AP Statistics Topic 5.3: Linear Regression Models

01 · Start here

The concept in 60 seconds

A linear regression model uses an explanatory variable to predict a response with a straight-line equation. The general form is:

\hat{y}=a+bx.

$x$ is the explanatory value you supply. $\hat{y}$ , read “y-hat,” is the response the model predicts. The coefficient $a$ is the intercept and $b$ is the slope.

Prediction is a model estimate. It is not a guaranteed result for an individual. An observed response is written $y$ ; the predicted response is written $\hat{y}$ .

For example, suppose delivery time in minutes is predicted from distance in kilometres by $\hat{y}=12+1.8x$ . A distance of $10\,\mathrm{km}$ gives $\hat{y}=12+1.8(10)=30\,\mathrm{min}$ . The model predicts a delivery time of $30$ minutes.

Quick check: what goes into the equation?

The explanatory value goes in for $x$ . You calculate $\hat{y}$ , the predicted response. You do not substitute an already observed response for $x$ .

02 · One situation, deeper understanding

A real situation: predict delivery time

A local delivery company records distance and total delivery time for $9$ deliveries on similar routes. Distance is measured in kilometres and time in minutes. These are invented teaching data.

Distance and time belong to the same delivery in each row.
Delivery ID	Distance $x$ (km)	Observed time $y$ (min)
A	$5$	$22$
B	$7.5$	$24.5$
C	$10$	$29$
D	$12.5$	$35.5$
E	$15$	$39$
F	$17.5$	$44.5$
G	$20$	$47$
H	$22.5$	$51.5$
I	$25$	$58$

Software supplies the fitted model below. For now, use the given equation; how least squares determines the coefficients is developed in Topic 5.5.

\hat{y}=12+1.8x.

Follow an input up to the prediction line

Dots are observed deliveries. The square marks a model prediction at a new distance; dashed guides connect the input to its predicted response. All data are invented.

Question: What time does the model predict for a delivery travelling $16\,\mathrm{km}$ ?

\hat{y}=12+1.8(16)=40.8.

Answer in context: For a delivery travelling $16$ kilometres on a comparable route, the model predicts a total time of about $40.8$ minutes. It does not promise that the delivery will take exactly that long.

The recorded distances span $5\le x\le25$ . Since $16$ lies inside that interval, this prediction is interpolation—even though there is no recorded delivery at exactly $16$ kilometres.

Why are the dots not all on the fitted line?

The dots show actual observations; the line shows model predictions. Traffic, loading and other factors create variability. For delivery C, the observed time is $29$ minutes at $10$ kilometres, while the model predicts $30$ minutes. Measuring the difference between observation and prediction is the focus of Topic 5.4.

03 · Understand the language

Key ideas and notation

Explanatory value: $x$

The input used to make the prediction. In the delivery model, it is distance in kilometres.

Observed response: $y$

The response actually recorded for an individual. A dot on the scatterplot represents an observed pair.

Predicted response: $\hat{y}$

The output from the fitted equation at a specified $x$ . A point on the regression line represents a model prediction.

Intercept: $a$

The predicted response when $x=0$ . Its units are the response units. Whether it has a useful real-world meaning depends on the context.

Slope: $b$

The change in the predicted response for a one-unit increase in the explanatory value. Its units are response units per explanatory unit.

Observed range

The interval from the smallest to the largest explanatory value used to fit the model. Check it before trusting a prediction.

Watch the hat: $y$ is measured; $\hat{y}$ is calculated from the model. Writing the hat helps keep evidence and prediction separate.

04 · Use a reliable method

Make a prediction in four steps

Identify the variables and units. Confirm which measurement is the input and which is being predicted.
Choose the explanatory value. Express it in the same units used in the model.
Substitute and calculate. Multiply before adding, keep the sign of the coefficient, and retain useful precision.
Interpret and check the range. State the predicted response with units and assess whether the input lies within the observed interval.

Apply the method to the delivery model

The input is distance in kilometres and the output is predicted time in minutes. For $x=16$ :

\hat{y}=12+1.8(16)

\hat{y}=12+28.8=40.8.

So the model predicts about $40.8$ minutes. The input lies between $5$ and $25$ kilometres, making this interpolation.

Keep the original units

If the distance is given as $16{,}000\,\mathrm{m}$ , convert it to $16\,\mathrm{km}$ before using this equation. Substituting $16{,}000$ into a model built for kilometres would produce an incorrect prediction.

Round at the end

Use the supplied coefficients as given. If software provides more digits than the written equation, use that precision unless the question specifies the rounded model. Write “approximately” when appropriate; a decimal output does not make the prediction exact.

05 · Connect the equation and graph

Read the line and its units

In $\hat{y}=12+1.8x$ , the intercept is $a=12$ minutes and the slope is $b=1.8$ minutes per kilometre. Each extra kilometre corresponds to an increase of $1.8$ minutes in the predicted time.

The intercept gives the model’s prediction at $x=0$ : $\hat{y}=12$ . But zero kilometres is outside the recorded interval and a zero-distance delivery may not make sense in this setting. Treat that intercept as part of the equation; do not claim the data establish an actual zero-distance delivery time.

Negative slopes work the same way

Suppose an invented battery study provides $\hat{y}=98-6x$ , where $x$ is hours of use and $\hat{y}$ is predicted remaining charge as a percentage. The observed times span $0\le x\le10$ .

A decreasing model can still predict a positive response

The line is the supplied battery model over the observed time interval. The square marks the prediction after four hours; no individual measurements are shown.

\hat{y}=98-6(4)=74.

At $4$ hours, the model predicts $74\%$ remaining charge. A negative slope means predictions decrease as the input increases; it does not mean the predicted response must be negative.

This section helps you read a supplie[Truncated]