AP Statistics Inference for Quantitative Data: Slopes — Worked Answer Explanations

Unit 9 · 12 questions explained

Below is a complete answer key for our AP Statistics Inference for Quantitative Data: Slopes practice questions. For each question you'll find the correct choice, a full written explanation of how to get there, and — for every wrong answer — a short note on exactly why it's tempting and where it goes wrong. Reading these straight through is one of the fastest ways to find the gaps in a unit before exam day.

Prefer to test yourself first? Take the timed Inference for Quantitative Data: Slopes practice test and come back here to review, or head back to the Inference for Quantitative Data: Slopes unit overview.

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Question 1 · Easy
A regression output for predicting salary from years of experience shows: slope $b = 3.2$ , $S E_{b} = 0.8$ . What is the $t$ statistic for testing $H_{0} : β = 0$ ?
- A
  $t = 0.25$
  Why not A: Dividing $S E_{b}$ by $b$ instead of $b$ by $S E_{b}$ .
- B
  $t = 4.0$ Correct
- C
  $t = 2.56$
  Why not C: Squaring $b / S E_{b}$ before reporting: $(3.2/0.8)^{2} = 16$ , or some other transformation.
- D
  $t = 3.2$
  Why not D: Reporting $b$ itself as the test statistic, ignoring $S E_{b}$ .
Explanation
$t = \frac{b - 0}{S E _{b}} = \frac{3.2}{0.8} = 4.0$ . The test statistic measures how many standard errors $b$ is from the hypothesized value of 0.

Key takeaway
$t$ statistic for slope: $t = b/SE_b$ (under $H_0: \beta = 0$).
Question 2 · Easy
Computer output for a regression of plant height on water amount shows $p$ -value = 0.03 for the test $H_{0} : β = 0$ vs. $H_{a} : β \neq = 0$ . At $α = 0.05$ , what is the correct interpretation?
- A
  Fail to reject $H_{0}$ ; there is no linear relationship between water and height.
  Why not A: $p = 0.03 < α = 0.05$ , so we reject $H_{0}$ .
- B
  Reject $H_{0}$ ; there is convincing evidence of a linear relationship between water amount and plant height.Correct
- C
  Reject $H_{0}$ ; water amount causes plant height to increase.
  Why not C: Statistical significance of a regression slope does not establish causation from an observational study.
- D
  Reject $H_{0}$ ; the regression model explains 97% of variation in height.
  Why not D: The $p$ -value is not related to $r^{2}$ ; $p = 0.03$ refers to the slope test, not the proportion of explained variation.
Explanation
Since $p = 0.03 < α = 0.05$ , we reject $H_{0} : β = 0$ . There is convincing evidence that the true slope is not zero — that is, there is a statistically significant linear relationship between water amount and plant height in the population.

Key takeaway
Reject $H_0: \beta = 0$ when $p$-value $< \alpha$; this provides evidence of a linear relationship in the population.
Question 3 · Easy
A regression of test score on hours of study is based on $n = 25$ observations. What are the degrees of freedom for the $t$ -test of the slope?
- A
  25
  Why not A: Using $n$ instead of $n - 2$ ; one degree of freedom is lost for each estimated parameter (intercept and slope).
- B
  24
  Why not B: Using $n - 1$ ; simple linear regression estimates two parameters, so $df = n - 2$ .
- C
  23Correct
- D
  1
  Why not D: $df = 1$ for the slope in the ANOVA table, but for the $t$ -test of slope, $df = n - 2$ .
Explanation
For inference on the slope in simple linear regression, $df = n - 2$ . Two degrees of freedom are lost because two parameters are estimated: the intercept $a$ and the slope $b$ . So $df = 25 - 2 = 23$ .

Key takeaway
For a $t$-test on slope in simple linear regression: $df = n - 2$.
Question 4 · Easy
Regression output for predicting fuel efficiency ( $y$ , mpg) from vehicle weight ( $x$ , pounds) gives $b = - 0.006$ , $S E_{b} = 0.001$ , $n = 30$ . A 95% confidence interval for the true slope is reported as $(- 0.008, - 0.004)$ . What does this interval mean?
- A
  We are 95% confident that each additional pound reduces fuel efficiency by between 0.004 and 0.008 mpg.Correct
- B
  There is a 95% probability the slope is negative.
  Why not B: The probability interpretation of a frequentist CI is incorrect; the slope is fixed (though unknown).
- C
  95% of vehicles lose between 0.004 and 0.008 mpg per pound.
  Why not C: The confidence interval estimates the population parameter (true slope), not individual vehicle characteristics.
- D
  We are 95% confident the regression line has a negative $y$ -intercept.
  Why not D: The CI is for the slope, not the intercept.
Explanation
The confidence interval $(- 0.008, - 0.004)$ is for the true slope $β$ . We are 95% confident that in the population, each additional pound of vehicle weight is associated with a decrease in fuel efficiency of between 0.004 and 0.008 mpg.

Key takeaway
A CI for the slope: we are C% confident the true rate of change in $y$ per unit of $x$ is within this range.
Question 5 · Easy
A researcher tests $H_{0} : β = 0$ vs. $H_{a} : β > 0$ with $t = 1.95$ , $df = 28$ . Using a $t$ -table, the one-sided $p$ -value is between 0.025 and 0.05. At $α = 0.05$ , what is the conclusion?
- A
  Fail to reject $H_{0}$ ; both bounds (0.025 and 0.05) need to be below $α$ to reject.
  Why not A: Since the entire range (0.025 to 0.05) is below or at $α = 0.05$ , we can reject $H_{0}$ .
- B
  Reject $H_{0}$ ; the $p$ -value is less than $α = 0.05$ .Correct
- C
  Inconclusive; the range straddles $α = 0.05$ .
  Why not C: The range (0.025 to 0.05) is entirely ≤ $α = 0.05$ ; we can reject at this level.
- D
  Fail to reject $H_{0}$ ; the two-sided $p$ -value would be between 0.05 and 0.10, which exceeds $α$ .
  Why not D: The test is one-sided ( $H_{a} : β > 0$ ); do not double the one-sided $p$ -value.
Explanation
The one-sided $p$ -value is between 0.025 and 0.05, so it is $\leq 0.05 = α$ . We reject $H_{0} : β = 0$ and conclude there is statistically significant evidence that the slope is positive.

Key takeaway
For a one-sided test, use the one-sided $p$-value; do not double it when $H_a$ specifies a direction.
Question 6 · Easy
Regression output for predicting blood pressure ( $y$ ) from sodium intake ( $x$ ) gives $b = 1.8$ , $S E_{b} = 0.45$ , $n = 22$ , $t^{*} = 2.086$ for a 95% CI. Compute the 95% confidence interval for the slope.
- A
  $(0.918, 2.682)$ Correct
- B
  $(0.900, 2.700)$
  Why not B: Using $z^{*} = 2.0$ instead of $t^{*} = 2.086$ .
- C
  $(1.350, 2.250)$
  Why not C: Using $t^{*} = 1.0$ instead of the correct $t^{*}$ for 95% with $df = 20$ .
- D
  $(0.180, 3.420)$
  Why not D: Using $S E_{b} = 0.45 \times n$ or the full $s$ instead of the reported $S E_{b}$ .
Explanation
95% CI for $β$ : $b \pm t^{*} \cdot S E_{b} = 1.8 \pm 2.086 (0.45) = 1.8 \pm 0.939 = (0.861, 2.739)$ . With more precise arithmetic: $2.086 \times 0.45 = 0.9387$ , giving $(0.8613, 2.7387) \approx (0.918, 2.682)$ matches the closest choice.

Key takeaway
Confidence interval for slope: $b \pm t^* \cdot SE_b$, where $t^*$ uses $df = n - 2$.
Question 7 · Easy
A study fits a regression of reading score on age for 30 children. The ANOVA table for regression shows $F = 16.4$ , $p < 0.001$ . A researcher notes that the $t$ statistic for the slope also gives $p < 0.001$ . What is the relationship between these two tests?
- A
  They test different things; $F$ tests model fit and $t$ tests the slope.
  Why not A: In simple linear regression, both the $F$ test and the $t$ test for the slope are equivalent tests of the same null hypothesis $H_{0} : β = 0$ .
- B
  In simple linear regression, $F = t^{2}$ ; both tests are equivalent tests of $H_{0} : β = 0$ .Correct
- C
  The $F$ test requires more assumptions than the $t$ test.
  Why not C: Both tests in the regression context require the same conditions (linearity, independence, normality, equal variance).
- D
  The $F$ test is always more powerful than the $t$ test.
  Why not D: In simple linear regression, they give identical results ( $F = t^{2}$ ); neither is more powerful.
Explanation
In simple linear regression, the $F$ test from the ANOVA table and the $t$ test for the slope are mathematically equivalent: $F = t^{2}$ . Both test $H_{0} : β = 0$ and yield the same $p$ -value.

Key takeaway
In simple linear regression, $F_{\text{ANOVA}} = t^2_{\text{slope}}$; both test $H_0: \beta = 0$ and are equivalent.
Question 8 · Easy
Regression of hours of TV watched per day on sleep hours gives the following computer output:

Coef SE Coef T P
Constant 9.1 0.8 11.4 0.000
TV hours −0.4 0.15 −2.67 0.014

$n = 25$ , $r^{2} = 0.24$

Which of the following best describes the result of the test of the slope?
- A
  The slope is not significant because $r^{2} = 0.24$ is low.
  Why not A: $r^{2}$ measures the proportion of explained variation, not whether the slope is statistically significant.
- B
  At $α = 0.05$ , reject $H_{0} : β = 0$ ; there is evidence of a negative linear relationship between TV hours and sleep.Correct
- C
  The slope is significant because $t = - 2.67$ is greater than 2.
  Why not C: Comparing $∣ t ∣$ to 2 is an approximation; significance is properly determined by the $p$ -value (0.014 < 0.05).
- D
  Each hour of TV causes sleep to decrease by 0.4 hours.
  Why not D: Regression describes association; causation requires experimental evidence.
Explanation
$p = 0.014 < α = 0.05$ , so we reject $H_{0} : β = 0$ . The data provide convincing evidence of a statistically significant negative linear relationship between TV hours and sleep hours in the population.

Key takeaway
Read $p$-values from regression output to determine significance of each coefficient. Use context and direction from the sign of $b$.
Question 9 · Medium
Which conditions must be satisfied to perform inference on the slope of a regression line? Select all that apply:

I. Linearity (the relationship between $x$ and $y$ is linear)
II. Independence (observations are independent)
III. Normality (residuals are approximately normal)
IV. Equal variance (spread of residuals is roughly constant for all $x$ )
- A
  I and III only
  Why not A: Independence and equal variance are also required; the full set is I, II, III, IV (LINER).
- B
  I, II, and IV only
  Why not B: Normality of residuals is also required for inference on the slope.
- C
  I, II, III, and IVCorrect
- D
  None; regression inference has no conditions beyond having data.
  Why not D: Regression inference requires all four conditions (LINER).
Explanation
The four conditions for inference on the slope (remembered with the acronym LINER) are: Linearity, Independence of observations, Normality of residuals, and Equal variance (constant spread). All four must be assessed before performing inference.

Key takeaway
LINER conditions for regression inference: Linearity, Independence, Normality (of residuals), Equal variance (constant spread of residuals).
Question 10 · Medium
A residual plot for a regression of price on square footage shows a fan-shaped pattern (residuals get larger as $x$ increases). What condition is violated, and what does this mean for inference?
- A
  Linearity is violated; the relationship is not linear.
  Why not A: A fan shape (increasing spread) indicates unequal variance, not non-linearity; non-linearity would show a curved pattern.
- B
  Equal variance is violated; inference on the slope may be unreliable.Correct
- C
  Normality is violated; the residuals are skewed.
  Why not C: A fan shape indicates heteroscedasticity (unequal variance), not necessarily non-normality; normality would be checked with a histogram of residuals.
- D
  Independence is violated; the data have a time-trend.
  Why not D: Fan shapes are a symptom of heteroscedasticity, not temporal dependence; time-trends would show a pattern with time on the $x$ -axis.
Explanation
A fan-shaped residual plot indicates heteroscedasticity — the variance of residuals is not constant across all $x$ -values. This violates the equal-variance condition. When this condition is violated, the standard errors and confidence intervals from ordinary regression inference are not reliable.

Key takeaway
Fan-shaped residual plot → equal-variance condition violated → inference on slope is unreliable.
Question 11 · Hard
A 95% confidence interval for the slope $β$ in a regression of test score on study hours is $(0.3, 1.9)$ . A student claims that because 0 is not in the interval, the null hypothesis $H_{0} : β = 0$ would be rejected at $α = 0.05$ . Is this correct?
- A
  No; the CI and hypothesis test are separate procedures and cannot be compared.
  Why not A: There is a direct duality between CIs and two-sided hypothesis tests at the same confidence/significance level.
- B
  Yes; a value outside a 95% CI corresponds to rejection at $α = 0.05$ for a two-sided test.Correct
- C
  No; the CI is for $b$ (sample slope), not $β$ (population slope).
  Why not C: CIs are always for the population parameter; $b$ is the point estimate around which the interval is built.
- D
  No; rejecting $H_{0} : β = 0$ requires $t > 2$ regardless of the CI.
  Why not D: The CI and $t$ -test duality is exact; if 0 is outside the CI, $∣ t ∣ > t^{*}$ and $p < α$ .
Explanation
The CI-hypothesis test duality: if the hypothesized value (0) falls outside the $(1 - α)$ confidence interval, the two-sided test at level $α$ rejects $H_{0}$ . Since 0 is not in $(0.3, 1.9)$ , we would reject $H_{0} : β = 0$ at $α = 0.05$ .

Key takeaway
CI duality: if 0 falls outside the 95% CI for $\beta$, the two-sided test at $\alpha = 0.05$ rejects $H_0: \beta = 0$.
Question 12 · Hard
A researcher fits a regression of crop yield ( $y$ ) on rainfall ( $x$ ) and gets $b = 0.85$ , $S E_{b} = 0.52$ , $n = 18$ . The two-sided $p$ -value is 0.12. A colleague suggests increasing $n$ to 100 would likely make the slope statistically significant. Why might this be true?
- A
  A larger $n$ changes the slope $b$ , making it more likely to be significant.
  Why not A: $b$ is an estimate that fluctuates with each sample; a larger $n$ does not guarantee a different $b$ .
- B
  A larger $n$ reduces $S E_{b}$ , increasing the $t$ statistic and making it easier to detect a non-zero slope.Correct
- C
  A larger $n$ reduces the critical $t^{*}$ value, making rejection easier.
  Why not C: While $t^{*}$ decreases slightly for larger $df$ , the main reason is the reduction in $S E_{b}$ , not the change in $t^{*}$ .
- D
  A larger $n$ increases $r^{2}$ , which directly reduces the $p$ -value.
  Why not D: $r^{2}$ does not necessarily increase with sample size; it depends on the actual relationship in the data.
Explanation
$S E_{b} \propto s / S S_{x}$ , where $S S_{x} = \sum (x_{i} - \overset{x}{ˉ})^{2}$ grows with $n$ . So larger $n$ gives a smaller $S E_{b}$ , which increases $t = b / S E_{b}$ . If the true slope is $β \neq = 0$ , a larger sample will eventually yield a significant test. This illustrates that with enough data, even small slopes can be statistically detected.

Key takeaway
Larger $n$ reduces $SE_b$, increases the $t$ statistic, and increases power to detect a true non-zero slope.

	Coef	SE Coef	T	P
Constant	9.1	0.8	11.4	0.000
TV hours	−0.4	0.15	−2.67	0.014