Inference for Regression Slope - College Statistics

Which of the following is a required condition for the calculated p-value from a t-test for the slope of a least-squares regression line to be valid?
Answer: The true relationship between the explanatory and response variables in the population must be linear. The inference procedures for linear regression (t-test and CI for the slope) are based on the model $y = \beta_0 + \beta_1 x + \epsilon$ . This model assumes that the underlying relationship between the variables in the population is linear. If the true relationship is non-linear, the slope of the best-fit line ( $\beta_1$ ) is not a meaningful parameter to describe the association, and inference on it is not valid for its intended purpose. Choice A is a common misconception; the condition is that the residuals (or errors, $\epsilon$ ) are normally distributed, not the variables themselves. Choice C is not required; the methods are valid for observational studies as long as the sample is representative (e.g., a random sample) and other conditions hold. Choice D is a rule of thumb, but not a strict requirement; if the errors are known to be normally distributed, the test is valid for any sample size (n > 2).
Two different labs studied the same linear relationship between a chemical's concentration (X) and its absorbance (Y). Lab 1 used $n=25$ samples and calculated a 95% confidence interval for the slope to be (1.2, 2.8). Lab 2 used $n=100$ samples and calculated a 95% confidence interval for the slope to be (1.7, 2.3). Assuming both studies were well-conducted, which is the most appropriate conclusion?
Answer: Lab 2's estimate of the slope is more precise than Lab 1's estimate. When comparing confidence intervals from different studies, you're essentially comparing the precision and reliability of statistical estimates. The key insight is understanding what the width of a confidence interval tells you about the quality of an estimate. Why A is correct: Precision in statistics refers to how narrow or wide your confidence interval is. Lab 1's interval spans from 1.2 to 2.8 (width = 1.6), while Lab 2's interval spans from 1.7 to 2.3 (width = 0.6). Lab 2's much narrower interval indicates greater precision in their slope estimate. This makes sense because larger sample sizes (n=100 vs n=25) reduce the standard error, leading to more precise estimates. Why the other options are wrong: Option B incorrectly assumes that a narrower confidence interval indicates stronger correlation (higher $R^2$ ). However, confidence interval width reflects precision of the slope estimate, not the strength of the linear relationship. Option C misunderstands confidence intervals—both intervals have the same confidence level (95%), so the "true slope" isn't more likely to be in either specific interval. The larger sample makes the estimate more precise, not more probable to contain the true value. Option D incorrectly suggests contradiction. Since both intervals overlap substantially (from 1.7 to 2.3), the results are actually consistent with each other. Study tip: Remember that larger sample sizes always lead to more precise estimates (narrower confidence intervals), but don't confuse precision with accuracy, strength of relationship, or probability of containing the true parameter.
A researcher investigating the relationship between hours of sleep and cognitive performance score finds a p-value of 0.34 for a t-test on the regression slope. Which of the following is the most appropriate conclusion?
Answer: The data provide insufficient evidence to conclude that a linear relationship exists between hours of sleep and cognitive performance score. A p-value of 0.34 is larger than any standard significance level (e.g., 0.01, 0.05, 0.10). Therefore, we fail to reject the null hypothesis. The correct interpretation of failing to reject $H_0$ is that there is insufficient evidence to support the alternative hypothesis. Choice C accurately states this. Choice A is incorrect because it makes a definitive statement about the population by accepting the null hypothesis, which is statistically improper. Choice B is a common misinterpretation of the p-value; the p-value is the probability of observing the data (or more extreme) assuming $H_0$ is true, not the probability that $H_0$ is true. Choice D is a possibility, but the hypothesis test result alone doesn't confirm it; the lack of significance could be due to a small sample size, large variability, or genuinely no relationship of any kind.
A study is conducted to examine the relationship between the time a patient waits for a doctor's appointment and their satisfaction rating. A 90% confidence interval for the slope of the regression line is found to be (–2.5, –0.5). Which of the following statements is a necessary consequence of this interval?
Answer: The p-value for the two-sided test of $H_0: \beta_1 = 0$ is less than 0.10. When you encounter a confidence interval for a regression slope, you're looking at the range of plausible values for the true relationship between your variables. The key insight is understanding the connection between confidence intervals and hypothesis testing. The 90% confidence interval (-2.5, -0.5) tells us that we're 90% confident the true slope falls within this range. Crucially, since this interval doesn't contain zero, we can reject the null hypothesis $H_0: \beta_1 = 0$ at the 0.10 significance level. This makes choice A correct—the p-value for testing whether the slope equals zero must be less than 0.10. Choice B incorrectly assumes a direct relationship between the confidence interval for the slope and the correlation coefficient. While both are negative here, you cannot determine the correlation's exact range from the slope's confidence interval alone—the relationship depends on the standard deviations of both variables. Choice C makes a common error about significance levels. Since we can reject at α = 0.10 (corresponding to our 90% confidence interval), this doesn't guarantee we can reject at the more stringent α = 0.05 level. We'd need a 95% confidence interval to make that determination. Choice D reverses the relationship between confidence level and interval width. A 95% confidence interval would be wider, not narrower, than the 90% interval because higher confidence requires casting a wider net. Remember: if a confidence interval for a parameter excludes zero, you can reject the null hypothesis that the parameter equals zero at the corresponding significance level.
Suppose a regression model is fit to predict Y from X. If the sample size is quadrupled, while the sample slope $b_1$ and the standard deviation of the residuals $s_e$ remain the same, how would the new value of the t-statistic for the slope compare to the original t-statistic? Assume the standard deviation of X also remains constant.
Answer: It would be approximately two times the original value. The t-statistic is given by $t = \frac{b_1}{SE_{b_1}}$ . The standard error of the slope can be expressed as $SE_{b_1} = \frac{s_e}{\sqrt{\sum(x_i - \bar{x})^2}} = \frac{s_e}{S_x\sqrt{n-1}}$ . If $b_1$ , $s_e$ , and $S_x$ remain constant, the only part of the expression that changes is $\sqrt{n-1}$ . When the sample size $n$ is quadrupled to $4n$ , the new standard error becomes $SE_{new} = \frac{s_e}{S_x\sqrt{4n-1}}$ . For large $n$ , $\sqrt{4n-1} \approx \sqrt{4n} = 2\sqrt{n}$ , while $\sqrt{n-1} \approx \sqrt{n}$ . Thus, the new standard error is approximately half the original standard error. Since the t-statistic is inversely proportional to the standard error ( $t = b_1 / SE_{b_1}$ ), halving the standard error will approximately double the t-statistic.
A regression analysis is performed to predict crop yield (in kilograms per hectare) from the amount of a specific fertilizer applied (in grams per square meter). A test of significance for the slope results in a t-statistic of $t=3.5$ and a p-value of 0.002. The fertilizer amounts were then converted to kilograms per hectare (1 gram per square meter = 10 kilograms per hectare). How will the t-statistic and p-value for the slope change after this conversion of the explanatory variable?
Answer: The t-statistic and the p-value will both remain unchanged. Changing the units of the explanatory variable (X) will change the value of the sample slope ( $b_1$ ) and its standard error ( $SE_{b_1}$ ) by the same factor. If $X_{new} = c \cdot X_{old}$ , then $b_{1,new} = b_{1,old} / c$ and $SE_{b_{1,new}} = SE_{b_{1,old}} / c$ . The t-statistic is the ratio of the slope to its standard error: $t = b_1 / SE_{b_1}$ . When both the numerator and the denominator are divided by the same constant $c$ , the ratio remains unchanged. Since the t-statistic and the degrees of freedom ( $n-2$ ) do not change, the p-value also remains unchanged. The statistical significance of the relationship is invariant to linear transformations of the variables.
An ecologist fits a least-squares regression line to predict tree height from trunk diameter. A 95% confidence interval for the slope of the regression line is calculated to be (2.1, 4.5). If the standard error of the slope ( $SE_{b_1}$ ) was 0.6, which of the following is the most plausible sample size used for the study?
Answer: 62 This is a multi-step problem. First, find the margin of error (ME) and the sample slope ( $b_1$ ) from the confidence interval. The width of the interval is $4.5 - 2.1 = 2.4$ , so the margin of error is $ME = 2.4 / 2 = 1.2$ . Second, use the formula for the margin of error, $ME = t^* \cdot SE_{b_1}$ , to find the critical t-value, $t^*$ . We have $1.2 = t^* \cdot 0.6$ , which gives $t^* = 2.0$ . Third, we need to find the degrees of freedom (df) that correspond to a $t^*$ value of approximately 2.0 for 95% confidence. Looking at a t-distribution table, a $t^*$ value of 2.000 for 95% confidence occurs at $df = 60$ . Finally, use the formula for degrees of freedom in regression, $df = n - 2$ . Solving for $n$ , we get $60 = n - 2$ , which means $n = 62$ .
An industrial engineer hypothesizes that the time to complete a manufacturing task (Y) decreases by exactly 2 minutes for every 1 degree Celsius increase in ambient temperature (X). From a sample of 27 trials, she fits a linear regression model and obtains a sample slope of $b_1 = -2.4$ minutes/degree with a standard error of $SE_{b_1} = 0.25$ . What are the appropriate test statistic and degrees of freedom for evaluating the engineer's hypothesis?
Answer: t = –1.6, with df = 25 The engineer's hypothesis is that the true slope $\beta_1$ is -2. This corresponds to a null hypothesis of $H_0: \beta_1 = -2$ . The question asks for the test statistic to evaluate this specific claim, not the usual test against a slope of zero. The formula for the t-statistic is $t = \frac{b_1 - \beta_0}{SE_{b_1}}$ . Plugging in the values: $t = \frac{-2.4 - (-2)}{0.25} = \frac{-0.4}{0.25} = -1.6$ . The degrees of freedom for a simple linear regression t-test are $df = n - 2$ . With a sample size of $n=27$ , the degrees of freedom are $df = 27 - 2 = 25$ . Distractors are based on testing against a null hypothesis of $\beta_1=0$ (choices A and D) or using the wrong degrees of freedom (choice A).
A researcher performs a t-test for the slope of a regression line with a sample of size $n=25$ and obtains a test statistic of $t=2.1$ . A second researcher repeats the study with a new sample of size $n=102$ and, by coincidence, obtains the exact same sample slope $b_1$ and standard error of the slope $SE_{b_1}$ . How will the p-value for the second study's test compare to the p-value from the first study?
Answer: The p-value will be smaller because the test statistic is evaluated on a t-distribution with more degrees of freedom. The p-value for a t-test depends on both the test statistic and the degrees of freedom. In the first study, $df = 25 - 2 = 23$ . In the second study, $df = 102 - 2 = 100$ . Although the test statistic $t=2.1$ is the same for both, it is evaluated on different t-distributions. As the degrees of freedom increase, the t-distribution becomes less spread out and more similar to the standard normal distribution. This means the tails of the distribution become thinner. For a fixed positive t-value, the area in the tail to the right of that value is smaller for a distribution with more degrees of freedom. Therefore, the p-value for the second study will be smaller than the p-value for the first study.
A 99% confidence interval for the slope of a regression of a car's highway fuel efficiency (in mpg) on its weight (in pounds) is (–0.008, –0.004). Which of the following is NOT a valid conclusion based on this interval?
Answer: There is a 99% probability that the true slope of the regression line is between –0.008 and –0.004. This question tests the formal interpretation of a confidence interval. Choice D represents a common misconception. In the frequentist framework, the true population slope $\beta_1$ is a fixed, unknown constant. It is not a random variable, so it does not make sense to assign a probability to it being in a certain range. The confidence level (99%) refers to the long-run success rate of the method used to construct the interval, not the probability that a specific interval contains the parameter. Choices A, B, and C are all valid statements. A is the correct contextual interpretation. B is valid because –0.005 is within the interval. C is the correct frequentist definition of confidence.
A researcher constructs a 95% confidence interval for the slope of a regression line, $\beta_1$ , and finds it to be (0.55, 0.95). What can be concluded about a hypothesis test of $H_0: \beta_1 = 0$ versus $H_a: \beta_1 > 0$ ?
Answer: The null hypothesis would be rejected at $\alpha = 0.025$ . A 95% confidence interval corresponds to a two-sided test with $\alpha = 0.05$ . Since the interval (0.55, 0.95) does not contain 0, we can reject $H_0: \beta_1 = 0$ in favor of $H_a: \beta_1 \neq 0$ at the $\alpha = 0.05$ level. For a one-sided test, the corresponding significance level is $\alpha/2$ . Since the entire interval is positive, this provides evidence for $H_a: \beta_1 > 0$ . The strength of this evidence corresponds to rejecting the null at $\alpha = 0.05/2 = 0.025$ . Therefore, we can reject $H_0$ at the $\alpha = 0.025$ significance level. Choice A is incorrect because rejection at 0.025 implies rejection at 0.05. Choice C is incorrect because the CI provides exactly the information needed to draw a conclusion. Choice D is incorrect because we only know the p-value is less than 0.025, not its exact value.
A marketing analyst studies the relationship between weekly sales of a product and the amount spent on radio advertising. With a very large sample of $n=5,000$ weeks, the analyst finds a p-value of less than 0.001 for the slope of the regression line, but an $R^2$ value of only 0.03. What is the best interpretation of these results?
Answer: The relationship is statistically significant, but its practical importance is minimal as advertising explains very little of the variation in sales. When you encounter regression analysis problems, you need to distinguish between statistical significance and practical significance - two completely different concepts that often confuse students. Here, the p-value less than 0.001 indicates strong statistical significance, meaning you can confidently reject the null hypothesis that there's no relationship between radio advertising and sales. With such a large sample size (n=5,000), even very small effects become statistically detectable. However, the $R^2 = 0.03$ means only 3% of the variation in sales is explained by radio advertising - the other 97% comes from other factors entirely. This makes answer D correct: the relationship is real (statistically significant) but explains very little of what drives sales success (minimal practical importance). Answer A is wrong because there's no requirement that low p-values accompany high $R^2$ values - with large samples, you can detect tiny but real effects. Answer B reverses the concepts entirely - we have statistical significance but limited practical significance. Answer C misinterprets the results; low $R^2$ doesn't necessarily indicate non-linearity, just that the predictor variable doesn't explain much variance regardless of the relationship's form. Remember this key distinction: statistical significance tells you whether an effect exists, while $R^2$ tells you how much that effect matters in explaining your outcome. Large sample sizes can make trivial relationships statistically significant, so always examine both the p-value and the effect size measures together.
For a simple linear regression model based on a sample of $n$ observations, which of the following statistics is NOT directly used in the formula for the t-statistic $t = b_1 / SE_{b_1}$ for the slope?
Answer: The correlation coefficient, $r$ . When you encounter questions about t-statistics for regression slopes, focus on understanding what components actually build the standard error formula. The t-statistic $t = \frac{b_1}{SE_{b_1}}$ tests whether the slope significantly differs from zero. The standard error of the slope has the formula: $SE_{b_1} = \frac{s_e}{S_x\sqrt{n-1}}$ , where $s_e$ is the standard deviation of residuals and $S_x$ is the standard deviation of the x-values. This shows that three components are directly involved in calculating the t-statistic. Option A is incorrect because $s_e$ appears directly in the numerator of the standard error formula. The residual standard deviation measures how much the observed y-values scatter around the regression line, making it essential for determining the slope's precision. Option B is incorrect because $S_x$ appears in the denominator of the standard error formula. Greater spread in x-values gives us more information about the slope, reducing the standard error and making our slope estimate more precise. Option C is incorrect because sample size $n$ appears in the formula as $\sqrt{n-1}$ . Larger samples provide more reliable slope estimates, directly affecting the standard error calculation. Option D is correct because the correlation coefficient $r$ doesn't appear anywhere in the t-statistic formula. While $r$ measures the strength of linear association between variables, it's not needed to calculate the standard error of the slope. Remember: When analyzing regression t-statistics, focus on the three building blocks of standard error—residual variation, x-variable spread, and sample size—rather than correlation measures.
In a linear regression analysis with $n=12$ , a researcher calculates a test statistic $t=4.1$ for the slope. The researcher wishes to find the p-value for a two-sided test. Which of the following is true about the p-value?
Answer: The p-value is between 0.002 and 0.01. The degrees of freedom for the test are $df = n - 2 = 12 - 2 = 10$ . We need to find the two-sided p-value corresponding to $t=4.1$ with $df=10$ . We can consult a t-distribution table. For $df=10$ , the critical value for a one-tail probability of 0.005 (two-tail probability of 0.01) is $t^* = 3.169$ . The critical value for a one-tail probability of 0.001 (two-tail probability of 0.002) is $t^* = 4.144$ . Our test statistic $t=4.1$ falls between these two critical values. Therefore, the two-sided p-value must be between 0.002 and 0.01. Choice A is incorrect because $t=4.1$ is less than the critical value for a p-value of 0.002. Choices C and D correspond to smaller t-statistics.
In a study with 40 participants, a regression of blood pressure on daily sodium intake yields an $R^2$ value of 0.20. Assuming a positive relationship, what is the value of the t-statistic for testing whether the slope is significantly different from zero?
Answer: 3.08 There is a direct formula connecting the t-statistic for the slope test to $R^2$ and the sample size $n$ : $t = \sqrt{\frac{R^2(n-2)}{1-R^2}}$ . We are given $n=40$ and $R^2=0.20$ . Plugging these values into the formula: $t = \sqrt{\frac{0.20(40-2)}{1-0.20}} = \sqrt{\frac{0.20(38)}{0.80}} = \sqrt{\frac{7.6}{0.80}} = \sqrt{9.5} \approx 3.08$ . Distractors are based on common errors: using $n$ or $n-1$ instead of $n-2$ in the numerator, or misplacing terms in the formula.

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