Teacher Guide: Correlation Word Problems

Solve real correlation problems: compute r step by step from a small data set, interpret it in context, judge causal claims and see how an outlier changes r.

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Class quiz

10 questions on Correlation. Students join with a name, you see everyone's score.

For Teachers

Learning Objectives

Compute the correlation coefficient $r$ by hand for a data set of 4 to 6 points
Interpret the direction and strength of $r$ in the context of a word problem
Decide whether a causal claim is justified and name a possible confounding variable
Describe how an outlier changes $r$ and decide when it may be removed
Check calculations with the zero-sum of deviations and the range $-1 \le r \le 1$

Prerequisites

• Introduction to Correlation: positive, negative and no correlation in scatter plots
• Correlation Coefficient: the formula for $r$ and the strength scale
• Computing the mean of a small data set
• Squares and square roots with a calculator

Discussion Starters

1. The correlation between umbrella sales and accidents is $r = 0.95$ . How could you check whether the rain, and not the umbrellas, explains it?
2. If you measured study time in minutes instead of hours, would $r$ change? Why or why not?
3. When is it honest to remove an outlier, and when is it cheating?
4. Why could a correlation be very strong in one town and weak in another for the same two variables?

Common Misconceptions

A bigger $S_{xy}$ always means a stronger correlation

Remediation: Show two data sets with the same $r$ but different units: $S_{xy}$ grows with the units, $r$ does not. Only the ratio $\frac{S_{xy}}{\sqrt{S_{xx} S_{yy}}}$ measures strength.

If $r$ is close to $1$ , the claim of cause and effect must be true

Remediation: Collect classic examples (umbrellas and traffic accidents, TVs and life expectancy, shoe size and reading) and let students find the confounding variable in each.

One point cannot matter much in a data set

Remediation: Use the fertilizer example: compute $r$ for four points ( $r = 1$ ), then add the fifth point and recompute ( $r = 0$ ).

Differentiation Ideas

For Struggling Students:

• Give a deviation table with the columns already labeled: $x_i - \bar{x}$ , $y_i - \bar{y}$ , product, squares
• Use data sets whose means are whole numbers
• Let students check each step with the calculator before moving on

For On-Level Students:

• Compute $r$ for 4 to 6 points and round to 2 decimal places
• Write a full context sentence for each $r$ (direction, strength, variables)
• Judge causal claims and name a confounding variable

For Advanced Students:

• Find a fifth point that turns a positive $r$ negative
• Prove that adding a constant to every $y$ value leaves $r$ unchanged
• Design a controlled experiment that could test a causal claim from the news

Standards Alignment

HSS.ID.C.8 (CCSS.MATH.CONTENT.HSS.ID.C.8)
Compute (using technology) and interpret the correlation coefficient of a linear fit
HSS.ID.C.9 (CCSS.MATH.CONTENT.HSS.ID.C.9)
Distinguish between correlation and causation
HSS.ID.B.6 (CCSS.MATH.CONTENT.HSS.ID.B.6)
Represent data on two quantitative variables on a scatter plot, and describe how the variables are related

Lesson Resources

visualScatter Plots with Trend Line
The study-hours data and the outlier data set, with the trend line drawn
activityDeviation Table Race
Pairs fill in a table of deviations, products and squares for a 5-point data set and compare their $r$
worksheetHeadline Detective
Students judge six news headlines based on correlations and name a confounding variable for each

Lesson Content

Everything students see: definition, examples, common mistakes, applications. Tap to open.

Definition

A correlation word problem gives you paired data from a real situation and asks you to measure, interpret and judge the relationship. You already know the formula for the correlation coefficient

r

from the lesson *Correlation Coefficient*. Here you use it as a full method.

The five-step method for a small data set:

1.Find the means $\bar{x}$ and $\bar{y}$ .
2.Find each deviation $x_i - \bar{x}$ and $y_i - \bar{y}$ .
3.Add the products: $S_{xy} = \sum(x_i - \bar{x})(y_i - \bar{y})$ .
4.Add the squares: $S_{xx} = \sum(x_i - \bar{x})^2$ and $S_{yy} = \sum(y_i - \bar{y})^2$ .
5.Divide:
$r = \frac{S_{xy}}{\sqrt{S_{xx} \cdot S_{yy}}}$

Then answer the question in context:

Direction: the sign of $r$ (positive: both rise together; negative: one rises while the other falls).
Strength: the size of $|r|$ (scale below).
Causation: a strong $r$ alone never proves that one variable causes the other. Look for a confounding variable (also called a lurking variable) and ask whether the data come from a controlled experiment.
Outliers: one unusual point can make $r$ much stronger, much weaker or even change its sign.

	r		Strength
$0.00$ to $0.19$	very weak
$0.20$ to $0.39$	weak
$0.40$ to $0.59$	moderate
$0.60$ to $0.79$	strong
$0.80$ to $1.00$	very strong

Worked Examples

Five students recorded their study hours and quiz scores:

(1, 64)

(2, 63)

(3, 77)

(4, 71)

(5, 75)

. Compute

r

and interpret it.

Find the means

\bar{x} = \frac{1+2+3+4+5}{5} = 3

\bar{y} = \frac{64+63+77+71+75}{5} = \frac{350}{5} = 70

→

\bar{x} = 3

\bar{y} = 70

Find the deviations

x_i - \bar{x}

-2, -1, 0, 1, 2

y_i - \bar{y}

-6, -7, 7, 1, 5

→ Each deviation sum is

0

(a quick check)

Multiply the deviations and add

(-2)(-6) + (-1)(-7) + (0)(7) + (1)(1) + (2)(5) = 12 + 7 + 0 + 1 + 10

→

S_{xy} = 30

Square the deviations and add

S_{xx} = 4 + 1 + 0 + 1 + 4 = 10

S_{yy} = 36 + 49 + 49 + 1 + 25 = 160

→

S_{xx} = 10

S_{yy} = 160

Apply the formula

r = \frac{30}{\sqrt{10 \cdot 160}} = \frac{30}{\sqrt{1600}} = \frac{30}{40}

→

r = 0.75

Interpret in context

Positive sign,

|r| = 0.75

is in the band

0.60

0.79

→ Strong positive correlation

Answer: $r = 0.75$ : a strong positive correlation. Students who studied longer tended to score higher, although the student who studied 1 hour scored better than the one who studied 2 hours, so the pattern is not perfect.

Common Mistakes

Multiplying the raw values $x_i y_i$ instead of the deviations

Why it's wrong: The formula $r = \frac{S_{xy}}{\sqrt{S_{xx} S_{yy}}}$ uses the distances from the means. Raw products are always large and positive for positive data and give a wrong $r$ .

Correct: Subtract the means first. A quick check: the deviations of each variable add up to $0$ .

Forgetting the square root or rounding too early

Why it's wrong: $\frac{30}{10 \cdot 160} \approx 0.019$ is not $r$ . Rounding $\sqrt{520}$ to $23$ gives $0.957$ instead of $0.965$ .

Correct: Take the square root of $S_{xx} \cdot S_{yy}$ and round only the final answer to 2 decimal places.

Treating a strong $r$ as proof that one variable causes the other

Why it's wrong: Observational data can show a strong correlation because a confounding variable drives both, as rainy weather drives umbrella sales and traffic accidents.

Correct: Say "is associated with". Only a randomized controlled experiment can justify a causal claim.

Ignoring an outlier or deleting it without a reason

Why it's wrong: One point can change $r$ from $1$ to $0$ . Ignoring it hides the problem; deleting it just because it is inconvenient is dishonest.

Correct: Plot the data, compute $r$ with and without the outlier and remove it only with a clear reason (a measuring error, a snail), stating that you did.

Why It Matters

Correlation numbers appear in news articles, school reports and science projects every day, and many of them are misread:

Health news: "People who drink coffee live longer" is usually a correlation from a survey, not proof.
Sports: coaches check whether training hours really go with better race times before changing a plan.
Business and energy: a shop compares temperature with sales, a family compares outdoor temperature with heating use.
Science projects: one broken measurement (an outlier) can ruin a result if nobody checks the scatter plot.

If you can compute

r

yourself, describe it in plain words and spot a false causal claim or a misleading outlier, you can judge statistics instead of just believing them.

Real World Applications

Sports Science

Coaches compare weekly training hours with race times to check whether more training goes with faster runs.

Example:

For a running club the sums are $S_{xy} = -45$ , $S_{xx} = 25$ and $S_{yy} = 100$ . A negative $r$ means: more training, lower (faster) times.

1Try It Yourself

A running club records training hours per week and 5 km times (minutes) for its members. The sums are $S_{xy} = -45$ , $S_{xx} = 25$ and $S_{yy} = 100$ .

What is the correlation coefficient $r$ ?

Step 1: Write the mathematical expression

Compute $r = \frac{S_{xy}}{\sqrt{S_{xx} \cdot S_{yy}}}$ :

Reading the News Critically

Journalists and readers must decide whether a reported correlation supports a claim about cause and effect.

Example:

A report finds $r = 0.83$ between the number of TVs per 100 people in a country and life expectancy, and a blog says TVs make people live longer.

2Try It Yourself

Across many countries, the number of TVs per 100 people and life expectancy have $r = 0.83$ . A blog concludes that buying more TVs makes people live longer.

Is the causal claim justified?

Step 1: Choose the right answer

Choose your verdict:

Quality Control of Sensors

Engineers check measurement data for faulty readings, because one bad value can hide a perfect relationship.

Example:

A temperature sensor is tested at five settings: $(2, 10)$ , $(4, 14)$ , $(6, 18)$ , $(8, 22)$ and a faulty reading $(10, 6)$ .

3Try It Yourself

A sensor gives the readings $(2, 10)$ , $(4, 14)$ , $(6, 18)$ , $(8, 22)$ , $(10, 6)$ . The last reading is a known error. The means are $\bar{x} = 6$ and $\bar{y} = 14$ .

What is $r$ once the faulty reading is removed?

Step 1: Write the mathematical expression

Compare $r$ with and without the error:

Key Takeaways

1To compute $r$ by hand: means, deviations, $S_{xy}$ , $S_{xx}$ , $S_{yy}$ , then $r = \frac{S_{xy}}{\sqrt{S_{xx} S_{yy}}}$
2Check your work: the deviations of each variable add up to $0$ , and $r$ must lie between $-1$ and $1$
3Round only the final value of $r$ , usually to 2 decimal places
4Interpret $r$ in context: the sign gives the direction, $|r|$ gives the strength
5A strong correlation from observational data does not justify a causal claim; look for confounding variables
6One outlier can strengthen, weaken or even reverse $r$ : plot the data and compare $r$ with and without it

Frequently Asked Questions

Why do the deviations always add up to zero?

The mean is the balance point of the data: the values above it and below it cancel out exactly. If your deviations do not add up to

0

, you have a mistake in the mean or in a subtraction.

Does adding the same number to every value change $r$ ?

No. Adding a constant moves all points together, so the deviations stay the same. Multiplying by a positive number (for example hours to minutes) does not change

r

either. Multiplying one variable by a negative number flips the sign of

r

When is it fine to remove an outlier?

When you have a clear reason that is not about the result: a measuring error, a typing mistake, a plant eaten by a snail. Always say that you removed it and, if possible, report

r

both with and without the point.

How can we ever show causation?

With a randomized controlled experiment: assign the treatment (for example fertilizer) at random to some subjects and not to others, keep everything else the same, and compare. Random assignment spreads confounding variables evenly over both groups.

Glossary

Deviation: The distance of a value from the mean, $x_i - \bar{x}$ ; it can be positive or negative
$S_{xy}$: The sum of the products of deviations, $\sum(x_i - \bar{x})(y_i - \bar{y})$ ; its sign is the sign of $r$
$S_{xx}$ and $S_{yy}$: The sums of squared deviations of $x$ and of $y$ ; they are never negative
Outlier: A data point far from the pattern of the others; it can change $r$ a lot
Confounding variable: A variable not included in the study that affects both variables and can create a misleading correlation
Controlled experiment: A study in which the researcher assigns the treatment at random; the only kind of study that can justify a causal claim

Formula Card

Correlation coefficient

r = \frac{S_{xy}}{\sqrt{S_{xx} \cdot S_{yy}}}

$S_{xy} = \sum(x_i - \bar{x})(y_i - \bar{y})$, $S_{xx} = \sum(x_i - \bar{x})^2$, $S_{yy} = \sum(y_i - \bar{y})^2$

Checks

\sum(x_i - \bar{x}) = 0, \quad -1 \le r \le 1

If a deviation sum is not $0$ or $|r| > 1$, there is a calculation error

Strength scale

|r|: \ 0.2 \ \text{weak}, \ 0.4 \ \text{moderate}, \ 0.6 \ \text{strong}, \ 0.8 \ \text{very strong}

Each word starts at its value; below $0.20$ the correlation is very weak

Teacher Guide: Correlation Word Problems

Use this lesson with your class

For Teachers

For Struggling Students:

For On-Level Students:

For Advanced Students:

Lesson Content

Definition

Worked Examples

Common Mistakes

Why It Matters

Real World Applications

Sports Science

Reading the News Critically

Quality Control of Sensors

Key Takeaways

Frequently Asked Questions

Glossary

Formula Card

Correlation coefficient

Checks

Strength scale

More in This Topic

Introduction to Correlation

Correlation Coefficient

Interpreting Correlation

Correlation vs Causation