illness cases</th></tr><tr><td>Lake</td><td>\(50{,}000\)</td><td>\(2\)</td><td>\(2500\)</td></tr><tr><td>Metro</td><td>\(300{,}000\)</td><td>\(8\)</td><td>\(12{,}000\)</td></tr></table> a) Which region has more hospitals and more reported illness cases? b) Find the reported illness rate in each region. c) Explain why the totals do not support the conclusion that hospitals cause illness.

Hints

- Compare the raw totals first. - Divide reported cases by population to compare per-person rates. - Distinguish an association between totals from evidence that one variable causes the other.

Solution

1. Metro has \(8\) hospitals and \(12{,}000\) reported cases, both larger totals than Lake’s \(2\) hospitals and \(2500\) cases. 2. Lake’s illness rate is \(\frac{2500}{50{,}000}=0.05=5\%\). 3. Metro’s illness rate is \(\frac{12{,}000}{300{,}000}=0.04=4\%\). 4. Metro’s larger population can produce both more hospitals and more total cases. The larger totals therefore do not show that hospitals cause illness; the per-person illness rate is actually lower in Metro.

Answer

a) Metro b) Lake: \(5\%\). Metro: \(4\%\). c) Population size affects both totals. More hospitals and more total cases do not establish causation, and Metro has the lower illness rate.

5441789

Identical products are randomly assigned to two packing methods before being shipped on the same route. Method A has a lower damage rate. What conclusion is supported, and why is shipping route control important?

Hints

- Identify the treatment and the outcome. - Ask what the common shipping route keeps from varying. - Limit the causal statement to the tested conditions.

Solution

1. Random assignment balances product differences between the packing groups. 2. Using the same route prevents route roughness or distance from being mixed with packing method. 3. The experiment supports that Method A caused a lower damage rate under the tested shipping conditions.

Answer

Method A caused a lower damage rate under the tested conditions. Keeping the route the same prevents transportation differences from confounding the result.

54418012

A school wants to test whether a reflective roof coating lowers classroom temperature. Design a randomized experiment using similar portable classrooms. State the treatment, controls, and response variable.

Hints

- Use classrooms that are as similar as possible before treatment. - Randomly assign the coating rather than choosing the hottest rooms. - Measure the same temperature outcome on the same schedule.

Solution

1. Randomly assign similar portable classrooms to receive the reflective coating or remain uncoated. 2. Keep thermostat settings, measurement times, occupancy schedules, and window use consistent. 3. Record interior temperature with the same calibrated sensors over comparable weather periods. 4. Compare average temperatures to estimate the causal effect of the coating under the study conditions.

Answer

Randomly assign similar classrooms to coated and uncoated groups, hold operating conditions constant, and compare their measured interior temperatures.

5221019

A social-science survey asks \(500\) families whether the parent exercises regularly (event \(P\)) and whether the child exercises regularly (event \(C\)). <table> <tr><td></td><td>\(C\)</td><td>\(\overline{C}\)</td><td>Total</td></tr> <tr><td>\(P\)</td><td>\(120\)</td><td>\(80\)</td><td>\(200\)</td></tr> <tr><td>\(\overline{P}\)</td><td>\(60\)</td><td>\(240\)</td><td>\(300\)</td></tr> <tr><td>Total</td><td>\(180\)</td><td>\(320\)</td><td>\(500\)</td></tr> </table> a) Use conditional relative frequencies to determine whether the parent's and child's exercise habits are associated. b) Explain why the observed association does not prove that the parent's example is the sole cause of the child's exercise habits.

Hints

- Compare the child-exercise rate within each parent group. - Different conditional relative frequencies indicate an association. - Consider variables that could influence both the parent and the child. - Distinguish an observational pattern from a demonstrated cause.

Solution

1. Among families in which the parent exercises regularly, the child-exercise rate is \(P(C\mid P)=\frac{120}{200}=0.60\). 2. Among families in which the parent does not exercise regularly, the rate is \(P(C\mid\overline{P})=\frac{60}{300}=0.20\). 3. Because the conditional relative frequencies differ, the variables are associated and the corresponding events are not independent. 4. The survey is observational. Other variables, such as access to recreation, family income, neighborhood resources, health, or shared preferences, could affect both the parent's and child's behavior. Therefore, the association alone does not establish a single causal explanation.

Answer

a) The variables are associated because \(P(C\mid P)=0.60\neq 0.20=P(C\mid\overline{P})\). b) The survey shows an association, not causation. Shared family or environmental factors could influence both exercise habits.

5221029

A study examines whether using a vocabulary app (event \(A\)) is associated with passing a language exam (event \(E\)). The results for \(800\) participants are shown. <table> <tr><td></td><td>\(E\)</td><td>\(\overline{E}\)</td><td>Total</td></tr> <tr><td>\(A\)</td><td>\(240\)</td><td>\(80\)</td><td>\(320\)</td></tr> <tr><td>\(\overline{A}\)</td><td>\(120\)</td><td>\(360\)</td><td>\(480\)</td></tr> <tr><td>Total</td><td>\(360\)</td><td>\(440\)</td><td>\(800\)</td></tr> </table> a) Use the multiplication rule for independent events to determine whether app use and passing the exam are independent. b) Evaluate the claim, “Using the app causes students to pass the exam.” Explain the difference between association and causation.

Hints

- Find the two marginal probabilities and their product. - Compare the product with the observed joint probability. - Ask whether participants were randomly assigned to use the app. - Identify a variable that could influence both app use and exam performance.

Solution

1. From the margins, \(P(A)=\frac{320}{800}=0.40\) and \(P(E)=\frac{360}{800}=0.45\). 2. If the events were independent, \(P(A\cap E)\) would equal \(P(A)P(E)=0.40\cdot 0.45=0.18\). 3. The observed joint probability is \(P(A\cap E)=\frac{240}{800}=0.30\), so the events are dependent and positively associated. 4. The table does not show that participants were randomly assigned to use the app. A confounding variable, such as motivation, study time, or prior language skill, could increase both app use and the chance of passing. The association therefore does not prove causation.

Answer

a) The events are dependent because \(P(A\cap E)=0.30\neq 0.18=P(A)P(E)\). b) App use and passing are positively associated, but the data do not prove that the app caused the result. Other variables could affect both.

5221099

A school surveys \(200\) students about whether they regularly eat breakfast (event \(B\)) and whether they show high concentration during first period (event \(C\)). <table> <tr><td></td><td>\(C\) (high concentration)</td><td>\(\overline{C}\) (lower concentration)</td><td>Total</td></tr> <tr><td>\(B\) (eats breakfast)</td><td>\(72\)</td><td>\(48\)</td><td>\(120\)</td></tr> <tr><td>\(\overline{B}\) (does not eat breakfast)</td><td>\(18\)</td><td>\(62\)</td><td>\(80\)</td></tr> <tr><td>Total</td><td>\(90\)</td><td>\(110\)</td><td>\(200\)</td></tr> </table> a) Show that \(B\) and \(C\) are dependent. b) Find \(P(B\cap C)\) and \(P(C\mid B)\). Describe the association, and give one plausible explanation or confounding variable without claiming that the survey proves a cause.

Hints

- Compare the joint probability with the product of the marginal probabilities. - For \(P(C\mid B)\), use students who eat breakfast as the denominator. - Compare the conditional rate with the overall concentration rate. - Give a plausible explanation while recognizing that an observational survey cannot isolate a cause.

Solution

1. The marginal probabilities are \(P(B)=\frac{120}{200}=0.60\) and \(P(C)=\frac{90}{200}=0.45\). 2. The joint probability is \(P(B\cap C)=\frac{72}{200}=0.36\). 3. Since \(P(B)P(C)=0.60\cdot 0.45=0.27\neq 0.36\), the events are dependent. 4. The conditional probability is \(P(C\mid B)=\frac{72}{120}=0.60\). This is higher than the overall high-concentration rate of \(0.45\), so breakfast and concentration are positively associated in this sample. 5. Breakfast could plausibly contribute to concentration, but variables such as sleep, household routines, stress, or access to food could influence both. The survey alone cannot identify the cause.

Answer

a) The events are dependent because \(P(B\cap C)=0.36\neq 0.27=P(B)P(C)\). b) \(P(B\cap C)=0.36\), and \(P(C\mid B)=0.60\). The variables are positively associated, but the survey does not establish causation; sleep or household routines are possible confounding variables.

5221119

A study across several high schools reports that students who play a musical instrument have higher average mathematics grades. A music teacher then argues, “Every student should be required to learn an instrument so that our school's overall mathematics performance will improve.” Evaluate the teacher's argument. Distinguish association from causation, and identify one possible confounding variable that could explain the observed association.

Hints

- Does observing a difference between two groups prove that the group characteristic caused the difference? - Identify a factor that could affect both access to music lessons and school performance. - Consider what kind of study would provide stronger causal evidence.

Solution

1. The study reports a positive association: students who play an instrument have higher average mathematics grades than students who do not. 2. The teacher assumes a causal relationship, but an association does not show that learning an instrument produces higher mathematics grades. 3. A possible confounding variable is family educational support or socioeconomic resources. Such factors could make music lessons more available and also support academic achievement. 4. Without a well-designed experiment or stronger causal evidence, the proposed requirement is not justified by the reported association alone.

Answer

The argument confuses association with causation. Playing an instrument may be associated with higher mathematics grades without causing them. Family educational support or socioeconomic resources could influence both, so the study alone does not justify requiring every student to learn an instrument.

5221359

In a health and exercise survey, \(40\%\) of respondents say they exercise regularly (event \(E\)), \(55\%\) say they are satisfied with their physical fitness (event \(F\)), and \(35\%\) report both. a) Find \(P(F)\) and \(P(F\mid E)\). b) Determine whether \(E\) and \(F\) are independent. c) Interpret the association and explain what the survey does and does not show about causation.

Hints

- Distinguish the overall fitness-satisfaction rate from the rate within the exercise group. - Under independence, a conditional probability equals the corresponding marginal probability. - Consider reverse causation and variables that could influence both responses.

Solution

1. The overall probability is given: \(P(F)=0.55\). 2. The conditional probability is \(P(F\mid E)=\frac{P(E\cap F)}{P(E)}=\frac{0.35}{0.40}=0.875\). 3. Since \(P(F\mid E)=0.875\neq 0.55=P(F)\), the events are dependent. Equivalently, \(P(E)P(F)=0.40\cdot 0.55=0.22\neq 0.35=P(E\cap F)\). 4. Regular exercise and fitness satisfaction are positively associated in the survey. Exercise could affect satisfaction, but reverse causation or confounding variables such as health status, time, or general health awareness could also explain part of the association.

Answer

a) \(P(F)=0.55\), and \(P(F\mid E)=0.875\), or \(87.5\%\). b) The events are dependent because \(P(F\mid E)\neq P(F)\). c) The survey shows a positive association, but it does not prove that exercise alone causes greater fitness satisfaction.

5221369

An online learning platform reports that \(60\%\) of students completed all optional practice materials (event \(M\)). Of all students, \(10\%\) did not complete the materials but still passed the final exam (event \(E\)). Overall, \(65\%\) passed. a) Find \(P(E\mid M)\). b) Determine whether passing the exam is independent of completing the materials. c) Explain why the result does not prove that the materials were the sole cause of passing.

Hints

- Subtract the passing students who did not complete the materials from the overall passing proportion. - Use the material-completion group as the denominator for the conditional probability. - Compare the observed joint probability with the product expected under independence. - Identify a variable that could influence both behaviors.

Solution

1. The probability of completing the materials and passing is \(P(M\cap E)=P(E)-P(\overline{M}\cap E)=0.65-0.10=0.55\). 2. Therefore, \(P(E\mid M)=\frac{0.55}{0.60}=\frac{11}{12}\approx 0.9167\). 3. If the events were independent, \(P(M\cap E)\) would equal \(P(M)P(E)=0.60\cdot 0.65=0.39\). Since \(0.55\neq 0.39\), the events are dependent. 4. Students who are more motivated or have stronger prior knowledge may be more likely both to complete optional materials and to pass. Without random assignment or other causal evidence, the association does not prove that the materials alone caused the higher pass rate.

Answer

a) \(P(E\mid M)=\frac{11}{12}\approx 0.9167\), or about \(91.67\%\). b) The events are dependent because \(P(M\cap E)=0.55\neq 0.39=P(M)P(E)\). c) The data show an association, but motivation, prior knowledge, or other variables could affect both material completion and exam success.

5440829

A survey compared participation in music lessons with honor-roll status. <table><tr><th>Music lessons</th><th>Honor roll</th><th>Not on honor roll</th><th>Total</th></tr><tr><td>Yes</td><td>\(24\)</td><td>\(16\)</td><td>\(40\)</td></tr><tr><td>No</td><td>\(36\)</td><td>\(44\)</td><td>\(80\)</td></tr><tr><td>Total</td><td>\(60\)</td><td>\(60\)</td><td>\(120\)</td></tr></table> a) Among students taking music lessons, what percentage are on the honor roll? b) Among honor-roll students, what percentage take music lessons? c) Explain why the two percentages are different, and compare honor-roll rates for students with and without music lessons.

Hints

- Name the group described after the word “among”; that group supplies the denominator. - Do not reverse the conditioning group when forming a percentage. - For the final comparison, calculate the same outcome rate within both lesson groups.

Solution

1. Among students taking music lessons, the honor-roll percentage is \(\frac{24}{40}=60\%\). 2. Among honor-roll students, the music-lesson percentage is \(\frac{24}{60}=40\%\). 3. The percentages use different conditioning groups and therefore different denominators. 4. The honor-roll rate without music lessons is \(\frac{36}{80}=45\%\). The rate is \(15\) percentage points higher among students taking music lessons, showing an association in this survey.

Answer

a) \(60\%\) b) \(40\%\) c) The denominators differ: part a conditions on music lessons, while part b conditions on honor-roll status. Honor-roll rates are \(60\%\) with music lessons and \(45\%\) without, a \(15\)-percentage-point difference.

5441739

An observational study finds that students from homes with more books tend to earn higher standardized test scores. Would placing \(100\) more books in every home necessarily raise scores? Explain why or why not, and give one possible confounder.

Hints

- Identify whether households were assigned different numbers of books. - Think about family characteristics related to both variables. - Separate access to objects from the behaviors and resources surrounding them.

Solution

1. The number of books was observed rather than randomly assigned. 2. Parent education, household income, or time spent reading with children could influence both the home library and test performance. 3. These factors can create the observed correlation without the physical presence of additional books being the direct cause.

Answer

The causal claim is not justified. Parent education is one plausible confounder because it may be related to both the number of books at home and student test scores.

5441769

An observational study finds that adults who own pets report lower average stress. Give one self-selection explanation and one possible confounder. Explain why the association does not prove that adopting a pet reduces stress for everyone.

Hints

- Ask who is able and willing to own a pet. - Look for a resource related to both ownership and stress. - Notice that the claim applies to every person, while the data show an average association.

Solution

1. People who enjoy animals and have stable routines may be more likely to choose pet ownership, creating self-selection. 2. Housing security or household income may make pet ownership easier and may also reduce stress. 3. These alternative explanations mean the observational association does not prove a universal causal benefit.

Answer

People with lower-stress lifestyles may be more likely to adopt pets, and housing security could affect both pet ownership and stress. The study does not prove adopting a pet reduces stress for everyone.

5441779

A teacher randomly assigns students to use either retrieval practice or rereading for the same amount of study time. Everyone takes the same quiz under the same conditions. The retrieval-practice group has a higher average score. What causal conclusion is supported, and to whom does it apply most directly?

Hints

- Identify how students entered the two groups. - Check which conditions were held constant. - Separate causal validity for participants from broader generalization.

Solution

1. Random assignment tends to balance prior ability, motivation, and other student characteristics between the groups. 2. The study holds study time and quiz conditions constant, isolating the review method. 3. The score difference supports a causal effect of retrieval practice for students similar to those who participated.

Answer

The experiment supports that retrieval practice caused a higher average quiz score than rereading for students like those in the study.

5441799

Patients treated by specialists sometimes have lower recovery rates than patients treated by general practitioners. Explain the selection effect that makes a causal conclusion unreliable.

Hints

- Ask which patients are referred to specialists. - Compare the groups before treatment begins. - Look for selection based on the seriousness of the case.

Solution

1. Specialists often receive the most severe or complicated cases. 2. Those cases have lower recovery chances even before the specialist provides treatment. 3. Patient severity influences both the type of doctor and the outcome, so the observed association does not show that specialists cause worse recovery.

Answer

Specialists are more likely to treat severe cases. Case severity can explain the lower recovery rate, so the comparison does not establish harmful specialist care.

54418112

A sports league wants to compare two approved warm-up routines for reducing muscle strains. Design an ethical randomized experiment and identify an appropriate outcome that accounts for teams playing different numbers of athlete-hours.

Hints

- Compare only treatments that meet safety standards. - Randomize the routine rather than allowing teams to choose. - Use a rate that adjusts for unequal time at risk.

Solution

1. Randomly assign teams or athletes to approved Routine A or approved Routine B; do not assign an unsafe no-warm-up condition. 2. Keep training schedules, injury definitions, and reporting procedures consistent. 3. Record muscle strains per \(1000\) athlete-hours of participation so different exposure amounts are comparable. 4. Compare injury rates to estimate the causal difference between the two approved routines.

Answer

Randomly assign participants to the two approved routines and compare muscle-strain rates per \(1000\) athlete-hours using the same injury-reporting rules.

5441829

People who exercise more often also report having more daily energy. Give one reverse-causality explanation and one confounding explanation. Explain why the association does not identify a single cause.

Hints

- Reverse the proposed direction between the two measured variables. - Look for a third lifestyle factor related to both. - Notice the unsupported word “only.”

Solution

1. Reverse causality is possible because people who naturally feel more energetic may choose to exercise more often. 2. Sleep quality, overall health, or work schedule could influence both energy and exercise frequency. 3. The association therefore does not show that exercise is the only cause of the reported energy difference.

Answer

More energetic people may exercise more, and sleep quality could affect both exercise and energy. The correlation does not establish a single causal explanation.

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Correlation and causation

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