Math Core

In Unit 2 you fit a least-squares line to a set of data and interpreted its slope. But that slope came from one sample. A different random sample would give a different line and a different slope. In this lesson you'll build a confidence interval that estimates the slope of the true relationship in the whole population, using the same "statistic plus or minus margin of error" idea you used for proportions and means.

The population regression line

Imagine every individual in a population, not just the ones in your sample. For each value of $x$ , the responses $y$ have a distribution with some mean $\mu_y$ . If those means fall on a straight line, that line is the population regression line:

$\mu_y = \alpha + \beta x$

The slope $\beta$ (beta) and intercept $\alpha$ (alpha) are parameters. You never know them exactly. Your sample gives the least-squares line $\hat{y} = a + bx$ , and its slope $b$ is a statistic that estimates $\beta$ .

Definition

Population slope and sample slope

The population slope $\beta$ is the change in the mean response $\mu_y$ for each one-unit increase in $x$ , for the whole population. The sample slope $b$ is the slope of the least-squares line computed from a sample. It is a point estimate of $\beta$ .

How much does the sample slope vary?

If you took many random samples of the same size and computed $b$ each time, the values of $b$ would form a sampling distribution. When the conditions below are met, that distribution is approximately Normal, centered at $\beta$ (so $b$ is an unbiased estimator), with standard deviation

$\sigma_b = \frac{\sigma}{\sigma_x\sqrt{n}},$

where $\sigma$ is the standard deviation of the responses around the population line. Since $\sigma$ is unknown, you estimate it with $s$ , the standard deviation of the residuals. That gives the standard error of the slope:

$SE_b = \frac{s}{s_x\sqrt{n-1}}$

You will almost never compute $SE_b$ by hand. On the AP exam it is given in computer output, in the row for the explanatory variable, under a heading like "SE Coef."

The formula still tells you what makes a slope estimate precise. $SE_b$ is smaller when the points hug the line (small $s$ ), when the $x$ values are spread out (large $s_x$ ), and when the sample is large (large $n$ ).

Conditions: LINER

Before using any inference procedure for slope, check five conditions. The acronym LINER helps you remember them.

Conditions for regression inference (LINER)

Linear: The true relationship between $x$ and $y$ is linear. Check that the scatterplot looks linear and the residual plot shows no curved pattern.
Independent: Individual observations are independent. When sampling without replacement, check the 10% condition: $n \le 10\%$ of the population.
Normal: For any fixed $x$ , the responses vary Normally around the population line. Check that a dotplot, histogram or Normal probability plot of the residuals shows no strong skew or outliers.
Equal SD: The standard deviation of $y$ is the same for every value of $x$ . Check that the residual plot has roughly equal vertical spread everywhere, with no "fan" or "funnel" shape.
Random: The data come from a random sample or a randomized experiment.

Reading computer output

Here is the kind of output you'll see. A coffee cart owner takes a random sample of $10$ days and records the daily high temperature ( $x$ , in °F) and the number of iced drinks sold ( $y$ ).

x-axis: daily high temperature (°F). y-axis: iced drinks sold.Open in grapher →

Predictor	Coef	SE Coef	T	P
Constant	-26.0141	14.7730	-1.76	0.116
Temperature	2.3313	0.1944	11.99	0.000

S = 5.2974, R-Sq = 94.7%

The least-squares line is $\widehat{\text{drinks}} = -26.0141 + 2.3313(\text{temperature})$ . The slope $b = 2.3313$ is in the Temperature row, Coef column.
$SE_b = 0.1944$ is in the same row, SE Coef column. Don't grab $14.7730$ from the Constant row by mistake.
$s = 5.2974$ is the standard deviation of the residuals: actual sales typically differ from the predicted sales by about $5.3$ drinks.

The confidence interval

t interval for the slope

$b \pm t^* \cdot SE_b, \qquad df = n - 2$

Find $t^*$ from a $t$ distribution with $n - 2$ degrees of freedom. You lose two degrees of freedom because the sample is used to estimate two things: the slope and the intercept.

Worked example: Building the interval

Use the coffee cart output to build a $95\%$ confidence interval for the slope of the population regression line.

With $n = 10$ , $df = 10 - 2 = 8$ , and $t^* = 2.306$ .

$2.3313 \pm 2.306(0.1944) = 2.3313 \pm 0.4483$

The interval is about $(1.883,\ 2.780)$ .

Worked example: Interpreting the interval

Interpret the interval from the previous example, and decide whether it gives convincing evidence of a linear relationship.

Interpretation: We are $95\%$ confident that the interval from $1.883$ to $2.780$ captures the slope of the population regression line relating daily high temperature to iced drinks sold. That is, for each $1$ °F increase in the daily high temperature, the mean number of iced drinks sold increases by somewhere between about $1.9$ and $2.8$ drinks.

Evidence: Every value in the interval is positive, and $0$ is not a plausible value for $\beta$ . So the data give convincing evidence of a positive linear relationship between temperature and iced drink sales.

Common mistake

Interpret the slope as a change in the mean (or predicted) response, not in each individual's response. "Each extra degree increases sales by 2 drinks" is wrong: on any given day, sales bounce around the line. Also, the interval is about the population slope $\beta$ . You already know $b$ exactly; there is no need to be "confident" about it.

Checking conditions with a residual plot

Here is the residual plot for the coffee cart data.

x-axis: daily high temperature (°F). y-axis: residual (drinks).Open in grapher →

Worked example: Checking LINER

Check the conditions for the coffee cart interval. A dotplot of the residuals (not shown) is roughly symmetric with no outliers.

Linear: The residual plot shows no leftover curved pattern; points are scattered above and below $0$ .
Independent: $10$ days is less than $10\%$ of all days the cart operates.
Normal: The dotplot of residuals shows no strong skew or outliers.
Equal SD: The vertical spread of the residuals is about the same from left to right. There is no fan shape.
Random: The days were a random sample.

All conditions are met, so the $t$ interval is appropriate.

Worked example: Working from output only

A random sample of $22$ used cars of one model gives mileage $x$ (thousands of miles) and price $y$ (thousands of dollars). The output shows $b = -0.0852$ and $SE_b = 0.0118$ . Construct and interpret a $99\%$ confidence interval for $\beta$ . Assume the conditions are met.

$df = 22 - 2 = 20$ , so $t^* = 2.845$ .

$-0.0852 \pm 2.845(0.0118) = -0.0852 \pm 0.0336$

The interval is about $(-0.1188,\ -0.0516)$ . We are $99\%$ confident that for each additional $1{,}000$ miles, the mean price of this model decreases by between about $52 and $119.

Tip

If you're given an interval and need to recover the pieces: the slope $b$ is the midpoint, and the margin of error is half the width. Dividing the margin of error by $t^*$ gives $SE_b$ .

Practice

Practice 1

A researcher computes a confidence interval for the slope using a random sample of $15$ individuals. How many degrees of freedom should she use for $t^*$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

For a random sample of $25$ individuals, which critical value should be used for a $90\%$ confidence interval for the slope?

Practice 3

A random sample of $18$ homes in a city gives the output below, where $x$ is floor area (hundreds of square feet) and $y$ is price (thousands of dollars).

Predictor	Coef	SE Coef	T	P
Constant	41.37	22.85	1.81	0.089
Area	8.62	1.35	6.39	0.000

S = 28.41, R-Sq = 71.8%

Assuming the conditions are met, find the lower endpoint of a $95\%$ confidence interval for the population slope. Round to the nearest hundredth.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

A $95\%$ confidence interval for the slope relating a student's hours of weekly practice ( $x$ ) to typing speed in words per minute ( $y$ ) is $(1.4,\ 3.1)$ . Which is the best interpretation?

Practice 5

A residual plot for a regression shows residuals close to $0$ for small values of $x$ and spreading out more and more as $x$ increases, in a fan shape. Which condition for inference is most clearly violated?

Practice 6

A $90\%$ confidence interval for a population slope is $(1.2,\ 3.4)$ . What is the sample slope $b$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

A $95\%$ confidence interval for a population slope, based on a random sample of $20$ individuals, is $(0.84,\ 1.56)$ . Find the standard error of the slope, $SE_b$ . Round to the nearest thousandth.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

A scientist plans a study to estimate the slope relating fertilizer amount ( $x$ ) to plant height ( $y$ ). Which change to her design would tend to produce a narrower confidence interval for the slope, keeping the confidence level the same?