What is the difference between y and \hat{y}?

y is an actual observed value from your data, while \hat{y} is the predicted value calculated from the regression line. The difference y - \hat{y} is called a residual and represents how far off the prediction was for that particular data point.

Can I use the regression line to predict values outside the range of my original data?

You can use the equation to calculate predictions for any x (called extrapolation), but you should be cautious—the linear relationship observed in your data may not hold far outside the range you measured, so predictions become increasingly unreliable.

What does a negative slope in the regression equation mean?

A negative slope means that as x increases, \hat{y} decreases; this reflects a negative correlation, such as selling price decreasing as house age increases. The regression line slopes downward from left to right.

How do I know if my regression line is actually a good predictor?

Check the correlation coefficient r: values very close to -1 or +1 indicate a strong linear relationship and more reliable predictions, while values of r close to 0 indicate a weak relationship where the regression line may not predict well.

Least Squares Regression Line

The line of best fit that minimizes squared prediction errors, used to predict one variable from another in a linear relationship.

\hat{y} = b_0 + b_1 x, \quad b_1 = r\frac{s_y}{s_x}

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What each symbol means

What Least Squares Regression Line takes

Symbol	Meaning
$\hat{y}$	The predicted or fitted value of the response variable for a given $x$ —confusing it with actual $y$ makes you think the line passes through every data point when it actually minimizes overall error.
$b_0$	The y-intercept, the predicted value when $x = 0$ , calculated as $b_0 = \bar{y} - b_1 \bar{x}$ —forgetting to include it gives you only a slope with no complete equation.
$b_1$	The slope of the regression line, the change in $\hat{y}$ per unit increase in $x$ , calculated from $b_1 = r\frac{s_y}{s_x}$ —it always has the same sign as $r$ .
$r$	The correlation coefficient (Pearson's $r$ ), ranging from -1 to 1, measuring how closely data follow a linear pattern—it is not the slope, and confusing the two is a major error.
$x$	The predictor or independent variable used as input to make predictions—predictions outside the range of observed data become increasingly unreliable.

When to use it

When you have bivariate data with a linear trend and want to predict one variable from another.

Level

Usually taught in: Statistics · Appears on: AP Statistics

Worked examples

1. Clean regression line with positive correlation

Problem

A student recorded study time (hours) and test score for 10 classmates. The data shows

r = 0.9

s_x = 2

hours,

s_y = 6

points, mean study time

\bar{x} = 5

hours, and mean test score

\bar{y} = 70

points. Write the regression equation and predict the test score for a student who studies 7 hours.

$b_1 = r \frac{s_y}{s_x} = 0.9 \cdot \frac{6}{2}$
Apply the formula for slope using the correlation coefficient and the ratio of standard deviations.
$b_1 = 0.9 \cdot 3 = 2.7$
Simplify the fraction $\frac{6}{2} = 3$ , then multiply by 0.9 to get the slope.
$b_0 = \bar{y} - b_1 \bar{x} = 70 - 2.7(5) = 70 - 13.5 = 56.5$
Use the means and the slope to find the y-intercept so the line passes through $(\bar{x}, \bar{y}) = (5, 70)$ .
$\hat{y} = 56.5 + 2.7x$
Write the complete regression equation in the form $\hat{y} = b_0 + b_1 x$ .
$\hat{y} = 56.5 + 2.7(7) = 56.5 + 18.9 = 75.4$
Substitute $x = 7$ to predict the test score for a student who studied 7 hours.

Answer: $\hat{y} = 56.5 + 2.7x \text{; predicted score} = 75.4 \text{ points}$

We first calculate the slope using the formula $b_1 = r\frac{s_y}{s_x}$ , then find the y-intercept so the line passes through the center of the data. Finally, we substitute the given study time to make a prediction.

2. Regression line with negative correlation

Problem

A real estate agent analyzed 20 homes, finding that age (years) and selling price (thousands of dollars) have

r = -0.8

s_x = 2.5

years,

s_y = 25

thousand dollars, mean age

\bar{x} = 18

years, and mean price

\bar{y} = 300

thousand. Find the regression line and predict the selling price of a 25-year-old house.

$b_1 = r \frac{s_y}{s_x} = (-0.8) \cdot \frac{25}{2.5}$
Apply the slope formula, being careful to keep the negative sign from the correlation coefficient.
$b_1 = (-0.8) \cdot 10 = -8$
Divide $\frac{25}{2.5} = 10$ , then multiply by $-0.8$ to get a negative slope (older houses are cheaper).
$b_0 = \bar{y} - b_1 \bar{x} = 300 - (-8)(18) = 300 + 144 = 444$
Substitute carefully: subtracting a negative is addition, so $-(-8)(18) = +144$ .
$\hat{y} = 444 - 8x$
The negative slope appears as a minus sign in the equation; the regression line slopes downward.
$\hat{y} = 444 - 8(25) = 444 - 200 = 244$
Substitute $x = 25$ years to predict that a 25-year-old house sells for 244 thousand dollars.

Answer: $\hat{y} = 444 - 8x \text{; predicted price} = 244 \text{ thousand dollars}$

The negative correlation means older homes have lower prices, so the slope is negative. We carefully handle the negative sign when computing the y-intercept, then use the equation to predict a price for a specific age.

3. Word problem: coffee shop sales and temperature

Problem

A coffee shop recorded daily high temperature (°F) and iced coffee sales (units sold) for 30 days. The data shows

r = 0.9

, standard deviation of temperature

s_x = 4

°F, standard deviation of sales

s_y = 8

units, mean temperature

\bar{x} = 70

°F, and mean sales

\bar{y} = 140

units. Find the regression equation and estimate how many iced coffees the shop will sell on a day when the high temperature reaches 80°F.

$b_1 = r \frac{s_y}{s_x} = 0.9 \cdot \frac{8}{4}$
Use the correlation and standard deviations to find how much sales increase per degree of temperature.
$b_1 = 0.9 \cdot 2 = 1.8$
Simplify $\frac{8}{4} = 2$ , then multiply by 0.9; each degree Fahrenheit corresponds to 1.8 more units sold.
$b_0 = \bar{y} - b_1 \bar{x} = 140 - 1.8(70) = 140 - 126 = 14$
Calculate the y-intercept using the mean values so the line passes through the center of the data.
$\hat{y} = 14 + 1.8x$
The complete regression equation; the shop sells 14 units at a temperature of 0°F (extrapolation) plus 1.8 units per degree.
$\hat{y} = 14 + 1.8(80) = 14 + 144 = 158$
Substitute the temperature of 80°F to predict the number of iced coffees the shop will sell that day.

Answer: $\hat{y} = 14 + 1.8x \text{; predicted sales at 80°F} = 158 \text{ units}$

Higher temperatures correlate with more iced coffee sales, so we use the regression line to predict sales based on the forecast temperature. The slope 1.8 tells us the shop sells about two more iced coffees per degree increase.

Common mistakes

Where Least Squares Regression Line usually goes wrong

The mistake	Why it is wrong	The fix
Using $r$ directly as the slope: writing $\hat{y} = 0.8x + 5$ when the correlation is $r = 0.8$ .	The correlation coefficient $r$ measures the strength of the relationship, not the slope; you must convert it to slope using the standard deviations with the formula $b_1 = r\frac{s_y}{s_x}$ .	Always calculate $b_1 = r\frac{s_y}{s_x}$ before writing the equation.
Writing only the slope as the regression line: claiming the regression line is $\hat{y} = 2.5x$ without a y-intercept term.	The least squares regression line must have the form $\hat{y} = b_0 + b_1 x$ and must pass through the point $(\bar{x}, \bar{y})$ ; the y-intercept $b_0$ is essential and must be calculated as $b_0 = \bar{y} - b_1 \bar{x}$ .	Always compute $b_0 = \bar{y} - b_1 \bar{x}$ and include it in the equation as a separate term.
Assuming all data points lie on the regression line: stating 'the regression line passes through every point in the data set' or using actual $y$ values and predicted $\hat{y}$ values interchangeably.	The regression line is a model that minimizes the sum of squared residuals, not a curve that fits every point exactly; most real data points do not fall on the line, and that is by design.	Distinguish between observed values $y$ (from the data) and predicted values $\hat{y}$ (from the equation); the regression line balances errors across all points rather than fitting any single point perfectly.

Tips and when to use something else

The regression line always passes through the point $(\bar{x}, \bar{y})$ —verify this by substituting these values into your equation to check your work.
The slope $b_1$ always has the same sign as the correlation $r$ : positive correlation gives positive slope, negative correlation gives negative slope.
A strong correlation (close to ±1) produces a steep regression line and more reliable predictions, while a weak correlation (close to 0) produces a flatter line and less trustworthy forecasts.
If the correlation $r$ is very close to zero (weak linear relationship), the regression line may not be a good predictive tool—consider instead predicting using the sample mean $\bar{y}$ for all values of $x$ .

Related formulas

Correlation CoefficientCorrelation coefficient measures the strength and direction of a linear relationship between two quantitative variables, ranging from -1 to 1.Coefficient of DeterminationCoefficient of Determination (R²) indicates how much of the variation in your outcome variable is explained by a regression model's predictors.Arithmetic MeanArithmetic Mean is the average value found by adding all numbers in a dataset and dividing by how many numbers there are.Standard DeviationStandard deviation measures how spread out data points are from the average—use it to understand variability in a dataset.

Frequently asked questions

What is the difference between $y$ and $\hat{y}$ ?: $y$ is an actual observed value from your data, while $\hat{y}$ is the predicted value calculated from the regression line. The difference $y - \hat{y}$ is called a residual and represents how far off the prediction was for that particular data point.
Can I use the regression line to predict values outside the range of my original data?: You can use the equation to calculate predictions for any $x$ (called extrapolation), but you should be cautious—the linear relationship observed in your data may not hold far outside the range you measured, so predictions become increasingly unreliable.
What does a negative slope in the regression equation mean?: A negative slope means that as $x$ increases, $\hat{y}$ decreases; this reflects a negative correlation, such as selling price decreasing as house age increases. The regression line slopes downward from left to right.
How do I know if my regression line is actually a good predictor?: Check the correlation coefficient $r$ : values very close to -1 or +1 indicate a strong linear relationship and more reliable predictions, while values of $r$ close to 0 indicate a weak relationship where the regression line may not predict well.

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Reviewed 2026-09-18