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How do you construct a confidence interval for the slope of a regression model?

Topic 9.2 Confidence Intervals for the Slope of a Regression Model: check the regression conditions and construct a t-interval for the population slope using the sample slope, its standard error, and n minus 2 degrees of freedom.

A focused answer to AP Statistics Topic 9.2, on building a t-interval for the population slope - checking the regression conditions, reading the slope and its standard error from computer output, and using n minus 2 degrees of freedom - with a full worked interval.

Generated by Claude Opus 4.811 min answer

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  1. What this topic is asking
  2. The interval and the standard error
  3. The regression conditions (LINER)
  4. Interpreting the interval
  5. Try this

What this topic is asking

The College Board (Topic 9.2) wants you to construct a confidence interval for the population slope β\beta: check the regression conditions, read the sample slope bb and its standard error SEbSE_b (usually from computer output), and use the tt-distribution with n2n - 2 degrees of freedom.

The interval and the standard error

The structure is the familiar "estimate ±\pm critical value ×\times standard error." The estimate is the sample slope bb; the standard error SEbSE_b measures how much bb varies from sample to sample (it shrinks with more data, a wider spread of xx-values, and a tighter fit). On the exam, bb and SEbSE_b are typically given in computer output, where the slope row lists the coefficient, its standard error, a t-statistic, and a P-value. You read bb and SEbSE_b from that row. The degrees of freedom are n2n - 2, because estimating both the intercept and the slope uses up two.

The regression conditions (LINER)

You check these mainly from residual plots and a description of the data: a residual plot with no pattern supports linearity, roughly constant vertical spread supports equal variance, and a histogram or normal plot of residuals supports normality. State each condition and the evidence for it; "the residual plot shows no pattern and constant spread" addresses two conditions at once. These conditions earn the tt-model for the slope, just as the normal/large-sample conditions earned it for a mean.

Interpreting the interval

Interpret a slope interval in the units of the relationship: "We are C%C\% confident that the true slope, the mean change in yy for each one-unit increase in xx, is between [low] and [high]." The decisive feature is whether the interval contains 00. A slope of 00 means no linear relationship (changing xx does not change the predicted yy). So if the interval excludes 00, there is evidence of a real linear association; if it contains 00, no relationship is plausible. This zero-check is the slope analogue of the difference-interval zero-check and previews the test of H0:β=0H_0: \beta = 0. As with all intervals, higher confidence widens it and a larger sample (or a wider spread of xx) narrows it.

Try this

Q1. A regression with n=30n = 30 gives b=2.0b = 2.0, SEb=0.6SE_b = 0.6. Find the degrees of freedom and a rough 95%95\% margin of error (use t2.05t^{*} \approx 2.05). [2 points]

  • Cue. df=n2=28df = n - 2 = 28; margin of error 2.05×0.6=1.23\approx 2.05 \times 0.6 = 1.23, interval about (0.77,3.23)(0.77, 3.23).

Q2. What does it mean if a slope interval contains 00? [1 point]

  • Cue. A slope of 00 (no linear relationship) is plausible, so there is not convincing evidence of a linear association between xx and yy.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2018 (style)1 marksSection I (multiple choice). A regression of yy on xx uses n=22n = 22 data points. A t-interval for the slope uses degrees of freedom (A) 2222 (B) 2121 (C) 2020 (D) 22
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The correct answer is (C).

Inference for a regression slope uses df=n2=222=20df = n - 2 = 22 - 2 = 20 (two degrees of freedom are spent estimating the intercept and slope).

(A) uses nn. (B) uses n1n - 1 (the one-mean rule). (D) is incorrect. The slope uses n2n - 2.

AP 2022 (style)4 marksSection II (free response). Computer output for a regression of fuel use (yy) on speed (xx) from n=25n = 25 cars gives slope estimate b=0.045b = 0.045 with standard error SEb=0.012SE_b = 0.012. Residual plots show no pattern, roughly constant spread, and the residuals are approximately normal. (a) State the conditions and confirm they are met. (b) Construct a 95%95\% confidence interval for the population slope (use t=2.069t^{*} = 2.069 for df=23df = 23). (c) Interpret the interval in context.
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A 4-point slope interval.

(a) (1 point) Conditions (LINER): Linear (residual plot shows no curved pattern), Independent observations, Normal residuals (stated approximately normal), Equal variance (constant spread of residuals), Random sample. All stated as met.
(b) (2 points) df=n2=23df = n - 2 = 23, t=2.069t^{*} = 2.069. Interval =b±tSEb=0.045±2.069(0.012)=0.045±0.0248=(0.020, 0.070)= b \pm t^{*} \cdot SE_b = 0.045 \pm 2.069(0.012) = 0.045 \pm 0.0248 = (0.020,\ 0.070).
(c) (1 point) We are 95%95\% confident that the true slope, the mean change in fuel use per unit increase in speed, is between 0.0200.020 and 0.0700.070.

Markers reward stating and checking the regression conditions, the tSEbt^{*} \cdot SE_b interval with df=n2df = n - 2, and a slope interpretation in context.

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