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.

Advanced35 min18 problems
Practice Problems

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 rr from the lesson *Correlation Coefficient*. Here you use it as a full method.
The five-step method for a small data set:
  1. 1.Find the means xˉ\bar{x} and yˉ\bar{y}.
  2. 2.Find each deviation xi−xˉx_i - \bar{x} and yi−yˉy_i - \bar{y}.
  3. 3.Add the products: Sxy=∑(xi−xˉ)(yi−yˉ)S_{xy} = \sum(x_i - \bar{x})(y_i - \bar{y}).
  4. 4.Add the squares: Sxx=∑(xi−xˉ)2S_{xx} = \sum(x_i - \bar{x})^2 and Syy=∑(yi−yˉ)2S_{yy} = \sum(y_i - \bar{y})^2.
  5. 5.Divide:
    r=SxySxx⋅Syyr = \frac{S_{xy}}{\sqrt{S_{xx} \cdot S_{yy}}}
Then answer the question in context:
  • Direction: the sign of rr (positive: both rise together; negative: one rises while the other falls).
  • Strength: the size of ∣r∣|r| (scale below).
  • Causation: a strong rr 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 rr much stronger, much weaker or even change its sign.
rStrength
0.000.00 to 0.190.19very weak
0.200.20 to 0.390.39weak
0.400.40 to 0.590.59moderate
0.600.60 to 0.790.79strong
0.800.80 to 1.001.00very strong

Try it now

A survey of teenagers finds r=−0.85r = -0.85 between daily screen time and hours of sleep. Which description is correct?

Worked Examples

Five students recorded their study hours and quiz scores: (1,64)(1, 64), (2,63)(2, 63), (3,77)(3, 77), (4,71)(4, 71), (5,75)(5, 75). Compute rr and interpret it.
1

Find the means

xˉ=1+2+3+4+55=3\bar{x} = \frac{1+2+3+4+5}{5} = 3, yˉ=64+63+77+71+755=3505=70\bar{y} = \frac{64+63+77+71+75}{5} = \frac{350}{5} = 70 → xˉ=3\bar{x} = 3, yˉ=70\bar{y} = 70
2

Find the deviations

xi−xˉx_i - \bar{x}: −2,−1,0,1,2-2, -1, 0, 1, 2 yi−yˉy_i - \bar{y}: −6,−7,7,1,5-6, -7, 7, 1, 5 → Each deviation sum is 00 (a quick check)
3

Multiply the deviations and add

(−2)(−6)+(−1)(−7)+(0)(7)+(1)(1)+(2)(5)=12+7+0+1+10(-2)(-6) + (-1)(-7) + (0)(7) + (1)(1) + (2)(5) = 12 + 7 + 0 + 1 + 10 → Sxy=30S_{xy} = 30
4

Square the deviations and add

Sxx=4+1+0+1+4=10S_{xx} = 4 + 1 + 0 + 1 + 4 = 10 Syy=36+49+49+1+25=160S_{yy} = 36 + 49 + 49 + 1 + 25 = 160 → Sxx=10S_{xx} = 10, Syy=160S_{yy} = 160
5

Apply the formula

r=3010⋅160=301600=3040r = \frac{30}{\sqrt{10 \cdot 160}} = \frac{30}{\sqrt{1600}} = \frac{30}{40} → r=0.75r = 0.75
6

Interpret in context

Positive sign, ∣r∣=0.75|r| = 0.75 is in the band 0.600.60 to 0.790.79 → Strong positive correlation

Common Mistakes

Multiplying the raw values xiyix_i y_i instead of the deviations

Why it's wrong: The formula r=SxySxxSyyr = \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 rr.

Correct: Subtract the means first. A quick check: the deviations of each variable add up to 00.

Forgetting the square root or rounding too early

Why it's wrong: 3010⋅160≈0.019\frac{30}{10 \cdot 160} \approx 0.019 is not rr. Rounding 520\sqrt{520} to 2323 gives 0.9570.957 instead of 0.9650.965.

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

Treating a strong rr 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 rr from 11 to 00. Ignoring it hides the problem; deleting it just because it is inconvenient is dishonest.

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

Interactive Visual

Scatter Plot

Study hours and quiz scores (r = 0.75)

123456065707580Hours studiedQuiz score

ŷ = 3x + 61

r = 0.75 · strong positive correlation

Tap a point to see its coordinates.

Scatter Plot

One outlier: r drops from 1 to 0

1234502468Fertilizer (g)Growth (cm)Outlier

ŷ = 4

r = 0 · no linear correlation

Tap a point to see its coordinates.

Interactive Sandbox

Expression Calculator

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Practice Problems

18 problems
Problem 1 of 18
Easy
A survey of teenagers finds r=−0.85r = -0.85 between daily screen time and hours of sleep. Which description is correct?

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 rr 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 Sxy=−45S_{xy} = -45, Sxx=25S_{xx} = 25 and Syy=100S_{yy} = 100. A negative rr 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 Sxy=−45S_{xy} = -45, Sxx=25S_{xx} = 25 and Syy=100S_{yy} = 100.

What is the correlation coefficient rr?

Step 1: Write the mathematical expression

Compute r=SxySxx⋅Syyr = \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=0.83r = 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=0.83r = 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)(2, 10), (4,14)(4, 14), (6,18)(6, 18), (8,22)(8, 22) and a faulty reading (10,6)(10, 6).

3Try It Yourself

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

What is rr once the faulty reading is removed?

Step 1: Write the mathematical expression

Compare rr with and without the error:

Key Takeaways

  • 1To compute rr by hand: means, deviations, SxyS_{xy}, SxxS_{xx}, SyyS_{yy}, then r=SxySxxSyyr = \frac{S_{xy}}{\sqrt{S_{xx} S_{yy}}}
  • 2Check your work: the deviations of each variable add up to 00, and rr must lie between −1-1 and 11
  • 3Round only the final value of rr, usually to 2 decimal places
  • 4Interpret rr in context: the sign gives the direction, ∣r∣|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 rr: plot the data and compare rr with and without it

Frequently Asked Questions

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 00, you have a mistake in the mean or in a subtraction.
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 00, you have a mistake in the mean or in a subtraction.
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 rr either. Multiplying one variable by a negative number flips the sign of rr.
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 rr both with and without the point.
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, xi−xˉx_i - \bar{x}; it can be positive or negative
SxyS_{xy}
The sum of the products of deviations, ∑(xi−xˉ)(yi−yˉ)\sum(x_i - \bar{x})(y_i - \bar{y}); its sign is the sign of rr
SxxS_{xx} and SyyS_{yy}
The sums of squared deviations of xx and of yy; they are never negative
Outlier
A data point far from the pattern of the others; it can change rr 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=SxySxx⋅Syyr = \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

∑(xi−xˉ)=0,−1≤r≤1\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 weak, 0.4 moderate, 0.6 strong, 0.8 very strong|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

Today's challengeOne problem, three tries

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