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Correlation and causation

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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 correlation, 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 correlation 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 correlated, 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 significantly 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 correlation from causation, and identify one possible confounding variable that could explain the observed association.

Hints

- Does observing two variables together prove that one causes the other? - 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 correlation: instrument playing and higher mathematics grades occur together more often than expected. 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 correlation alone.

Answer

The argument confuses correlation 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 correlation, 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 correlation, but motivation, prior knowledge, or other variables could affect both material completion and exam success.

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