Wize University Statistics Textbook > Inference for Linear Regression

Prediction Intervals for Y

Prediction Interval of Y

We know how to use the regression line to make point predictions. Specifically, we get a predicted value $\hat{y}_i$ for every $x_i$ . We take this a step further by making prediction intervals.

When we make a point estimate of $\hat{y}$ for a given single observation $x_o$ we construct a prediction interval to determine how closely $\hat{y}$ matches the true value of $y$ . A prediction interval is a range that is likely to contain the true response variation given a single explanatory variable based on the linear model.

Image illustrating Prediction Interval of Y

Distance Value

\displaystyle\boxed{d=\frac{1}{n}+\frac{\left(x_o-\overline{x}\right)^2}{SS_{xx}}}

Prediction interval for an individual value of y

\displaystyle\boxed{\hat{y}\pm t_\frac{\alpha}{2}S_e\sqrt{1+ d}}

\displaystyle\boxed{\hat{y}\pm t_\frac{\alpha}{2}S_e\sqrt{1+ \frac{1}{n}+\frac{\left(x_o-\overline{x}\right)^2}{SS_{xx}}}}

Several things could cause a wider prediction interval

1) A large difference between $x_o$ and $\bar x$

2) A large residual standard deviation $S_e$

3) A larger prediction interval (e.g. a 95% PI is wider than a 90% PI)

How to construct a prediction interval

Step 1: Find $\hat{y}$ given $x_o$ where $\hat{y}=b_o+b_1x_o$

Step 2: Find $SS_{xx}$ and $S_e$

Step 3: Find $t_\frac{\alpha}{2}$ for a $100(1-a)$ prediction interval.

Step 4: Construct a prediction interval $\displaystyle{\hat{y}\pm t_\frac{\alpha}{2}S_e\sqrt{1+ \frac{1}{n}+\frac{\left(x_o-\overline{x}\right)^2}{SS_{xx}}}}$

Example: Prediction Interval for Y

Consider this regression line:

$\hat{y}=89.25+19.257x$

You are given the following summary:

$n=12$

$\sum x=35$

$\sum x^2=141$

$\sum_{ }^{ }y=1745$

$\sum_{ }^{ }y^2=286001$

$\sum xy=5839$

Construct a 95% prediction interval for $x_o=5$

Step 1: Find $\hat{y}$ given $x_o$ where $\hat{y}=b_o+b_1x_o$

$\hat{y}=b_o+b_1x_o$

$\hat{y}=89.25+19.257(5)=185.535$

Step 2: Find $SS_{xx}$ and $S_e$

$k=1$ (since there is one explanatory variable in a simple linear regression model)

$\displaystyle{SS_{xx}=\sum_{ }^{ }x_{ }^2-\frac{(\sum_{ }^{ }x_{ })^2}{n}}$

$\displaystyle{SS_{xx}=141-\frac{\left(35\right)^2}{12}=38.92}$

You will need SSE to find $S_e$ :

$SSE=\sum_{ }^{ }y_{ }^2-b_0\sum_{ }^{ }y_{ }-b_1\sum_{ }^{ }xy_{ }$

$SSE=286001-89.25\left(1745\right)-19.257\left(5839\right)$

$SSE=17818$

$\displaystyle{s_e=\sqrt{\frac{SSE}{n-k-1}}}$

$\displaystyle{S_e=\sqrt{\frac{17818}{12-1-1}}=42.21}$

Image illustrating Example

Step 3: Find $t^{\star}$ for a 95% prediction interval.

$df=n-k-1$

$df=12-1-1=10$

$t^{\star}=2.228$

Step 4: Construct a prediction interval

$\displaystyle\boxed{\hat{y}\pm t_\frac{\alpha}{2}S_e\sqrt{1+ \frac{1}{n}+\frac{\left(x_o-\overline{x}\right)^2}{SS_{xx}}}}$

Distance Value:

d=\frac{1}{n}+\frac{\left(x_o-\overline{x}\right)^2}{SS_{xx}}

$\displaystyle{\overline{x}=\frac{\sum_{ }^{ }x}{n}=\frac{35}{12}=2.9167}$

d=\frac{1}{12}+\frac{\left(5-2.9167\right)^2}{38.92}

d=0.1949

Prediction interval for an individual value of y:

\boxed{\hat{y}\pm t_\frac{\alpha}{2}S_e\sqrt{1+ d}}

185.535\pm (2.228)(42.21)\sqrt{1+ 0.1949}

185.535\pm102.801

[82.734, 288.336]

Given $x_o=5$ , we predict that $y$ is between these two values in the 95% prediction interval.

We want to predict how much money a visitor spends at a pet convention on a given day based on the number of venders.

Illustration for practice question: Confidence Intervals: Predictions

$\sum x=622$

$\sum x^2=36714$

$\sum_{ }^{ }y=1060$

$\sum_{ }^{ }y^2=137488$

$\sum xy=65655$

Construct a 90% prediction interval to estimate the spending of a visitor given that there are 65 vendors. [Enter whole numbers only without the $ sign. Example: 50]

[Click on 'HINT' for steps, formulas, and t-table.]

Given that there are $x_o=65$ venders, we predict that a visitor would spend between $ and $ on a given day.

I don't know

Practice: Prediction Interval for Y

Which of the following is NOT a reason that a prediction interval is wide?

The standard error is low.

You construct a 99% prediction interval instead of a 95% prediction interval.

The difference between the predictor value $x_o$ and the mean $\overline{x}$ is very large.

None of the above.

I don't know

Extra Practice

Prediction Intervals for Y

Extrapolation refers to

Prediction of Intervals for Y

Extrapolation refers to

Prediction Intervals for Y

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Prediction Interval of Y

Several things could cause a wider prediction interval

How to construct a prediction interval

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Example: Prediction Interval for Y

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Practice: Prediction Interval for Y

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Extra Practice