Biostatistics Quiz: Simple Linear Regression

20 questions · exam conditions

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Simple Linear RegressionQuestion 1 of 20

In a dose-response study, the regression equation is $\hat{Y} = 15.2 + 3.8X$ where Y is response level and X is drug dose (mg). If the 95% confidence interval for the slope is (2.1, 5.5), what can be concluded about the relationship between dose and response?

There is insufficient evidence of any relationship since the confidence interval is relatively wide compared to the point estimate

There is statistically significant evidence of a positive linear relationship since the entire confidence interval lies above zero

The relationship is practically significant since the slope estimate of 3.8 falls near the middle of the confidence interval

There is evidence of a relationship, but we cannot determine its direction since confidence intervals are two-sided

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Biostatistics Quiz

Biostatistics Quiz: Simple Linear Regression

Practice Simple Linear Regression 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 Simple Linear Regression, 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

There is insufficient evidence of any relationship since the confidence interval is relatively wide compared to the point estimate
There is statistically significant evidence of a positive linear relationship since the entire confidence interval lies above zero (correct answer)
The relationship is practically significant since the slope estimate of 3.8 falls near the middle of the confidence interval
There is evidence of a relationship, but we cannot determine its direction since confidence intervals are two-sided

Explanation: Since the entire 95% confidence interval (2.1, 5.5) lies above zero, we can conclude with 95% confidence that the true slope is positive, indicating a statistically significant positive linear relationship. Choice A incorrectly focuses on interval width rather than position relative to zero. Choice C confuses statistical significance with practical significance. Choice D incorrectly suggests uncertainty about direction when the interval clearly excludes negative values.

Question 2

In a simple linear regression with n = 30 observations, $SST = 450$ , $SSE = 180$ , and $\hat{y} = 12 + 0.8x$ . If the mean of x is 25, what is the mean of the observed y values?

$32.0$ (correct answer)
$37.0$
$20.0$
$25.0$
$12.8$

Explanation: When you encounter a simple linear regression problem, remember that the regression line must pass through the point

(\bar{x}, \bar{y})

- this is a fundamental property that connects the means of your variables to the fitted equation. Given the regression equation

\hat{y} = 12 + 0.8x

and

\bar{x} = 25

, you can find

\bar{y}

by substituting the mean of x into the equation:

\bar{y} = 12 + 0.8(25) = 12 + 20 = 32

This gives us answer choice A)

32.0

, which is correct. Let's examine why the other options are wrong. Choice B)

37.0

might result from incorrectly adding the slope to

\bar{x}

instead of multiplying:

12 + 0.8 + 25 = 37.8

(rounded to 37). Choice C)

20.0

could come from only calculating the slope times the mean:

0.8 \times 25 = 20

, forgetting the intercept entirely. Choice D)

25.0

simply uses

\bar{x}

as the answer, ignoring the regression relationship altogether. Notice that the given information about

SST = 450

and

SSE = 180

are red herrings in this particular question - they're not needed to find the mean of y values, though they could help you calculate

R^2

or other regression statistics. Study tip: Always remember that the regression line passes through

(\bar{x}, \bar{y})

. When you know the regression equation and one mean, you can always find the other by direct substitution.

Question 3

A simple linear regression yields $\hat{y} = 25 - 0.4x$ with $s_e = 3.2$ , where $s_e$ is the standard error of the estimate. If a new observation has $x = 10$ , what is the predicted value and its interpretation?

$21.0$ ; approximately 68% of observations with $x = 10$ fall between $17.8$ and $24.2$ (correct answer)
$21.0$ ; approximately 95% of observations with $x = 10$ fall between $14.6$ and $27.4$
$29.0$ ; approximately 68% of observations with $x = 10$ fall between $25.8$ and $32.2$
$25.0$ ; approximately 68% of observations with $x = 10$ fall between $21.8$ and $28.2$
$21.0$ ; the observation will fall exactly at this value with 95% confidence

Explanation: When you encounter a linear regression problem asking for predictions and interpretation, you need to distinguish between point predictions and prediction intervals. The regression equation gives you the point estimate, while the standard error helps construct intervals for where future observations might fall. To find the predicted value, substitute

x = 10

into the regression equation:

\hat{y} = 25 - 0.4(10) = 25 - 4 = 21.0

. The standard error of estimate

s_e = 3.2

tells you about the typical deviation of actual observations from the regression line. For prediction intervals, approximately 68% of observations fall within one standard error of the predicted value:

21.0 \pm 3.2

, giving the interval

(17.8, 24.2)

. Choice A correctly identifies both the predicted value of 21.0 and the proper 68% prediction interval. Choice B has the right predicted value but incorrectly uses

\pm 2s_e

(which would be for a 95% interval) while claiming it's 68%. Choice C makes a calculation error, getting 29.0 instead of 21.0 - this suggests incorrectly adding

0.4 \times 10

instead of subtracting it. Choice D uses

x = 0

(the y-intercept of 25) instead of

x = 10

, showing a failure to substitute the given x-value. Remember that in regression problems, always substitute the given x-value carefully, and match the confidence level with the correct multiple of the standard error:

\pm 1s_e

for 68%,

\pm 2s_e

for 95%.

Question 4

In a simple linear regression analysis, the total sum of squares is $SST = 200$ and the regression sum of squares is $SSR = 150$ . A new data point is added that lies exactly on the regression line. How does this affect the $R^2$ value?

$R^2$ will increase because SSR increases while SSE remains proportionally smaller
$R^2$ will remain exactly the same at 0.75 since the point contributes no residual error (correct answer)
$R^2$ will decrease because adding any point reduces the explained variance proportion
$R^2$ will increase to exactly 1.0 since there is now a perfect fit for all points
$R^2$ cannot be determined without recalculating the entire regression with the new point

Explanation: When you encounter questions about

R^2

and adding data points, focus on what

R^2

actually measures: the proportion of total variance explained by the regression model. Let's establish what we know. Initially,

R^2 = \frac{SSR}{SST} = \frac{150}{200} = 0.75

, meaning 75% of the variance is explained by the regression line. When you add a point that lies exactly on the regression line, this point has zero residual error (the distance between the actual and predicted value is zero). Here's the key insight: adding a point on the regression line increases both

SSR

and

SST

proportionally, leaving

R^2

unchanged. The new point contributes to the total variability (

SST

) but also contributes equally to the explained variability (

SSR

) since it's perfectly predicted. The error sum of squares (

SSE

) increases only by the amount that

SST

increases beyond the

SSR

increase, maintaining the same proportion. Answer A is wrong because while

SSR

does increase,

SSE

doesn't remain "proportionally smaller" — the ratios stay identical. Answer C incorrectly assumes any additional point reduces explained variance, but points on the line actually maintain the explanation quality. Answer D makes the critical error of thinking one perfect point creates perfect fit for all points — the original data still has the same residual errors relative to the line. Remember:

R^2

measures the model's explanatory power, not individual point accuracy. Adding points that conform perfectly to the existing model preserves that explanatory proportion.

Question 5

A simple linear regression of cholesterol level (y) on age (x) yields $\hat{y} = 120 + 2.5x$ . The mean age in the sample is 45 years with a standard deviation of 12 years. What is the predicted cholesterol level for a patient who is 1.5 standard deviations above the mean age?

$277.5$ mg/dL (correct answer)
$255.0$ mg/dL
$232.5$ mg/dL
$165.0$ mg/dL
$157.5$ mg/dL

Explanation: This question tests your ability to work with standardized scores in the context of linear regression. When you encounter problems involving "standard deviations above or below the mean," you need to convert that standardized description back to the actual value before using your regression equation. First, find the actual age for someone 1.5 standard deviations above the mean. Since the mean age is 45 years with a standard deviation of 12 years:

x = 45 + 1.5(12) = 45 + 18 = 63

years. Now substitute this age into the regression equation:

\hat{y} = 120 + 2.5(63) = 120 + 157.5 = 277.5

mg/dL. Looking at the wrong answers: Answer B (

255.0

) likely results from calculating

120 + 2.5(54)

, which suggests incorrectly subtracting 1.5 standard deviations instead of adding them. Answer C (

232.5

) might come from using

x = 45

(the mean) instead of the adjusted value, giving

120 + 2.5(45) = 232.5

. Answer D (

165.0

) could result from various calculation errors, possibly confusing the standard deviation value directly in the equation. The key insight is recognizing this as a two-step process: first convert the standardized score to an actual x-value, then apply the regression equation. Remember that "above the mean" means you add standard deviations, while "below the mean" means you subtract them. Always double-check your arithmetic when working through both conversion steps.

Question 6

A simple linear regression model predicts house price (thousands of dollars) from square footage: $\hat{y} = 45 + 0.12x$ . The average house in the sample is 1800 square feet. If the model explains 49% of the variance in house prices, what is the correlation between square footage and price?

$r = 0.70$ (correct answer)
$r = 0.49$
$r = 0.24$
$r = -0.70$
$r = 0.35$

Explanation: When you encounter questions about linear regression and correlation, remember that the coefficient of determination (

R^2

) and the correlation coefficient (

r

) are directly related through a simple square root relationship. The key insight here is that

R^2 = r^2

. Since the model explains 49% of the variance, we know

R^2 = 0.49

. To find the correlation coefficient, we take the square root:

r = \pm\sqrt{0.49} = \pm 0.70

. The sign of the correlation matches the sign of the regression slope. Since the slope is

+0.12

(positive), the correlation must also be positive. Therefore,

r = +0.70

, making choice A correct. Let's examine why the other options are wrong. Choice B (

r = 0.49

) represents a common mistake of confusing

R^2

with

r

directly—students sometimes think the correlation equals the percentage of variance explained rather than its square root. Choice C (

r = 0.24

) appears to come from incorrectly calculating

\sqrt{0.49/2}

or some other mathematical error. Choice D (

r = -0.70

) gets the magnitude right but misses the sign—since the regression slope is positive (house prices increase with square footage), the correlation must also be positive. Study tip: Always remember that correlation and

R^2

are related by

R^2 = r^2

, and the sign of

r

always matches the sign of the slope in simple linear regression. When you see "explains X% of variance," immediately think

R^2 = X\%

Question 7

A researcher fits $\hat{y} = 15 + 4x$ to predict exam scores (y) from study hours (x). The regression is based on n = 25 students. If the sum of squared deviations of x from its mean is $\sum(x_i - \bar{x})^2 = 100$ , and the standard error of the slope is 0.8, what is the estimated error variance?

$s^2 = 64$ (correct answer)
$s^2 = 80$
$s^2 = 100$
$s^2 = 6.4$
$s^2 = 8$

Explanation: When you encounter regression problems involving standard errors, remember that the standard error of the slope connects the error variance to the variability in your predictor variable through a specific formula. The standard error of the slope coefficient is given by

SE(\hat{\beta_1}) = \sqrt{\frac{s^2}{\sum(x_i - \bar{x})^2}}

, where

s^2

is the error variance we need to find. Given that

SE(\hat{\beta_1}) = 0.8

and

\sum(x_i - \bar{x})^2 = 100

, you can solve for

s^2

. Starting with

0.8 = \sqrt{\frac{s^2}{100}}

, square both sides:

0.64 = \frac{s^2}{100}

. Multiplying both sides by 100 gives

s^2 = 64

, confirming answer A is correct. Let's examine why the other options are wrong. Option B (

s^2 = 80

) would result from incorrectly multiplying 0.8 by 100 instead of squaring first. Option C (

s^2 = 100

) represents the sum of squared deviations of x, which is given data, not the error variance. Option D (

s^2 = 6.4

) comes from multiplying 0.8 by 8, showing confusion about the relationship between the standard error and error variance. Remember this key pattern: the standard error of the slope is the square root of error variance divided by the sum of squared deviations of x. When solving for error variance, always square the standard error first, then multiply by the sum of squared deviations. Don't confuse the error variance with other given quantities in the problem.

Question 8

A researcher reports a simple linear regression with $\hat{y} = 30 - 1.5x$ and states that when x increases by 10 units, y decreases by 15 units on average. A colleague questions whether the intercept is meaningful for interpretation. Under what condition would the intercept of 30 be meaningless for practical interpretation?

When x = 0 is outside the range of observed data or represents an impossible value (correct answer)
When the correlation coefficient is negative, making the intercept unreliable
When the slope coefficient is negative, contradicting the positive intercept
When the sample size is too small to provide stable estimates of parameters
When the R-squared value is below 0.5, indicating poor model fit

Explanation: When interpreting linear regression results, you need to understand what each parameter represents and when those interpretations are practically meaningful. The intercept represents the predicted value of y when x equals zero, while the slope tells you how much y changes for each unit increase in x. The intercept of 30 in

\hat{y} = 30 - 1.5x

means that when x = 0, the predicted value of y is 30. However, this interpretation is only meaningful if x = 0 represents a realistic or observable scenario within your data context. Answer A correctly identifies that the intercept becomes meaningless when x = 0 falls outside the range of your observed data or represents an impossible value in the real world. For example, if x represents age in years, then x = 0 (birth) might be outside your study range of adults, making the intercept practically meaningless even though it's mathematically valid. Answer B incorrectly suggests that negative correlations make intercepts unreliable - the sign of the correlation doesn't affect the validity of the intercept. Answer C wrongly implies there's a mathematical contradiction between a negative slope and positive intercept - these can absolutely coexist and are independent parameters. Answer D focuses on sample size affecting parameter stability, but even with small samples, the intercept's meaningfulness depends on whether x = 0 is interpretable, not on estimation precision. Remember: always check whether x = 0 makes practical sense in your study context before interpreting the intercept. Mathematical validity doesn't guarantee practical meaningfulness in regression analysis.

Question 9

A pharmaceutical company fits a simple linear regression of drug concentration (μg/mL) on time (hours): $\hat{y} = 50e^{-0.1x}$ . A statistician points out that this is not a valid simple linear regression model. What would be the appropriate transformation to make this a linear regression?

Take the natural logarithm of y: $\ln(y) = \ln(50) - 0.1x$ (correct answer)
Take the natural logarithm of x: $y = 50 - 0.1\ln(x)$
Square both sides: $y^2 = 2500e^{-0.2x}$
Take the square root of y: $\sqrt{y} = \sqrt{50}e^{-0.05x}$
Reciprocal transformation: $\frac{1}{y} = \frac{1}{50}e^{0.1x}$

Explanation: When you encounter an exponential equation like

\hat{y} = 50e^{-0.1x}

, recognize that this isn't linear in its current form—the relationship between x and y follows a curve, not a straight line. Linear regression requires the model to be linear in its parameters, following the form

y = a + bx

. The key insight is using logarithmic transformation to linearize exponential relationships. Taking the natural logarithm of both sides of

y = 50e^{-0.1x}

gives you:

\ln(y) = \ln(50e^{-0.1x}) = \ln(50) + \ln(e^{-0.1x}) = \ln(50) - 0.1x

This creates a linear relationship between

\ln(y)

and

x

, where

\ln(50)

is the intercept and

-0.1

is the slope. Answer A correctly applies this logarithmic transformation. Answer B incorrectly transforms the independent variable (x) instead of the dependent variable (y)—this doesn't linearize the exponential relationship. Answer C squares both sides, which actually makes the equation more complex rather than simpler:

y^2 = 2500e^{-0.2x}

is still exponential. Answer D takes the square root of y, yielding

\sqrt{y} = \sqrt{50}e^{-0.05x}

, which remains exponential in form. Remember this pattern: when you see exponential models (with

e^x

terms), use natural logarithm transformation on the dependent variable to linearize. This is a fundamental technique in biostatistics for analyzing phenomena like drug decay, population growth, and dose-response relationships.

Question 10

In a clinical trial, the relationship between dosage (mg) and therapeutic response is modeled as $\hat{y} = 15 + 2.5x$ . The standard error of the slope is 0.6, and the study used $\alpha = 0.05$ with $df = 18$ . The critical t-value is 2.101. What conclusion can be drawn about the slope coefficient?

The slope is significantly different from zero because $|t| = 4.17 > 2.101$ (correct answer)
The slope is not significantly different from zero because the confidence interval includes zero
The slope is significantly different from zero because $|t| = 1.5 < 2.101$
Cannot determine significance without knowing the sample correlation coefficient
The slope is significantly different from 2.0 because $|t| = 0.83 < 2.101$

Explanation: When you encounter regression analysis questions, you're testing whether the relationship between variables is statistically significant. This requires calculating a t-statistic to compare against the critical value. To test if the slope coefficient is significantly different from zero, you calculate:

t = \frac{\text{slope}}{\text{standard error}} = \frac{2.5}{0.6} = 4.17

. Since the absolute value of this t-statistic (4.17) exceeds the critical t-value (2.101), you reject the null hypothesis that the slope equals zero. This means the dosage has a statistically significant effect on therapeutic response. Looking at the wrong answers: Option B incorrectly assumes the confidence interval includes zero without actually calculating it. When

|t| > t_{\text{critical}}

, the confidence interval will not include zero. Option C uses an incorrect t-statistic calculation (1.5 instead of 4.17) and reaches the wrong conclusion based on this error. Option D suggests you need the correlation coefficient, but you already have all necessary information—the slope estimate and its standard error—to perform the significance test. The key insight is that statistical significance in regression depends on whether your calculated t-statistic exceeds the critical value, not on additional parameters like correlation coefficients. Study tip: For regression significance tests, always remember the formula

t = \frac{\text{coefficient}}{\text{standard error}}

. If

|t| > t_{\text{critical}}

, the coefficient is significantly different from zero. You don't need extra information beyond the coefficient estimate and its standard error.

Question 11

A researcher fits two simple linear regression models to the same dataset. Model 1: y vs. x gives $R^2 = 0.64$ . Model 2: x vs. y gives $R^2 = 0.81$ . What can be concluded about these results?

There is an error because both models should yield the same $R^2$ value (correct answer)
Model 2 is better because it has higher $R^2$ , indicating better predictive ability
Model 1 explains 64% of variance in y, while Model 2 explains 81% of variance in x
The correlation coefficient between x and y is $r = 0.72$ based on the average $R^2$
Model 2 should be used because the higher $R^2$ indicates less measurement error

Explanation: When you encounter regression problems involving swapped variables, remember that the coefficient of determination (

R^2

) measures the proportion of variance in the dependent variable explained by the independent variable. This relationship is fundamental and creates a key constraint. In simple linear regression, there's a crucial mathematical relationship: if you fit y vs. x and get

R^2_1

, then fit x vs. y and get

R^2_2

, both

R^2

values must equal

r^2

, where

r

is the correlation coefficient between x and y. Since correlation is symmetric (

r_{xy} = r_{yx}

), both models must yield identical

R^2

values. The correct answer is A because having different

R^2

values (0.64 and 0.81) for swapped variables violates this mathematical principle. One or both calculations contain an error. Answer B is wrong because you can't directly compare these models' predictive ability—they're predicting different variables (y vs. x), making comparison meaningless. Answer C correctly describes what each

R^2

represents but misses the fundamental error. While the interpretation is technically accurate, it ignores that these values are mathematically impossible to obtain simultaneously. Answer D attempts to average the

R^2

values, which is statistically invalid. You can't determine the correlation coefficient by averaging different

R^2

values from swapped regressions. Study tip: Always remember that

R^2

equals

r^2

in simple linear regression, regardless of which variable you designate as dependent. If you see different

R^2

values for swapped variables, there's definitely a calculation error.

Question 12

In a study of medication adherence, a simple linear regression of adherence percentage (y) on number of daily doses (x) yields $\hat{y} = 95 - 8x$ . If the mean number of daily doses is 2.5, what does the slope coefficient indicate about the clinical relationship?

Each additional daily dose is associated with an 8% decrease in adherence on average (correct answer)
Patients taking 8 doses daily will have 0% adherence to their medication regimen
The baseline adherence rate without any medication doses would be 95%
Each additional daily dose is associated with an 8% increase in adherence on average
The model predicts perfect adherence when patients take fewer than 2.5 doses daily

Explanation: When interpreting simple linear regression equations, focus on understanding what each component tells you about the relationship between variables. The equation

\hat{y} = 95 - 8x

reveals how adherence percentage changes as the number of daily doses changes. The slope coefficient (-8) represents the change in the dependent variable (adherence percentage) for each one-unit increase in the independent variable (daily doses). Since the slope is negative, this indicates an inverse relationship: as daily doses increase, adherence decreases. Specifically, each additional daily dose is associated with an 8 percentage point decrease in adherence on average. Answer A correctly identifies this interpretation. The negative slope of -8 means adherence drops by 8% with each additional daily dose, which makes clinical sense since more frequent dosing often reduces patient compliance. Answer B misinterprets the equation by treating the slope as if it were the x-value. While

95 - 8(8) = 31

, this doesn't mean patients taking 8 doses have 0% adherence. Answer C confuses the y-intercept with a meaningful clinical baseline. The intercept (95) represents predicted adherence when x = 0, but zero daily doses isn't clinically meaningful in a medication adherence study. Answer D gets the direction wrong, claiming adherence increases with more doses. This contradicts the negative slope coefficient. Remember: In regression equations, the slope coefficient always tells you the direction and magnitude of change in y per unit change in x. Pay attention to the sign—negative slopes indicate inverse relationships, which often make intuitive sense in healthcare contexts like medication adherence.

Question 13

In a simple linear regression, the slope coefficient is $b_1 = -0.8$ and the correlation coefficient is $r = -0.6$ . If the variance of x is $\text{Var}(x) = 25$ , what is the variance of y?

$\text{Var}(y) = 44.44$ (correct answer)
$\text{Var}(y) = 20.00$
$\text{Var}(y) = 16.00$
$\text{Var}(y) = 31.25$
$\text{Var}(y) = 15.00$

Explanation: When you encounter a linear regression problem involving the relationship between slope, correlation, and variances, remember that these three quantities are mathematically connected through a fundamental formula. The key relationship is:

b_1 = r \cdot \frac{\sqrt{\text{Var}(y)}}{\sqrt{\text{Var}(x)}}

Given

b_1 = -0.8

r = -0.6

, and

\text{Var}(x) = 25

, you can solve for

\text{Var}(y)

-0.8 = -0.6 \cdot \frac{\sqrt{\text{Var}(y)}}{\sqrt{25}}

-0.8 = -0.6 \cdot \frac{\sqrt{\text{Var}(y)}}{5}

0.8 = 0.6 \cdot \frac{\sqrt{\text{Var}(y)}}{5}

\frac{0.8 \times 5}{0.6} = \sqrt{\text{Var}(y)}

\frac{4}{0.6} = 6.67 = \sqrt{\text{Var}(y)}

\text{Var}(y) = (6.67)^2 = 44.44

This confirms answer A is correct. Answer B (20.00) might result from incorrectly using

\text{Var}(y) = r^2 \times \text{Var}(x) = 0.36 \times 25

, but this formula doesn't exist. Answer C (16.00) could come from mistakenly calculating

b_1^2 \times \text{Var}(x) = 0.64 \times 25

, which is also incorrect. Answer D (31.25) might result from using

\frac{b_1^2}{\text{Var}(x)} = \frac{0.64}{25}

in some confused calculation. Remember: the slope coefficient bridges the correlation and the ratio of standard deviations. Always use the fundamental relationship

b_1 = r \cdot \frac{\text{SD}(y)}{\text{SD}(x)}

to solve these problems systematically.

Question 14

A simple linear regression of weight (kg) on height (cm) gives $\hat{y} = -100 + 1.1x$ with $n = 50$ observations. The sum of squared residuals is $\sum(y_i - \hat{y}_i)^2 = 480$ . What is the standard error of the estimate?

$3.16$ kg (correct answer)
$3.08$ kg
$9.60$ kg
$10.00$ kg
$96.00$ kg

Explanation: When you encounter a linear regression problem asking for the "standard error of the estimate," you're being asked to find the standard deviation of the residuals - essentially how much the actual data points typically deviate from the regression line. The standard error of the estimate uses the formula:

s_e = \sqrt{\frac{\sum(y_i - \hat{y}_i)^2}{n-2}}

The denominator is

n-2

because we lose two degrees of freedom when estimating the slope and intercept in simple linear regression. With the given information:

\sum(y_i - \hat{y}_i)^2 = 480

and

n = 50

, we calculate:

s_e = \sqrt{\frac{480}{50-2}} = \sqrt{\frac{480}{48}} = \sqrt{10} = 3.16

This confirms answer A)

3.16

kg is correct. Looking at the wrong answers: B)

3.08

kg likely results from incorrectly using

n-1

in the denominator (

\sqrt{480/49} = 3.13

, close to 3.08). C)

9.60

kg appears to come from forgetting the square root entirely (

480/50 = 9.6

). D)

10.00

kg represents the intermediate calculation

480/48 = 10

without taking the square root. Remember this key distinction: for regression problems, always use

n-2

degrees of freedom for the standard error of the estimate, not

n-1

like you would for a sample standard deviation. The "2" accounts for estimating both regression parameters.

Question 15

Two simple linear regressions are fit to related datasets. Dataset A: $\hat{y} = 20 + 3x$ with $s_e = 4$ . Dataset B contains the same x values, but each y value is doubled. What is the regression equation and standard error for Dataset B?

$\hat{y} = 40 + 6x$ with $s_e = 8$ (correct answer)
$\hat{y} = 40 + 6x$ with $s_e = 4$
$\hat{y} = 20 + 6x$ with $s_e = 8$
$\hat{y} = 40 + 3x$ with $s_e = 8$
$\hat{y} = 20 + 3x$ with $s_e = 8$

Explanation: When you encounter linear transformation problems in regression, think about how scaling affects both the regression coefficients and the error terms. This tests your understanding of how linear transformations propagate through regression equations. Since each y value in Dataset B is doubled, we apply the transformation

y_{new} = 2y_{old}

. In linear regression, when you multiply all y values by a constant, both the intercept and slope get multiplied by that same constant, and the standard error also gets multiplied by that constant. Starting with Dataset A's equation

\hat{y} = 20 + 3x

and

s_e = 4

New intercept: $2 \times 20 = 40$
New slope: $2 \times 3 = 6$
New standard error: $2 \times 4 = 8$

This gives us

\hat{y} = 40 + 6x

with

s_e = 8

, which is answer choice A. Answer B incorrectly keeps the standard error unchanged at 4, missing that error terms scale with the y-transformation. Answer C only transforms the slope coefficient, incorrectly keeping the intercept at 20. Answer D only transforms the intercept, incorrectly keeping the slope at 3. Study tip: Remember the "all or nothing" rule for y-transformations in regression: when you multiply all y values by a constant, that constant multiplies everything in the regression output—intercept, slope, and standard error. This uniform scaling preserves the relative relationships in your model while adjusting for the new scale of measurement.

Question 16

A regression model predicting weight loss (kg) from weeks in program gives $\hat{Y} = 2.1 + 0.8X$ . The researcher claims this proves the program causes weight loss. In the context of regression analysis, what is the primary issue with this causal interpretation?

The positive intercept suggests people gain weight before starting, invalidating any causal claims about the program
Causal interpretation requires R² > 0.8, and this value cannot be determined from the regression equation alone
The slope coefficient is too small to represent meaningful weight loss, making causal claims irrelevant regardless of statistical significance
Linear regression can only establish association, not causation; confounding variables and study design determine causal validity (correct answer)

Explanation: When you encounter regression equations in biostatistics, remember that statistical models reveal relationships between variables, but the strength of causal claims depends entirely on study design and methodology, not the mathematical relationship itself. The regression equation

\hat{Y} = 2.1 + 0.8X

shows that for each additional week in the program, predicted weight loss increases by 0.8 kg. However, this association alone cannot prove causation. Linear regression is a descriptive tool that quantifies how variables move together, but correlation never implies causation. To make valid causal claims, you need proper experimental design (like randomized controlled trials) that controls for confounding variables, establishes temporal sequence, and rules out alternative explanations. Option A incorrectly focuses on the intercept interpretation. The positive intercept (2.1 kg) represents the predicted weight loss at week 0, which could reflect initial program effects or measurement timing—it doesn't invalidate causal inference by itself. Option B creates a false rule about R². No specific R² threshold determines causal validity. Even perfect correlation (R² = 1.0) doesn't prove causation if confounding variables aren't controlled. Option C misunderstands the relationship between effect size and causation. A 0.8 kg weekly weight loss is actually quite substantial. More importantly, effect magnitude doesn't determine whether a relationship is causal—proper study design does. Remember this key principle: statistical significance and strong associations are necessary but not sufficient for causal claims. Always evaluate the study design, potential confounders, and whether alternative explanations have been ruled out before accepting causal interpretations.

Question 17

A study examines the relationship between hours of sleep (X) and reaction time in milliseconds (Y). The fitted model is $\hat{Y} = 520 - 15X$ with $s_e = 25$ (residual standard error). Assuming normality, what is the approximate 95% prediction interval for reaction time when sleep hours = 7?

(365, 465) (correct answer)
(375, 455)
(385, 445)
(395, 435)

Explanation: Predicted value:

\hat{Y} = 520 - 15(7) = 520 - 105 = 415

ms. For a 95% prediction interval, we use approximately ±2 standard errors: 415 ± 2(25) = 415 ± 50, giving (365, 465). Prediction intervals are wider than confidence intervals because they account for individual prediction uncertainty plus model uncertainty. Other choices represent inappropriately narrow intervals.

Question 18

In a study of bone density, a simple linear regression yields $\hat{Y} = 2.1 - 0.03X$ , where Y is bone density (g/cm²) and X is age in years. The correlation coefficient is r = -0.67. What percentage of the variation in bone density is explained by the linear relationship with age?

67%
45% (correct answer)
33%
82%

Explanation: The coefficient of determination R² represents the percentage of variation explained by the regression. R² = r² = (-0.67)² = 0.4489 ≈ 0.45 = 45%. Choice A incorrectly uses the absolute value of r. Choice C incorrectly calculates 100% - 67%. Choice D incorrectly uses some other calculation not related to r².

Question 19

A regression of heart rate (Y) on exercise duration in minutes (X) gives $\hat{Y} = 72 + 2.5X$ with standard error of the slope $SE(\hat{\beta_1}) = 0.8$ . For a 95% confidence interval of the slope (assuming df = 28), which t-value should be used and what does the slope coefficient represent in context?

Use t = 1.96; slope represents the baseline heart rate when exercise duration is zero minutes
Use t = 2.048; slope represents the expected increase in heart rate per additional minute of exercise (correct answer)
Use t = 1.96; slope represents the expected increase in heart rate per additional minute of exercise
Use t = 2.048; slope represents the total heart rate after any given exercise duration

Explanation: With df = 28, the critical t-value for 95% confidence is approximately 2.048 (not 1.96 which is for large samples/normal distribution). The slope coefficient (2.5) represents the expected change in Y for each unit increase in X, i.e., the expected increase in heart rate per additional minute of exercise. Choice A uses wrong t-value and confuses slope with intercept. Choice C uses wrong t-value. Choice D misinterprets the slope as total rather than rate of change.

Question 20

A researcher fits $\hat{Y} = 42.5 - 1.8X$ to model the relationship between cognitive test scores (Y) and age (X) in elderly patients. The sum of squared errors (SSE) is 2,840 and the total sum of squares (SST) is 7,200. If a new 75-year-old patient is tested, what is the predicted cognitive score and the coefficient of determination?

Predicted score = -92.5; R² = 0.394
Predicted score = 177.5; R² = 0.606
Predicted score = -92.5; R² = 0.606 (correct answer)
Predicted score = 177.5; R² = 0.394

Explanation: Predicted score:

\hat{Y} = 42.5 - 1.8(75) = 42.5 - 135 = -92.5

. R² = 1 - SSE/SST = 1 - 2840/7200 = 1 - 0.394 = 0.606. The negative prediction indicates the model may not be appropriate for this age range, but mathematically this is the correct calculation. Choice A uses wrong R² formula. Choice B incorrectly adds rather than subtracts. Choice D combines both computational errors.