Correlation vs Causation
Grade 11 ยท Statistics ยท Worksheet 2
- A study of 14 coastal towns finds a Pearson correlation coefficient of r = 0.91 between the number of beach umbrellas rented (x) and the number of jellyfish stings reported (y) in August. The mean of x is 120 with standard deviation 15, and the mean of y is 45 with standard deviation 9. Calculate the slope of the least-squares regression line for predicting jellyfish stings from beach umbrellas rented, then explain why this strong correlation does not imply that renting beach umbrellas causes jellyfish stings. Answer: ______________
- Matiu collects data from 14 coastal towns on the number of beach umbrellas rented (x) and the number of jellyfish stings reported (y) in July. The Pearson correlation coefficient is r = 0.82. The mean of x is 62 with standard deviation 11, and the mean of y is 38 with standard deviation 9. Calculate the slope of the least-squares regression line for predicting jellyfish stings from beach umbrellas rented. Then, explain why this strong correlation does not imply that renting beach umbrellas causes jellyfish stings. Answer: ______________
- Emma is a researcher studying the relationship between the number of hours students spend practicing a musical instrument per week and their scores on a standardized mathematics test. She collects data from 150 high school students and calculates a Pearson correlation coefficient of r = 0.55. The school board, upon seeing this result, proposes a policy requiring all students to spend at least 5 hours per week practicing an instrument, arguing that music practice directly improves math performance. As a critical thinker, explain why the school board's causal conclusion is flawed, and identify a plausible confounding variable that could explain the observed correlation. Answer: ______________
- A pharmaceutical company is testing a new medication and claims it reduces blood pressure by an average of 15 mmHg. In their clinical trial with 200 participants, they calculate a 95% confidence interval for the mean reduction as (12.8, 17.2) mmHg. The margin of error for this interval is 2.2 mmHg. If the company wants to reduce this margin of error to 1.1 mmHg while maintaining the same confidence level, how many participants would they need in their new trial? Answer: ______________
- Hana collects data from 22 coastal towns on the number of beach umbrellas (x) and the number of jellyfish stings (y) reported in July. The Pearson correlation coefficient is r = 0.83, with sx = 15 and sy = 9. Calculate the slope of the least-squares regression line for predicting jellyfish stings from beach umbrellas, then explain why this strong correlation does not imply that beach umbrellas cause jellyfish stings. Answer: ______________
Answer Key & Explanations
Correlation vs Causation ยท Grade 11 ยท Worksheet 2
- A study of 14 coastal towns finds a Pearson correlation coefficient of r = 0.91 between the number of beach umbrellas rented (x) and the number of jellyfish stings reported (y) in August. The mean of x is 120 with standard deviation 15, and the mean of y is 45 with standard deviation 9. Calculate the slope of the least-squares regression line for predicting jellyfish stings from beach umbrellas rented, then explain why this strong correlation does not imply that renting beach umbrellas causes jellyfish stings. Answer: 0.546 Solution: Identify given values: r = 0.91, sx = 15, sy = 9. Slope formula: b = r * (sy / sx) = 0.91 * (9 / 15) = 0.91 * 0.6 = 0.546.
Full step-by-step solution
Step 1: Identify given values: r = 0.91, sx = 15, sy = 9.
Step 2: Slope formula: b = r * (sy / sx) = 0.91 * (9 / 15) = 0.91 * 0.6 = 0.546.
Step 3: The slope means that for each additional beach umbrella rented, the predicted number of jellyfish stings increases by about 0.546.
Step 4: Correlation does not equal causation. A lurking variable, such as hot summer weather, causes more people to go to the beach (increasing umbrella rentals) and also more people to swim in the ocean, leading to more jellyfish stings. The umbrellas themselves do not cause stings; both are associated with the same external factor.
The answer is slope = 0.546.
- Matiu collects data from 14 coastal towns on the number of beach umbrellas rented (x) and the number of jellyfish stings reported (y) in July. The Pearson correlation coefficient is r = 0.82. The mean of x is 62 with standard deviation 11, and the mean of y is 38 with standard deviation 9. Calculate the slope of the least-squares regression line for predicting jellyfish stings from beach umbrellas rented. Then, explain why this strong correlation does not imply that renting beach umbrellas causes jellyfish stings. Answer: 0.6709 Solution: Identify given values: r = 0.82, sx = 11, sy = 9. Slope formula: b = r * (sy / sx) = 0.82 * (9 / 11). Calculate 9 / 11 = 0.81818...
Full step-by-step solution
Step 1: Identify given values: r = 0.82, sx = 11, sy = 9.
Step 2: Slope formula: b = r * (sy / sx) = 0.82 * (9 / 11).
Step 3: Calculate 9 / 11 = 0.81818...
Step 4: Multiply: 0.82 * 0.81818 = 0.6709 (rounded to four decimal places).
Step 5: The slope means that for each additional beach umbrella rented, the number of jellyfish stings is predicted to increase by about 0.671.
Step 6: Correlation does not equal causation. A lurking variable, such as warmer ocean temperatures in July, causes more people to go to the beach (increasing umbrella rentals) and also increases jellyfish activity (leading to more stings). The umbrellas do not cause jellyfish stings; both are associated with the same external factor.
The answer is slope = 0.6709.
- Emma is a researcher studying the relationship between the number of hours students spend practicing a musical instrument per week and their scores on a standardized mathematics test. She collects data from 150 high school students and calculates a Pearson correlation coefficient of r = 0.55. The school board, upon seeing this result, proposes a policy requiring all students to spend at least 5 hours per week practicing an instrument, arguing that music practice directly improves math performance. As a critical thinker, explain why the school board's causal conclusion is flawed, and identify a plausible confounding variable that could explain the observed correlation. Answer: The correlation of r = 0.55 does not prove causation; a plausible confounding variable is that students who practice music may also come from families with higher socioeconomic status, which provides access to better math tutoring, educational resources, and more structured extracurricular activities, all of which could independently improve math scores. Solution: Understand what the correlation coefficient means. r = 0.55 indicates a moderate positive linear relationship between hours of music practice and math scores.
Full step-by-step solution
Step 1: Understand what the correlation coefficient means. r = 0.55 indicates a moderate positive linear relationship between hours of music practice and math scores. This means that, in the sample, students who practiced more tended to have higher math scores. Step 2: Recognize the difference between correlation and causation. Correlation measures the strength and direction of an association between two variables, but it does not establish that one variable causes changes in the other. For causation, we would need to rule out confounding variables, establish temporal precedence (practice before improvement), and ideally conduct a randomized controlled experiment. Step 3: Identify a plausible confounding variable. A strong candidate is socioeconomic status (SES). Students from higher SES families are more likely to have access to music lessons, instruments, and encouragement to practice. They also tend to have access to better math tutoring, educational materials, and less stressful home environments, all of which boost math scores. Thus, SES could cause both increased practice time and higher math scores, creating a correlation without music practice directly causing math improvement. Step 4: Conclude. The school board's conclusion is flawed because correlation does not equal causation. The observed r = 0.55 could be entirely due to the confounding variable of socioeconomic status, which affects both variables independently.
- A pharmaceutical company is testing a new medication and claims it reduces blood pressure by an average of 15 mmHg. In their clinical trial with 200 participants, they calculate a 95% confidence interval for the mean reduction as (12.8, 17.2) mmHg. The margin of error for this interval is 2.2 mmHg. If the company wants to reduce this margin of error to 1.1 mmHg while maintaining the same confidence level, how many participants would they need in their new trial? Answer: 800 Solution: We want ME2 = 1.1, same z and ฯ.
Full step-by-step solution
Let's go step by step.
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**Step 1: Understand the relationship between margin of error and sample size**
The margin of error (ME) for a confidence interval of the mean is:
ME = z * (ฯ / โn)
where:
- z is the z-score for the confidence level
- ฯ is the population standard deviation
- n is the sample size
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**Step 2: Find the standard deviation from the first trial**
We know:
- n1 = 200
- ME1 = 2.2
- 95% confidence level โ z โ 1.96
From ME1 = z * (ฯ / โn1):
2.2 = 1.96 * (ฯ / โ200)
Solve for ฯ:
ฯ = (2.2 * โ200) / 1.96
First, โ200 โ 14.1421
So ฯ โ (2.2 * 14.1421) / 1.96
2.2 * 14.1421 โ 31.11262
31.11262 / 1.96 โ 15.873
So ฯ โ 15.873 mmHg.
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**Step 3: Set up for the new margin of error**
We want ME2 = 1.1, same z and ฯ.
ME2 = z * (ฯ / โn2)
1.1 = 1.96 * (15.873 / โn2)
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**Step 4: Solve for n2**
First, 1.96 * 15.873 โ 31.11108
So 1.1 = 31.11108 / โn2
โn2 = 31.11108 / 1.1 โ 28.2828
n2 = (28.2828)^2 โ 799.71
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**Step 5: Round up**
Since sample size must be a whole number, n2 โ 800.
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**Step 6: Explanation of shortcut**
Notice: ME is proportional to 1/โn.
If ME is halved (2.2 โ 1.1), then:
ME1 / ME2 = โ(n2 / n1)
2.2 / 1.1 = โ(n2 / 200)
2 = โ(n2 / 200)
Square both sides:
4 = n2 / 200
n2 = 800
This avoids calculating ฯ explicitly.
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**Final answer:** 800 participants are needed.
- Hana collects data from 22 coastal towns on the number of beach umbrellas (x) and the number of jellyfish stings (y) reported in July. The Pearson correlation coefficient is r = 0.83, with sx = 15 and sy = 9. Calculate the slope of the least-squares regression line for predicting jellyfish stings from beach umbrellas, then explain why this strong correlation does not imply that beach umbrellas cause jellyfish stings. Answer: 0.498 Solution: Identify the given values: r = 0.83, sx = 15, sy = 9. Simplify the fraction: 9/15 = 3/5 = 0.6. Multiply: 0.83 * 0.6 = 0.498.
Full step-by-step solution
Step 1: Identify the given values: r = 0.83, sx = 15, sy = 9.
Step 2: Apply the slope formula: b = r * (sy / sx) = 0.83 * (9 / 15).
Step 3: Simplify the fraction: 9/15 = 3/5 = 0.6.
Step 4: Multiply: 0.83 * 0.6 = 0.498.
Step 5: The slope means that for each additional beach umbrella, the predicted number of jellyfish stings increases by 0.498.
Step 6: Correlation does not imply causation. A lurking variable, such as warmer ocean temperatures in July, causes more people to go to the beach (increasing umbrella count) and also increases jellyfish populations (leading to more stings). The umbrellas do not cause stings; both are associated with the same external factor.
The answer is slope = 0.498.