Correlation Word Problems

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 |

Study Hours and Quiz Scores

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 $0.60$ to $0.79$ = 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.

Outdoor Temperature and Heating

A family records the average outdoor temperature (°C) and the energy used for heating (kWh) on five days: $(0, 38)$, $(5, 34)$, $(10, 26)$, $(15, 30)$, $(20, 22)$. Compute $r$ and describe the relationship.

Find the means: $\bar{x} = \frac{0+5+10+15+20}{5} = 10$, $\bar{y} = \frac{38+34+26+30+22}{5} = \frac{150}{5} = 30$ = $\bar{x} = 10$, $\bar{y} = 30$

Find the deviations: $x_i - \bar{x}$: $-10, -5, 0, 5, 10$ $y_i - \bar{y}$: $8, 4, -4, 0, -8$ = Deviations found

Multiply the deviations and add: $(-10)(8) + (-5)(4) + (0)(-4) + (5)(0) + (10)(-8) = -80 - 20 + 0 + 0 - 80$ = $S_{xy} = -180$

Square the deviations and add: $S_{xx} = 100 + 25 + 0 + 25 + 100 = 250$ $S_{yy} = 64 + 16 + 16 + 0 + 64 = 160$ = $S_{xx} = 250$, $S_{yy} = 160$

Apply the formula: $r = \frac{-180}{\sqrt{250 \cdot 160}} = \frac{-180}{\sqrt{40000}} = \frac{-180}{200}$ = $r = -0.90$

Interpret in context: Negative sign, $|r| = 0.90$ is in the band $0.80$ to $1.00$ = Very strong negative correlation

Answer: $r = -0.90$: a very strong negative correlation. On warmer days the family used less energy for heating.

Is the Causal Claim Justified?

Over ten autumn weeks a city recorded umbrella sales and traffic accidents and found $r = 0.95$. A local newspaper writes: "Umbrellas cause accidents." Is this claim justified?

Describe direction and strength: $r = 0.95 > 0$ and $|r| = 0.95$ is in the band $0.80$ to $1.00$ = Very strong positive correlation

Ask how the data were collected: Nobody was told to buy or not buy an umbrella; the city only counted what happened = Observational data, not an experiment

Look for a confounding variable: Rainy weeks make people buy umbrellas, and wet roads and poor visibility also cause more accidents = Confounding variable: rainy weather

Give a verdict: The correlation is real, but the rain explains both variables = The causal claim is not justified

Answer: Not justified. The correlation is very strong ($r = 0.95$), but the data are observational and rainy weather drives both umbrella sales and accidents. Only a controlled experiment could test a cause.

One Outlier Changes Everything

A student measures fertilizer (g) and plant growth (cm): $(1, 2)$, $(2, 4)$, $(3, 6)$, $(4, 8)$, $(5, 0)$. The last plant was eaten by a snail. Compute $r$ with all five plants and compare it with $r$ for the first four.

Look at the first four points: $(1, 2), (2, 4), (3, 6), (4, 8)$ all lie on the line $y = 2x$ = Without the outlier: $r = 1$

Find the means of all five points: $\bar{x} = \frac{15}{5} = 3$, $\bar{y} = \frac{2+4+6+8+0}{5} = \frac{20}{5} = 4$ = $\bar{x} = 3$, $\bar{y} = 4$

Find the deviations: $x_i - \bar{x}$: $-2, -1, 0, 1, 2$ $y_i - \bar{y}$: $-2, 0, 2, 4, -4$ = Deviations found

Multiply the deviations and add: $(-2)(-2) + (-1)(0) + (0)(2) + (1)(4) + (2)(-4) = 4 + 0 + 0 + 4 - 8$ = $S_{xy} = 0$

Apply the formula: $S_{xx} = 10$, $S_{yy} = 4 + 0 + 4 + 16 + 16 = 40$, so $r = \frac{0}{\sqrt{10 \cdot 40}} = \frac{0}{20}$ = With the outlier: $r = 0$

Decide what to report: The snail is a known reason that has nothing to do with fertilizer = Report both values and explain why the outlier was removed

Answer: With the outlier $r = 0$, without it $r = 1$. A single point erased a perfect correlation. Because the outlier has a clear reason (the snail), it is fair to remove it, as long as you say so.

Sunshine and Pool Visitors (Rounding r)

An open-air pool compares hours of sunshine with visitors (in hundreds) on four days: $(2, 4)$, $(4, 5)$, $(6, 9)$, $(8, 10)$. Compute $r$ to 2 decimal places and interpret it.

Find the means: $\bar{x} = \frac{2+4+6+8}{4} = 5$, $\bar{y} = \frac{4+5+9+10}{4} = \frac{28}{4} = 7$ = $\bar{x} = 5$, $\bar{y} = 7$

Find the deviations: $x_i - \bar{x}$: $-3, -1, 1, 3$ $y_i - \bar{y}$: $-3, -2, 2, 3$ = Deviations found

Compute the three sums: $S_{xy} = 9 + 2 + 2 + 9 = 22$ $S_{xx} = 9 + 1 + 1 + 9 = 20$ $S_{yy} = 9 + 4 + 4 + 9 = 26$ = $S_{xy} = 22$, $S_{xx} = 20$, $S_{yy} = 26$

Apply the formula and round at the end: $r = \frac{22}{\sqrt{20 \cdot 26}} = \frac{22}{\sqrt{520}} \approx \frac{22}{22.80} \approx 0.9648$ = $r \approx 0.96$

Interpret in context: Positive sign, $|r| \approx 0.96$ is in the band $0.80$ to $1.00$ = Very strong positive correlation

Answer: $r \approx 0.96$: a very strong positive correlation. Sunnier days went with more visitors. It is believable that sunshine attracts swimmers, but four observed days still do not prove a cause.

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.

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

Why: 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 $0$.

Mistake: Forgetting the square root or rounding too early

Why: $\frac{30}{10 \cdot 160} \approx 0.019$ 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.

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

Why: 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.

Mistake: Ignoring an outlier or deleting it without a reason

Why: 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.

Sports Science

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

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.

Reading the News Critically

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

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.

Quality Control of Sensors

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

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

To compute $r$ by hand: means, deviations, $S_{xy}$, $S_{xx}$, $S_{yy}$, then $r = \frac{S_{xy}}{\sqrt{S_{xx} S_{yy}}}$

Check your work: the deviations of each variable add up to $0$, and $r$ must lie between $-1$ and $1$

Round only the final value of $r$, usually to 2 decimal places

Interpret $r$ in context: the sign gives the direction, $|r|$ gives the strength

A strong correlation from observational data does not justify a causal claim; look for confounding variables

One outlier can strengthen, weaken or even reverse $r$: plot the data and compare $r$ with and without it

Q: Why do the deviations always add up to zero?

A: 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.

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

A: 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 $r$.

Q: When is it fine to remove an outlier?

A: 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.

Q: How can we ever show causation?

A: 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.

Exit Presentation

Correlation Word Problems

1 / 14

Statistics / Correlation

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 min

Use arrow keys or spacebar to navigate