Why is checking linearity important?

Answer: Least-squares regression is meaningful only when the relationship is approximately linear Fitting a line to nonlinear data yields misleading interpretations and biased predictions.

Answer: An extrapolation and likely unreliable Predicting beyond the observed x-range is extrapolation; the model may not hold there.

What does the curved pattern suggest?

Answer: A linear model is not appropriate A clear curved (non-random) pattern in the residual plot indicates the linear model fails to capture the true relationship.

AP Statistics · Topic 2.9

Analyzing Departures from Linearity Practice

Part of Exploring Two-Variable Data.(DAT-1.F)

Practice questions

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Sample questions

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Sample 1difficulty 2/5
$Before fitting a least-squares line, an analyst examines the scatterplot.$
$Why is checking linearity important?$
- A
  $Linearity guarantees no outliers$
- B
  $Least-squares regression is meaningful only when the relationship is approximately linear$
  check_circle
- C
  $Linearity ensures r-squared = 100%$
- D
  $Without linearity, r equals zero$
Why
$Fitting a line to nonlinear data yields misleading interpretations and biased predictions.$
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Sample 2difficulty 2/5
$A model based on observed x-values from 5 to 20 is used to predict y at x = 50.$
$The prediction is:$
- A
  $An extrapolation and likely unreliable$
  check_circle
- B
  $Always equal to the mean of y$
- C
  $An interpolation and likely accurate$
- D
  $Independent of the range of x$
Why
$Predicting beyond the observed x-range is extrapolation; the model may not hold there.$
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Sample 3difficulty 3/5
$A residual plot is shown for a linear regression model.$
$What does the curved pattern suggest?$
- A
  $The data have no outliers$
- B
  $The correlation is exactly 0$
- C
  $A linear model is not appropriate$
  check_circle
- D
  $The model fits well$
Why
$A clear curved (non-random) pattern in the residual plot indicates the linear model fails to capture the true relationship.$
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Sample 4difficulty 3/5
$A linear model fits log(y-hat) = 1 + 0.30 x (log base 10).$
$Predict y at x = 2.$
- A
  $10^1.6 ~= 39.8$
  check_circle
- B
  $10 + 0.6$
- C
  $1.6$
- D
  $0.30 * 2$
Why
$log(y-hat) = 1 + 0.30(2) = 1.6, so y-hat = 10^1.6 ~= 39.8.$
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Sample 5difficulty 3/5
$A residual plot with random scatter (no pattern) suggests$
- A
  $Outlier present$
- B
  $r \approx 0$
- C
  $Linear model is appropriate$
  check_circle
- D
  $Nonlinear relationship$
Why
$No systematic pattern \to linear fit is suitable.$
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