Biostatistics Quiz: Inference For Regression Slope

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Inference For Regression SlopeQuestion 1 of 20

A researcher obtains a 98% confidence interval for the regression slope as $(-1.2, 3.8)$ with $n = 32$ . If she had instead calculated a 95% confidence interval, which statement would be true?

The 95% interval would be wider than

(-1.2, 3.8)

The 95% interval would be narrower than

(-1.2, 3.8)

The 95% interval would have the same width as

(-1.2, 3.8)

Cannot determine without knowing the point estimate and standard error

The 95% interval would not contain zero while the 98% interval does

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Biostatistics Quiz

Biostatistics Quiz: Inference For Regression Slope

Practice Inference For Regression Slope in Biostatistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Inference For Regression Slope, giving you a quick way to practice the rules, question types, and explanations that matter most for Biostatistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A researcher obtains a 98% confidence interval for the regression slope as $(-1.2, 3.8)$ with $n = 32$ . If she had instead calculated a 95% confidence interval, which statement would be true?

The 95% interval would be wider than $(-1.2, 3.8)$
The 95% interval would be narrower than $(-1.2, 3.8)$ (correct answer)
The 95% interval would have the same width as $(-1.2, 3.8)$
Cannot determine without knowing the point estimate and standard error
The 95% interval would not contain zero while the 98% interval does

Explanation: When you encounter confidence interval questions, focus on the fundamental relationship between confidence level and interval width. Higher confidence requires wider intervals to capture the true parameter more reliably. A confidence interval's width depends on the critical value (t-score) multiplied by the standard error. As confidence level decreases from 98% to 95%, you're allowing more risk of missing the true parameter, so you need less "safety margin." This means using a smaller critical t-value, which creates a narrower interval around the same point estimate. With

n = 32

, we have

df = 31

. The critical t-value for 98% confidence (

\alpha = 0.02

) is larger than for 95% confidence (

\alpha = 0.05

). Since the standard error remains constant, the 95% interval must be narrower than

(-1.2, 3.8)

, making B correct. Option A incorrectly suggests the opposite relationship—that lower confidence creates wider intervals. Option C is wrong because different confidence levels virtually always produce different interval widths (except in contrived scenarios). Option D represents a common misconception: while knowing the point estimate and standard error would let you calculate exact intervals, the directional relationship between confidence level and width is always predictable. Study tip: Remember the confidence-width trade-off: higher confidence = wider intervals, lower confidence = narrower intervals. This relationship holds for any parameter (means, proportions, slopes) and is testable across many biostatistics contexts.

Question 2

In a clinical trial, the regression of symptom improvement on drug dose yields $\hat{\beta}_1 = 0.6$ with $SE(\hat{\beta}_1) = 0.3$ and $n = 27$ . A competing drug claims their slope is 0.9. To test if the new drug is inferior (slope less than 0.9) at $\alpha = 0.05$ , what is the test statistic and critical value?

$t = \frac{0.6-0.9}{0.3} = -1.0$ ; critical value $= -t_{25,0.05} = -1.708$ (correct answer)
$t = \frac{0.6-0.9}{0.3} = -1.0$ ; critical value $= t_{25,0.05} = 1.708$
$t = \frac{0.6}{0.3} = 2.0$ ; critical value $= t_{25,0.05} = 1.708$
$t = \frac{0.6-0.9}{0.3} = -1.0$ ; critical value $= -t_{26,0.05} = -1.706$
$t = \frac{0.6-0.9}{0.3} = -1.0$ ; critical value $= -t_{25,0.025} = -2.060$

Explanation: When testing whether a regression slope is significantly different from a hypothesized value, you're performing a one-sample t-test on the slope coefficient. The key is setting up your null and alternative hypotheses correctly based on what you're trying to prove. Here, you want to test if the new drug is "inferior" (slope less than 0.9), so your alternative hypothesis is

H_a: \beta_1 < 0.9

and null hypothesis is

H_0: \beta_1 = 0.9

. This makes it a left-tailed test. The test statistic follows the standard formula:

t = \frac{\hat{\beta}_1 - \beta_{1,hypothesized}}{SE(\hat{\beta}_1)} = \frac{0.6 - 0.9}{0.3} = -1.0

. The degrees of freedom are

n - 2 = 27 - 2 = 25

(we lose 2 df in simple linear regression for estimating intercept and slope). For a left-tailed test at

\alpha = 0.05

, the critical value is

-t_{25,0.05} = -1.708

. Answer B incorrectly uses the positive critical value

+1.708

, which would be for a right-tailed test. Answer C completely ignores the hypothesis being tested by calculating

t = \frac{0.6}{0.3}

, essentially testing if the slope differs from zero rather than from 0.9. Answer D uses the wrong degrees of freedom (26 instead of 25), forgetting that regression analysis requires subtracting 2 from the sample size. Remember: for regression slope tests, always subtract 2 from your sample size for degrees of freedom, and make sure your critical value direction matches your alternative hypothesis (negative for "less than" tests).

Question 3

A researcher studying the relationship between daily exercise minutes (x) and systolic blood pressure (y) in adults obtains the following regression results: $\hat{y} = 140 - 0.3x$ with $SE(\hat{\beta}_1) = 0.12$ and $n = 25$ . If the researcher wants to test whether exercise significantly reduces blood pressure at $\alpha = 0.05$ , what is the appropriate test statistic?

$t = \frac{-0.3}{0.12} = -2.50$ (correct answer)
$t = \frac{0.3}{0.12} = 2.50$
$t = \frac{-0.3 - 0}{0.12} = -2.50$
$z = \frac{-0.3}{0.12} = -2.50$
$t = \frac{140 - 0.3}{0.12} = 1164.17$

Explanation: When testing whether a regression coefficient is significantly different from zero, you're conducting a one-sample t-test on the slope parameter. The key insight here is understanding what hypothesis you're actually testing and how to construct the appropriate test statistic. The researcher wants to test whether exercise significantly reduces blood pressure, meaning they're testing if the slope

\beta_1

is significantly less than zero (since negative slopes indicate reduction). The standard t-test formula for regression coefficients is:

t = \frac{\hat{\beta}_1 - \beta_{1,H_0}}{SE(\hat{\beta}_1)}

Since we're testing against the null hypothesis that

\beta_1 = 0

, this becomes:

t = \frac{-0.3 - 0}{0.12} = -2.50

. However, answer A shows the simplified version

t = \frac{-0.3}{0.12} = -2.50

, which is mathematically equivalent since subtracting zero doesn't change the value. Answer A is correct because it properly calculates the t-statistic using the negative slope coefficient. Answer B incorrectly uses the absolute value of the slope (0.3 instead of -0.3), which would give you the wrong direction for your test. Answer C shows the full formula including "- 0" but this is unnecessarily verbose when the simpler form in A is equivalent. Answer D incorrectly uses a z-statistic instead of a t-statistic, but with small samples (n=25) and unknown population variance, you must use the t-distribution. Study tip: In regression hypothesis testing, always preserve the sign of your coefficient in the test statistic—it tells you the direction of the relationship you're testing.

Question 4

A medical researcher tests $H_0: \beta_1 = 0$ versus $H_1: \beta_1 \neq 0$ for the slope of drug dosage predicting recovery time. With $n = 30$ patients, she obtains $t = -2.85$ . Using $\alpha = 0.01$ , what conclusion should she draw?

Reject $H_0$ because $|t| = 2.85 > t_{28,0.005} = 2.763$ (correct answer)
Fail to reject $H_0$ because $t = -2.85 < 0$
Reject $H_0$ because $|t| = 2.85 > t_{29,0.01} = 2.462$
Fail to reject $H_0$ because $|t| = 2.85 < t_{28,0.01} = 2.467$
Reject $H_0$ because $|t| = 2.85 > t_{28,0.01} = 2.467$

Explanation: When testing a regression slope coefficient, you're conducting a two-tailed t-test to determine if there's a significant linear relationship between your variables. Here, the researcher wants to know if drug dosage significantly predicts recovery time. For this hypothesis test, you need the correct degrees of freedom and critical value. With

n = 30

patients, the degrees of freedom are

df = n - 2 = 28

(we subtract 2 because we're estimating both the intercept and slope). For a two-tailed test with

\alpha = 0.01

, you need the critical value at

\alpha/2 = 0.005

in each tail, giving you

t_{28,0.005} = 2.763

. Since this is a two-tailed test, you compare the absolute value of your test statistic to the critical value:

|t| = |-2.85| = 2.85 > 2.763

. Therefore, you reject

H_0

and conclude there is a significant relationship between dosage and recovery time. Answer A is correct because it uses the proper degrees of freedom (28), the correct critical value for a two-tailed test (0.005), and applies the absolute value correctly. Answer B incorrectly focuses on the sign of the t-statistic, which is irrelevant in a two-tailed test. Answer C uses wrong degrees of freedom (29 instead of 28) and the wrong critical value (0.01 instead of 0.005). Answer D uses the correct degrees of freedom but wrong critical value, leading to an incorrect conclusion. Study tip: For regression slope tests, always remember

df = n - 2

and use

\alpha/2

for two-tailed critical values.

Question 5

In studying the effect of study hours on exam scores, a researcher obtains the 90% confidence interval for the slope as $(1.2, 4.8)$ . Based on this interval, what can be concluded about testing $H_0: \beta_1 = 0$ versus $H_1: \beta_1 \neq 0$ at $\alpha = 0.10$ ?

Reject $H_0$ because 0 is not contained in the confidence interval (correct answer)
Fail to reject $H_0$ because the interval contains only positive values
Cannot determine without knowing the sample size and standard error
Reject $H_0$ because the lower bound 1.2 is greater than zero
Fail to reject $H_0$ because this requires a 95% confidence interval

Explanation: When you encounter confidence intervals for regression slopes, you're dealing with the intimate connection between hypothesis testing and interval estimation. The key insight is that a confidence interval contains all the plausible values for the parameter at your chosen confidence level. The correct reasoning leads to answer A: since the 90% confidence interval (1.2, 4.8) does not contain zero, you can reject

H_0: \beta_1 = 0

\alpha = 0.10

. This works because there's a direct correspondence between confidence intervals and two-tailed hypothesis tests—if a

(1-\alpha) \times 100\%

confidence interval doesn't contain the null hypothesis value, you reject that null hypothesis at significance level

\alpha

. Answer B makes a logical error by suggesting that containing only positive values somehow supports failing to reject the null. The sign of the values isn't the issue—it's whether zero (the null value) is included. Answer C incorrectly assumes you need additional information. The confidence interval already incorporates the sample size and standard error in its construction, giving you everything needed for the hypothesis test conclusion. Answer D reaches the right conclusion but focuses on an incomplete justification. While it's true that the lower bound exceeds zero, the complete reasoning requires recognizing that zero falls outside the entire interval. Study tip: Remember the confidence interval-hypothesis test duality. If your null hypothesis value lies outside a

(1-\alpha) \times 100\%

confidence interval, reject the null at significance level

\alpha

. This relationship is your shortcut to hypothesis testing conclusions.

Question 6

A researcher wants to determine if there is evidence that each additional year of education increases annual income by more than $2000. The regression gives $ $\hat{\beta}_1 = 2400$ $(in dollars per year) with$ $SE(\hat{\beta}_1) = 300$ $and$ $n = 35$ $. What are the appropriate hypotheses and test statistic?

$H_0: \beta_1 \leq 2000$ , $H_1: \beta_1 > 2000$ ; $t = \frac{2400-2000}{300} = 1.33$ (correct answer)
$H_0: \beta_1 = 0$ , $H_1: \beta_1 > 0$ ; $t = \frac{2400}{300} = 8.0$
$H_0: \beta_1 = 2000$ , $H_1: \beta_1 \neq 2000$ ; $t = \frac{2400-2000}{300} = 1.33$
$H_0: \beta_1 \geq 2000$ , $H_1: \beta_1 < 2000$ ; $t = \frac{2400-2000}{300} = 1.33$
$H_0: \beta_1 \leq 2000$ , $H_1: \beta_1 > 2000$ ; $t = \frac{2400}{300\sqrt{35}} = 1.35$

Explanation: When testing whether a regression coefficient exceeds a specific threshold, you need to set up hypotheses that directly address the research question. The researcher wants evidence that each additional year of education increases income by more than $2000, which means they're testing whether $

\beta_1 > 2000

$. The correct approach uses

H_0: \beta_1 \leq 2000

and

H_1: \beta_1 > 2000

. The null hypothesis assumes the increase is

2000 or less, while the alternative tests for increases greater than

2000. The test statistic compares the estimated coefficient to the threshold:

t = \frac{2400-2000}{300} = 1.33

. This tells us how many standard errors our estimate is above the $2000 threshold. Option B tests whether education has any positive effect on income (

\beta_1 > 0

) rather than whether it exceeds the specific

2000 threshold. While this might seem relevant, it doesn't answer the researcher's actual question about exceeding

2000. Option C uses a two-tailed test (

H_1: \beta_1 \neq 2000

), which would test whether the effect differs from $2000 in either direction. However, the research question specifically asks for evidence of an effect greater than $2000, requiring a one-tailed test. Option D reverses the hypotheses, testing whether the effect is less than $2000 rather than greater than $2000, which contradicts the research objective. Study tip: Always translate the research question into mathematical hypotheses first. Words like "more than," "exceeds," or "greater than" signal one-tailed tests, and your null hypothesis should represent the opposite of what you're trying to prove.

Question 7

A researcher studying the relationship between hours of sleep and reaction time obtains a regression slope of $\hat{\beta}_1 = -15$ milliseconds per hour with $SE(\hat{\beta}_1) = 4$ ms/hour and $n = 20$ . What is the 90% confidence interval for the change in reaction time per additional hour of sleep?

$-15 \pm 1.734(4) = (-21.94, -8.06)$ milliseconds (correct answer)
$-15 \pm 1.645(4) = (-21.58, -8.42)$ milliseconds
$-15 \pm 1.729(4) = (-21.92, -8.08)$ milliseconds
$-15 \pm 2.101(4) = (-23.40, -6.60)$ milliseconds
$-15 \pm 1.725(4) = (-21.90, -8.10)$ milliseconds

Explanation: When constructing confidence intervals for regression coefficients, you need to identify the correct critical value based on your sample size and confidence level. This requires understanding which distribution to use and how degrees of freedom work in regression. The correct approach uses the formula:

\hat{\beta}_1 \pm t_{\alpha/2, df} \times SE(\hat{\beta}_1)

. With

n = 20

observations, you have

df = n - 2 = 18

degrees of freedom (subtracting 2 for the intercept and slope parameters). For a 90% confidence interval,

\alpha = 0.10

, so

\alpha/2 = 0.05

. The critical value

t_{0.05, 18} = 1.734

. Therefore:

-15 \pm 1.734(4) = (-21.94, -8.06)

. Answer A is correct because it uses the proper t-distribution critical value of 1.734 for 18 degrees of freedom at the 90% confidence level. Answer B incorrectly uses

z_{0.05} = 1.645

, the standard normal critical value. This is wrong because regression uses the t-distribution, not the normal distribution, especially with small samples. Answer C uses 1.729, which appears to be

t_{0.05, 19}

(incorrectly using

n-1

degrees of freedom instead of

n-2

). This is a common error when students forget that simple linear regression loses 2 degrees of freedom. Answer D uses 2.101, which is

t_{0.025, 18}

- the critical value for a 95% confidence interval, not 90%. Study tip: Always remember the "2" rule for simple linear regression: use

n-2

degrees of freedom and verify your confidence level matches your critical value. Double-check whether the problem calls for t or z distribution.

Question 8

A pharmaceutical company tests whether a new drug reduces cholesterol levels. They regress cholesterol reduction on drug dosage and obtain $\hat{\beta}_1 = 0.8$ mg/dL per mg dose with $t = 2.4$ and $p = 0.025$ for $n = 24$ . The company wants evidence that higher doses reduce cholesterol more. What conclusion is appropriate?

Reject the null hypothesis at $\alpha = 0.05$ because $p = 0.025 < 0.05$ using two-tailed test
Fail to reject the null hypothesis because the slope is positive, indicating no reduction
The evidence supports the company's claim at $\alpha = 0.05$ with one-tailed $p = 0.0125$ (correct answer)
Cannot conclude anything because cholesterol reduction should have a negative slope coefficient
Reject the null hypothesis at $\alpha = 0.01$ because $t = 2.4 > 2.074$ critical value

Explanation: When you encounter regression problems involving directional hypotheses, you need to carefully consider both the sign of the coefficient and whether to use one-tailed or two-tailed testing. The company wants evidence that "higher doses reduce cholesterol more." This suggests a positive relationship between dosage and cholesterol reduction (more dose = more reduction). The positive slope

\hat{\beta}_1 = 0.8

supports this—each additional mg of drug increases cholesterol reduction by 0.8 mg/dL. Since the company has a specific directional hypothesis (higher doses should be better), a one-tailed test is appropriate. For a one-tailed test, you divide the two-tailed p-value by 2:

p_{one-tailed} = 0.025/2 = 0.0125 < 0.05

, supporting the company's claim. Option A incorrectly uses a two-tailed test when the research question clearly has direction. While the conclusion about rejecting the null is correct, it's not the most powerful test for this scenario. Option B fundamentally misunderstands the variables—a positive slope here means more drug leads to more cholesterol reduction, which is exactly what the company wants. Option D makes the same conceptual error as B. The response variable is "cholesterol reduction" (a benefit), not "cholesterol level." A positive coefficient means higher doses produce greater reductions, which is the desired effect. Remember: Always identify what your response variable actually measures. When the outcome is already framed as a benefit (like "reduction" or "improvement"), a positive coefficient indicates the treatment is working in the desired direction.

Question 9

A regression analysis yields the following results: $\hat{\beta}_1 = 3.2$ , $SE(\hat{\beta}_1) = 1.1$ , $df = 16$ . If the 95% confidence interval for $\beta_1$ is $(0.87, 5.53)$ , what can be concluded about testing $H_0: \beta_1 = 6$ versus $H_1: \beta_1 \neq 6$ ?

Reject $H_0$ at $\alpha = 0.05$ because 6 is not in the confidence interval (correct answer)
Fail to reject $H_0$ at $\alpha = 0.05$ because 6 > 5.53
Cannot determine without calculating the test statistic
Reject $H_0$ because $\hat{\beta}_1 = 3.2 \neq 6$
Fail to reject $H_0$ because the interval contains positive values

Explanation: When you encounter regression problems involving hypothesis testing and confidence intervals, remember that these two methods are intimately connected. A 95% confidence interval contains all parameter values that would NOT be rejected at the

\alpha = 0.05

level in a two-tailed test. The key insight here is that if a hypothesized value falls outside the 95% confidence interval, you can immediately conclude it would be rejected in a hypothesis test at

\alpha = 0.05

. Since the confidence interval for

\beta_1

(0.87, 5.53)

and our null hypothesis claims

\beta_1 = 6

, we can see that 6 falls outside this interval (specifically, above the upper bound of 5.53). Answer A is correct because 6 is not contained in the confidence interval, which means we reject

H_0

\alpha = 0.05

. Answer B makes a logical error—while it correctly notes that 6 > 5.53, it incorrectly concludes we should fail to reject

H_0

. When the hypothesized value exceeds the confidence interval bounds, we reject, not fail to reject. Answer C is wrong because the confidence interval provides all the information needed; no additional test statistic calculation is required. Answer D reaches the right conclusion (reject) but for the wrong reason—simply having

\hat{\beta}_1 \neq 6

doesn't determine significance. Study tip: Always check whether hypothesized parameter values fall within the corresponding confidence interval. If outside the interval, reject; if inside, fail to reject. This shortcut works for any confidence level and saves calculation time.

Question 10

In a simple linear regression with 18 degrees of freedom for error, the slope estimate is $\hat{\beta}_1 = 2.4$ with standard error $SE(\hat{\beta}_1) = 0.8$ . What is the 95% confidence interval for the true slope?

$2.4 \pm 2.101(0.8) = (0.72, 4.08)$ (correct answer)
$2.4 \pm 1.96(0.8) = (0.83, 3.97)$
$2.4 \pm 2.086(0.8) = (0.73, 4.07)$
$2.4 \pm 2.048(0.8) = (0.76, 4.04)$
$2.4 \pm 1.734(0.8) = (1.01, 3.79)$

Explanation: When you're constructing a confidence interval for a regression slope, you need to use the t-distribution because you're estimating the population standard deviation from sample data. The key insight is identifying the correct degrees of freedom to find the appropriate t-critical value. The confidence interval formula is:

\hat{\beta}_1 \pm t_{\alpha/2,df} \times SE(\hat{\beta}_1)

. Here, you have

\hat{\beta}_1 = 2.4

and

SE(\hat{\beta}_1) = 0.8

. For a 95% confidence interval,

\alpha = 0.05

, so you need

t_{0.025,18}

. With 18 degrees of freedom, the critical t-value is 2.101. This gives you:

2.4 \pm 2.101(0.8) = 2.4 \pm 1.681 = (0.719, 4.081)

, which rounds to

(0.72, 4.08)

. Choice A is correct because it uses the proper t-critical value of 2.101 for 18 degrees of freedom. Choice B incorrectly uses 1.96, which is the z-critical value you'd use if you knew the population standard deviation—a common mistake when students forget that regression problems require the t-distribution. Choice C uses 2.086, which is the t-critical value for 19 degrees of freedom, suggesting confusion about how degrees of freedom are calculated in regression. Choice D uses 2.048, which corresponds to 20 degrees of freedom. Study tip: Always double-check your degrees of freedom in regression problems. For simple linear regression with n observations, error df = n - 2. Then look up the exact t-critical value rather than memorizing approximate values, as small differences significantly affect your interval bounds.

Question 11

A study of advertising spending and sales revenue gives $\hat{\beta}_1 = 4.2$ with $SE(\hat{\beta}_1) = 1.5$ and $n = 18$ . The researcher wants to test if each dollar of advertising increases revenue by more than $3. What is the p-value for this test?

$P(t_{16} > 0.8) \approx 0.22$ (correct answer)
$P(|t_{16}| > 0.8) \approx 0.44$
$P(t_{17} > 0.8) \approx 0.22$
$P(t_{16} > 2.8) \approx 0.007$
$P(t_{16} < 0.8) \approx 0.78$

Explanation: When testing whether a regression coefficient exceeds a specific value, you're conducting a one-tailed hypothesis test. Here, you want to test if advertising increases revenue by more than $3, so your null hypothesis is $

H_0: \beta_1 = 3

and alternative hypothesis is

H_1: \beta_1 > 3

$. The test statistic follows the formula:

t = \frac{\hat{\beta}_1 - \beta_0}{SE(\hat{\beta}_1)} = \frac{4.2 - 3}{1.5} = \frac{1.2}{1.5} = 0.8

Since you have

n = 18

observations, your degrees of freedom are

df = n - 2 = 16

(we subtract 2 because we estimate both the intercept and slope in simple linear regression). Because this is a one-tailed test asking if the coefficient is greater than 3, the p-value is

P(t_{16} > 0.8)

, which equals approximately 0.22. Answer A is correct because it uses the right degrees of freedom (16), the correct test statistic (0.8), and the proper one-tailed format. Answer B is wrong because it shows a two-tailed test with

|t_{16}|

, but you're only testing if the coefficient is greater than 3, not different from 3. Answer C uses the wrong degrees of freedom (17 instead of 16) - a common mistake when students forget that simple linear regression costs 2 degrees of freedom. Answer D incorrectly uses 2.8 as the test statistic, possibly from calculating

\frac{4.2}{1.5}

instead of

\frac{4.2-3}{1.5}

. Study tip: Always remember that hypothesis tests compare your estimate to the hypothesized value, not to zero, and simple linear regression always uses

n-2

degrees of freedom.

Question 12

A medical researcher tests the effect of a new treatment on recovery time. The regression gives $\hat{\beta}_1 = -3.5$ days per treatment unit with $SE(\hat{\beta}_1) = 1.2$ and $n = 28$ . She wants to show the treatment reduces recovery time. At $\alpha = 0.01$ , what should she conclude?

Reject $H_0: \beta_1 \geq 0$ because $t = -2.92 < -t_{26,0.01} = -2.479$ (correct answer)
Fail to reject $H_0: \beta_1 \geq 0$ because $t = -2.92 > -t_{26,0.01} = -2.479$
Reject $H_0: \beta_1 = 0$ because $|t| = 2.92 > t_{26,0.005} = 2.779$
Fail to reject $H_0: \beta_1 = 0$ because $|t| = 2.92 < t_{26,0.005} = 2.779$
Reject $H_0: \beta_1 \leq 0$ because the treatment reduces recovery time

Explanation: When you encounter regression analysis with hypothesis testing, you need to carefully identify what the researcher is trying to prove and set up the appropriate one-tailed or two-tailed test. Since the researcher wants to show the treatment reduces recovery time, she's testing whether the slope is negative. This requires a one-tailed test with

H_0: \beta_1 \geq 0

versus

H_1: \beta_1 < 0

. The test statistic is

t = \frac{\hat{\beta}_1}{SE(\hat{\beta}_1)} = \frac{-3.5}{1.2} = -2.92

with

df = n-2 = 26

. For a one-tailed test at

\alpha = 0.01

, the critical value is

-t_{26,0.01} = -2.479

. Since

t = -2.92 < -2.479

, we reject

H_0

and conclude the treatment significantly reduces recovery time. Choice A is correct because it uses the proper one-tailed hypothesis, calculates the test statistic correctly, and applies the right decision rule. Choice B uses the correct hypothesis and test statistic but makes a critical error in the decision rule—it says

-2.92 > -2.479

, which is mathematically wrong. Choice C incorrectly sets up a two-tailed test (

H_0: \beta_1 = 0

) when the researcher has a specific directional hypothesis. While the conclusion happens to be the same, the approach doesn't match the research question. Choice D also uses the wrong two-tailed setup and incorrectly concludes there's insufficient evidence, missing the significant effect entirely. Study tip: Always match your hypothesis test to the research question. Directional claims like "reduces," "increases," or "improves" require one-tailed tests with appropriately positioned critical regions.

Question 13

A researcher claims that each additional year of experience increases salary by at least $1500. With $ $n = 40$ $, she obtains$ $\hat{\beta}_1 = 1350$ $and$ $SE(\hat{\beta}_1) = 200$ $. To test her claim at$ $\alpha = 0.05$ $, what should she conclude?

Reject the claim because $t = -0.75$ and $p > 0.05$
Support the claim because $\hat{\beta}_1 = 1350 > 0$
Fail to support the claim because $t = -0.75 > -1.686$ (correct answer)
Support the claim because the confidence interval contains 1500
Reject the claim because $1350 < 1500$

Explanation: When you encounter hypothesis testing for regression coefficients, you need to set up the null and alternative hypotheses based on the researcher's claim. Here, the claim is that each additional year increases salary by "at least $1500," so you're testing $

H_0: \beta_1 = 1500

versus

H_1: \beta_1 < 1500

$ (one-tailed test). The test statistic is

t = \frac{\hat{\beta}_1 - \beta_0}{SE(\hat{\beta}_1)} = \frac{1350 - 1500}{200} = -0.75

. With

n = 40

, you have

df = 38

, and the critical value for a one-tailed test at

\alpha = 0.05

is approximately

-1.686

. Since

t = -0.75 > -1.686

, you fail to reject

H_0

, meaning you cannot support the claim that the increase is at least $1500. Answer C correctly identifies this reasoning: the test statistic

t = -0.75

is greater than the critical value

-1.686

, so we fail to support the claim. Answer A correctly calculates the t-statistic and notes

p > 0.05

, but incorrectly says "reject the claim" when you actually fail to reject the null hypothesis. Answer B commits a fundamental error by simply comparing

\hat{\beta}_1

to zero rather than to the claimed value of 1500. Answer D is wrong because a confidence interval around 1350 would likely not contain 1500 given the standard error. Remember: always set up your null hypothesis based on the specific claim being tested, not just whether the coefficient is different from zero. The null hypothesis should reflect the boundary of what's being claimed.

Question 14

A regression analysis yields $\hat{\beta}_1 = 1.8$ with $SE(\hat{\beta}_1) = 0.6$ and $n = 22$ . To test $H_0: \beta_1 = 2.5$ versus $H_1: \beta_1 \neq 2.5$ , what is the p-value?

$P(|t_{20}| > 1.17) \approx 0.26$ (correct answer)
$P(t_{20} < -1.17) \approx 0.13$
$P(|t_{21}| > 1.17) \approx 0.25$
$P(|z| > 1.17) \approx 0.24$
$P(|t_{20}| > 3.0) \approx 0.007$

Explanation: When testing hypotheses about regression coefficients, you're conducting a t-test using the sampling distribution of the coefficient estimate. The key framework here is setting up the test statistic and identifying the correct degrees of freedom. To test

H_0: \beta_1 = 2.5

versus

H_1: \beta_1 \neq 2.5

, you calculate the t-statistic as:

t = \frac{\hat{\beta}_1 - \beta_{1,0}}{SE(\hat{\beta}_1)} = \frac{1.8 - 2.5}{0.6} = \frac{-0.7}{0.6} = -1.17

Since this is a two-tailed test (

H_1: \beta_1 \neq 2.5

), the p-value is

P(|t| > 1.17)

. The degrees of freedom for a simple linear regression are

n - 2 = 22 - 2 = 20

, giving us

P(|t_{20}| > 1.17) \approx 0.26

. Answer A is correct because it uses the proper degrees of freedom (20) and correctly formulates the two-tailed test with the absolute value notation. Answer B incorrectly calculates a one-tailed p-value (

P(t_{20} < -1.17)

) instead of the two-tailed test specified by the alternative hypothesis. Answer C uses incorrect degrees of freedom (21 instead of 20). In simple linear regression, you always lose 2 degrees of freedom for the intercept and slope parameters. Answer D incorrectly uses the standard normal distribution (z) instead of the t-distribution. With a sample size of 22, the t-distribution is appropriate, especially since we're working with an estimated standard error. Remember: regression hypothesis tests use t-distributions with

n-2

degrees of freedom, and two-tailed tests require absolute value notation in the p-value calculation.

Question 15

For a regression model predicting blood pressure from age, the output shows $\hat{\beta}_1 = 0.8$ , $SE(\hat{\beta}_1) = 0.25$ , and $t = 3.2$ with $p = 0.003$ . If the researcher constructs a 99% confidence interval for $\beta_1$ , which statement is most likely true?

The interval will contain 0 because p > 0.001
The interval will not contain 0 because the test rejects at $\alpha = 0.01$ (correct answer)
The interval will contain 0 because 99% confidence requires $\alpha = 0.01$
Cannot determine without knowing the exact sample size and critical value
The interval will not contain 0 only if the sample size exceeds 30

Explanation: When you encounter regression output with hypothesis testing results, remember that confidence intervals and hypothesis tests are mathematically linked through the same underlying distribution and significance levels. The key insight is recognizing the relationship between p-values and confidence intervals. Since

p = 0.003 < 0.01

, we reject the null hypothesis

H_0: \beta_1 = 0

at the

\alpha = 0.01

significance level. This means there's sufficient evidence that the true slope is not zero. When a hypothesis test rejects

H_0: \beta_1 = 0

at significance level

\alpha

, the corresponding

(1-\alpha) \times 100\%

confidence interval will not contain zero. Since we reject at

\alpha = 0.01

, the 99% confidence interval will exclude zero. Choice A incorrectly suggests the interval contains zero because

p > 0.001

. While true that

p > 0.001

, the relevant comparison is

p < 0.01

, which leads to rejection and an interval excluding zero. Choice C misunderstands the relationship between confidence levels and hypothesis testing. A 99% confidence interval corresponds to

\alpha = 0.01

, and since we reject

H_0

at this level, the interval won't contain zero. Choice D unnecessarily complicates the situation. The p-value already incorporates the degrees of freedom and critical value information needed for this determination. Study tip: Remember this fundamental relationship: if a two-sided test rejects

H_0: \beta = 0

at level

\alpha

, then the

(1-\alpha) \times 100\%

confidence interval for

\beta

will not contain zero. This connection appears frequently on biostatistics exams.

Question 16

In a regression of weight loss (pounds) on exercise minutes per day, a researcher gets $\hat{\beta}_1 = 0.05$ with $SE(\hat{\beta}_1) = 0.02$ and $n = 30$ . To test if exercise has any effect on weight loss using $\alpha = 0.02$ , what is the rejection region?

$|t| > t_{28,0.01} = 2.467$ (correct answer)
$|t| > t_{29,0.02} = 2.150$
$|t| > t_{28,0.02} = 2.154$
$t > t_{28,0.02} = 2.154$
$|t| > 2.326$

Explanation: When testing whether a regression coefficient is significantly different from zero, you're conducting a two-tailed hypothesis test. The key elements you need to identify are the degrees of freedom and the appropriate critical value for your significance level. For this regression with

n = 30

observations, the degrees of freedom are

df = n - 2 = 28

(we subtract 2 because we estimate both the intercept and slope). Since we're testing "if exercise has any effect" (not specifically positive or negative), this is a two-tailed test with

H_0: \beta_1 = 0

versus

H_a: \beta_1 \neq 0

. With

\alpha = 0.02

for a two-tailed test, you split this between both tails:

\alpha/2 = 0.01

in each tail. Therefore, you need the critical value

t_{28,0.01} = 2.467

, and the rejection region is

|t| > 2.467

. This makes option A correct. Option B uses the wrong degrees of freedom (29 instead of 28) and fails to split alpha for the two-tailed test. Option C has the correct degrees of freedom but incorrectly uses

\alpha = 0.02

instead of

\alpha/2 = 0.01

for the critical value lookup. Option D makes the same alpha-splitting error as C and incorrectly specifies a one-tailed test (

t >

instead of

|t| >

). Remember: for regression coefficient tests, always use

df = n - 2

, and when testing "any effect" (two-tailed), split your alpha level in half when looking up critical values.

Question 17

A researcher studying crop yield versus fertilizer amount obtains the regression output: $\hat{\beta}_1 = 2.3$ kg/hectare per kg fertilizer, $SE(\hat{\beta}_1) = 0.7$ , $n = 21$ . The agricultural guidelines suggest the slope should be at least 2.0. What is the 90% confidence interval and what does it suggest about the guidelines?

$(1.09, 3.51)$ ; supports guidelines since interval is entirely above 2.0
$(1.09, 3.51)$ ; contradicts guidelines since interval contains values below 2.0 (correct answer)
$(1.09, 3.51)$ ; provides insufficient evidence about guidelines since 2.0 is in the interval
$(0.83, 3.77)$ ; supports guidelines since interval contains values above 2.0
$(1.12, 3.48)$ ; supports guidelines since the point estimate exceeds 2.0

Explanation: When evaluating whether data supports a threshold claim, you need to construct a confidence interval and examine whether it entirely exceeds the threshold value. To find the 90% confidence interval for the slope coefficient, use:

\hat{\beta}_1 \pm t_{\alpha/2} \times SE(\hat{\beta}_1)

. With

n = 21

, you have

df = 20

, so

t_{0.05} = 1.725

. The interval becomes:

2.3 \pm 1.725(0.7) = 2.3 \pm 1.21 = (1.09, 3.51)

. The key insight is interpreting what this interval means for the guidelines. Since the guidelines claim the slope should be "at least 2.0," you need the entire confidence interval to be above 2.0 to support this claim. However, this interval extends from 1.09 to 3.51, meaning plausible values include those below 2.0 (like 1.5 or 1.8). This contradicts the guidelines' assertion. Choice A incorrectly states the interval supports the guidelines - it doesn't because part of the interval falls below 2.0. Choice C mischaracterizes this as "insufficient evidence" when the interval actually provides evidence against the guidelines by including values below the threshold. Choice D uses an incorrect critical value (likely

t_{0.025} = 2.086

), yielding the wrong interval bounds. Study tip: For threshold claims, the entire confidence interval must be on the correct side of the threshold to provide support. If any part of the interval violates the claim, it contradicts rather than supports the hypothesis.

Question 18

In a regression analysis with $n = 15$ , a researcher obtains $\hat{\beta}_1 = -1.6$ and $SE(\hat{\beta}_1) = 0.4$ . For testing $H_0: \beta_1 \geq 0$ versus $H_1: \beta_1 < 0$ at $\alpha = 0.05$ , what is the critical value and decision?

Critical value is $-t_{13,0.05} = -1.771$ ; reject $H_0$ because $t = -4.0 < -1.771$ (correct answer)
Critical value is $t_{13,0.05} = 1.771$ ; reject $H_0$ because $|t| = 4.0 > 1.771$
Critical value is $-t_{14,0.05} = -1.761$ ; reject $H_0$ because $t = -4.0 < -1.761$
Critical value is $-t_{13,0.025} = -2.160$ ; fail to reject $H_0$ because $t = -4.0 > -2.160$
Critical value is $-1.96$ ; reject $H_0$ because $t = -4.0 < -1.96$

Explanation: When you encounter hypothesis testing for regression coefficients, you need to carefully identify the type of test (one-tailed vs. two-tailed) and determine the correct degrees of freedom and critical value accordingly. This is a one-tailed test since we're testing

H_0: \beta_1 \geq 0

versus

H_1: \beta_1 < 0

. The test statistic is

t = \frac{\hat{\beta}_1 - 0}{SE(\hat{\beta}_1)} = \frac{-1.6}{0.4} = -4.0

. For simple linear regression with

n = 15

, the degrees of freedom are

df = n - 2 = 13

. Since this is a left-tailed test (

H_1: \beta_1 < 0

), the critical value is

-t_{13,0.05} = -1.771

. We reject

H_0

because our test statistic

t = -4.0

falls in the rejection region (

t < -1.771

). Answer A is correct for these reasons. Answer B incorrectly treats this as a two-tailed test, using the positive critical value and absolute value of the test statistic. Answer C uses the wrong degrees of freedom (14 instead of 13) – remember that simple linear regression uses

n-2

degrees of freedom, not

n-1

. Answer D uses

\alpha/2 = 0.025

instead of

\alpha = 0.05

, which would be appropriate for a two-tailed test but not this one-tailed scenario. Study tip: Always identify whether the alternative hypothesis indicates a one-tailed or two-tailed test, and remember that simple linear regression uses

df = n-2

because you estimate two parameters (intercept and slope).

Question 19

A study of 26 patients examines the relationship between medication dose (mg) and symptom improvement score. The 95% confidence interval for the slope is $(-0.5, 2.1)$ . What can be concluded about the effectiveness of the medication?

The medication is effective because the interval contains positive values
There is insufficient evidence that dose affects symptoms at $\alpha = 0.05$ (correct answer)
The medication is harmful because the interval contains negative values
The medication is definitely effective because the upper bound is positive
The medication has no effect because the interval is centered near zero

Explanation: When interpreting confidence intervals for regression slopes, you're essentially testing whether there's a statistically significant relationship between your variables. The key question is: does the interval contain zero? Since the 95% confidence interval for the slope is (-0.5, 2.1), it includes zero. This means we cannot reject the null hypothesis that the true slope equals zero at

\alpha = 0.05

. In other words, there's insufficient statistical evidence that medication dose affects symptom improvement scores. Answer B correctly identifies this lack of statistical significance. The interval spanning zero indicates we cannot confidently conclude there's any relationship between dose and symptoms. Answer A makes a common error by focusing only on the positive values in the interval. While the interval does contain positive values, it also contains zero and negative values, so we cannot conclude effectiveness based solely on the positive portion. Answer C commits the opposite mistake, focusing only on the negative values. The presence of negative values doesn't prove harm any more than positive values prove benefit when zero is included in the interval. Answer D incorrectly interprets the upper bound as definitive proof of effectiveness. A positive upper bound alone means nothing when the interval includes zero – this suggests the student misunderstands how confidence intervals work. Remember this key principle: when a confidence interval for a slope contains zero, you lack evidence of a significant relationship. Always check whether zero falls within the interval before drawing conclusions about statistical significance.

Question 20

In studying patient recovery times, a researcher obtains $\hat{\beta}_1 = -2.5$ days per unit treatment with $SE(\hat{\beta}_1) = 0.8$ days and $n = 25$ . For the test $H_0: \beta_1 = -3$ versus $H_1: \beta_1 > -3$ at $\alpha = 0.10$ , what is the conclusion?

Reject $H_0$ because $t = 0.625 > t_{23,0.10} = 1.319$
Fail to reject $H_0$ because $t = 0.625 < t_{23,0.10} = 1.319$ (correct answer)
Reject $H_0$ because $-2.5 > -3$
Fail to reject $H_0$ because $t = -3.125$ is too negative
Reject $H_0$ because $|t| = 0.625 > 0$

Explanation: When you encounter hypothesis testing for regression coefficients, you need to systematically work through the t-test procedure rather than making intuitive comparisons. Here's the proper approach: First, calculate the test statistic using

t = \frac{\hat{\beta}_1 - \beta_{1,H_0}}{SE(\hat{\beta}_1)} = \frac{-2.5 - (-3)}{0.8} = \frac{0.5}{0.8} = 0.625

. Next, find the critical value. Since this is a one-tailed test (

H_1: \beta_1 > -3

) with

\alpha = 0.10

and

df = n-2 = 23

, you need

t_{23,0.10} = 1.319

. Finally, compare: since

t = 0.625 < 1.319

, you fail to reject

H_0

. Option A incorrectly states that

0.625 > 1.319

, which is mathematically wrong. Option C commits the fundamental error of comparing point estimates directly to hypothesized values without considering statistical significance—while

-2.5 > -3

numerically, this difference isn't statistically significant given the standard error. Option D uses an incorrect test statistic calculation, likely computing

\frac{-2.5}{0.8} = -3.125

by omitting the hypothesized value from the numerator. Option B correctly identifies both the proper test statistic (

t = 0.625

) and the right conclusion based on the comparison with the critical value. Study tip: Always follow the complete hypothesis testing framework: calculate the test statistic using the formula, find the appropriate critical value, then compare. Never rely on intuitive comparisons between estimates and hypothesized values—the standard error determines whether observed differences are statistically meaningful.