Mathfolis

Confidence Interval for the Slope

Unit 9 · Inference for Quantitative Data: Slopes

What AP Stats asks here

A slope CI takes the form b±t∗SEbb \pm t^* SE_b with df =n−2= n - 2 — two degrees of freedom 'spent' on estimating slope and intercept. The numbers bb and SEbSE_b come straight from the regression output. The recurring AP trap: confusing the slope CI (range for the rate β\beta) with a prediction interval for an individual yy.

Slope confidence interval

Estimate ± ME
b±t∗⋅SEbb \pm t^* \cdot SE_b
Degrees of freedom
df=n−2\text{df} = n - 2

Reading the output

Slope row 'Coef'
bb
Slope row 'SE Coef'
SEbSE_b
Residual SD
s (printed as ’S=’)s \text{ (printed as 'S=')}
AP Tip: Always write the slope CI with units AND with the 'per 1-unit increase in x' phrasing. 'Between 0.62 and 1.01 points of Final per 1-point increase in Midterm' beats 'between 0.62 and 1.01'.
Caution: The slope CI describes the rate β. It does NOT predict individual y values at a specific x. Prediction intervals for individual outcomes are wider and use a different formula.
Type 1

Construct b±t∗SEbb \pm t^* SE_b

Pick t∗t^* at df =n−2= n - 2, multiply by SEbSE_b, add and subtract from bb.

Example 1
A regression of Final on Midterm gives b=0.815b = 0.815, SEb=0.094SE_b = 0.094, n=28n = 28. Using t∗≈2.056t^* \approx 2.056 at df = 26, what is the 95% CI for β\beta? (A) (0.55,1.08)(0.55, 1.08) (B) (0.622,1.008)(0.622, 1.008) (C) (0.65,0.98)(0.65, 0.98) (D) (−0.19,1.82)(-0.19, 1.82)

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Type 2

Interpret slope CI with units

Always state the parameter as the change in predicted yy per 1-unit change in xx, with units.

Example 2
A 95% CI for β\beta in the Final-on-Midterm regression is (0.622,1.008)(0.622, 1.008). Which interpretation is best? (A) We are 95% confident the slope is between 0.622 and 1.008. (B) 95% of students gain between 0.622 and 1.008 points on the Final per point of Midterm. (C) We are 95% confident the true slope of the population regression of Final on Midterm is between 0.622 and 1.008 points of Final per 1-point increase in Midterm. (D) The probability the true slope is in this interval is 0.95.

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Type 3

Zero in the CI; not a prediction interval

Excludes 0 ⇒ significant linear association. The slope CI never predicts individual yy — prediction intervals are a different (wider) tool.

Example 3
A 95% slope CI for fish weight on length is (7.95,16.85)(7.95, 16.85) g/cm. A student writes: 'A 30-cm fish therefore weighs between 30⋅7.95=238.530 \cdot 7.95 = 238.5 g and 30⋅16.85=505.530 \cdot 16.85 = 505.5 g with 95% confidence.' Which critique is correct? (A) The student is right — slope × length = weight. (B) The slope CI is about the rate β\beta, not about individual yy values at x=30x = 30. Predicting individual weight requires a prediction interval, which is wider and uses a different formula. Also, the model's intercept matters when computing y^\hat{y} at a specific xx. (C) The student is wrong because r2r^2 was not used. (D) The student is right only for xx values inside the observed range.

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