Biostatistics Quiz: Binary Outcomes And Log Odds

20 questions · exam conditions

0:00

Binary Outcomes And Log OddsQuestion 1 of 20

A study reports that the log-odds coefficient for exercise (yes vs. no) in predicting cardiovascular disease is -0.8. If the baseline odds of disease for non-exercisers is 0.25, what are the odds of disease for exercisers?

0.25 \times e^{-0.8} = 0.112

0.25 + (-0.8) = -0.55

0.25 / e^{0.8} = 0.112

0.25 - 0.8 = -0.55

e^{-0.8} = 0.449

← Back to quizzes

Biostatistics Quiz

Biostatistics Quiz: Binary Outcomes And Log Odds

Practice Binary Outcomes And Log Odds 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 Binary Outcomes And Log Odds, 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

$0.25 \times e^{-0.8} = 0.112$ (correct answer)
$0.25 + (-0.8) = -0.55$
$0.25 / e^{0.8} = 0.112$
$0.25 - 0.8 = -0.55$
$e^{-0.8} = 0.449$

Explanation: When you encounter logistic regression coefficients, remember that they represent changes in log-odds, and you need exponential functions to convert between odds and log-odds scales. The log-odds coefficient of -0.8 tells you that exercise changes the log-odds of disease by -0.8 units. To find the actual odds for exercisers, you multiply the baseline odds by

e^{\text{coefficient}}

. Since the baseline odds for non-exercisers is 0.25, the odds for exercisers becomes:

0.25 \times e^{-0.8} = 0.25 \times 0.449 = 0.112

. Looking at the wrong answers: Option B incorrectly adds the log-odds coefficient directly to the odds, but you cannot add values from different scales (log-odds vs. odds). Option C uses division by

e^{0.8}

instead of multiplication by

e^{-0.8}

. While mathematically these are equivalent (

e^{-0.8} = 1/e^{0.8}

), this approach suggests confusion about the proper formula. The standard approach is to multiply baseline odds by

e^{\text{coefficient}}

. Option D makes the same error as B, attempting to subtract the log-odds coefficient from the odds directly. The negative coefficient confirms that exercise is protective (reduces disease odds), which makes biological sense. The result (0.112 < 0.25) confirms exercise reduces the odds compared to non-exercise. Study tip: In logistic regression, always remember the formula: New odds = Baseline odds ×

e^{\text{coefficient}}

. Never add or subtract coefficients directly to odds—they're on different mathematical scales.

Question 2

In a logistic regression model predicting the probability of disease occurrence, the log-odds coefficient for age is 0.05. If a patient's age increases from 40 to 60 years, what is the multiplicative change in the odds of disease?

$e^{1.0} = 2.72$ (correct answer)
$e^{0.05} = 1.05$
$e^{20} = 4.85 \times 10^8$
$0.05 \times 20 = 1.0$
$1 + (0.05 \times 20) = 2.0$

Explanation: When you encounter logistic regression questions involving changes in predictor variables, focus on how coefficients represent the log-odds change per unit increase in the predictor. In logistic regression, a coefficient tells you the change in log-odds for each one-unit increase in that variable. Here, the age coefficient is 0.05, meaning each additional year increases the log-odds by 0.05. When age increases from 40 to 60 years, that's a 20-unit change, so the total change in log-odds is

0.05 \times 20 = 1.0

. To find the multiplicative change in odds (not log-odds), you exponentiate this value:

e^{1.0} = 2.72

. This means the odds of disease are 2.72 times higher for a 60-year-old compared to a 40-year-old. Answer A is correct because it properly calculates

e^{1.0} = 2.72

after finding the total log-odds change. Answer B (

e^{0.05} = 1.05

) gives the odds ratio for just a one-year increase in age, not the full 20-year change. Answer C (

e^{20}

) incorrectly uses 20 as the exponent, forgetting to multiply by the coefficient first. Answer D (

0.05 \times 20 = 1.0

) stops at the log-odds change without exponentiating to get the actual odds ratio. Remember: coefficients in logistic regression are in log-odds units. To get interpretable odds ratios, always exponentiate the coefficient times the change in your predictor variable.

Question 3

A logistic regression model yields the equation: log-odds = -2.5 + 0.8X, where X represents treatment status (0 = control, 1 = treatment). What is the probability of the outcome for a patient in the treatment group?

$\frac{1}{1 + e^{-(-1.7)}} = 0.845$ (correct answer)
$\frac{1}{1 + e^{-(-2.5)}} = 0.924$
$\frac{1}{1 + e^{-(0.8)}} = 0.311$
$\frac{e^{-1.7}}{1 + e^{-1.7}} = 0.155$
$\frac{0.8}{1 + 0.8} = 0.444$

Explanation: When you encounter a logistic regression equation, you're working with a model that predicts the probability of a binary outcome. The key is understanding how to transform the log-odds equation into an actual probability using the logistic function:

P = \frac{1}{1 + e^{-(\text{log-odds})}}

. Given the equation log-odds = -2.5 + 0.8X, you need to substitute X = 1 for the treatment group. This gives you: log-odds = -2.5 + 0.8(1) = -1.7. Now apply the logistic transformation:

P = \frac{1}{1 + e^{-(-1.7)}} = \frac{1}{1 + e^{1.7}} = 0.845

. This matches option A perfectly. Option B incorrectly uses only the intercept (-2.5), ignoring the treatment effect entirely. This would give you the probability for the control group, not the treatment group. Option C uses only the coefficient (0.8) without the intercept, which misses the baseline log-odds. Option D uses the correct log-odds calculation (-1.7) but applies the wrong probability formula. The expression

\frac{e^{-1.7}}{1 + e^{-1.7}}

actually calculates the probability of the opposite outcome (failure rather than success). Remember this two-step process: first substitute all variable values into the log-odds equation, then transform using the standard logistic function. The most common error is forgetting to include all terms in the equation or mixing up the probability formula.

Question 4

A logistic regression analysis shows that smoking status has a coefficient of 0.693. The baseline probability of lung disease for non-smokers is 0.1. What is the probability of lung disease for smokers?

$\frac{1}{1 + e^{-(\ln(0.1/0.9) + 0.693)}} = 0.182$ (correct answer)
$0.1 + 0.693 = 0.793$
$0.1 \times e^{0.693} = 0.2$
$\frac{0.1 \times 2}{1 - 0.1 + 0.1 \times 2} = 0.154$
$0.1 \times (1 + e^{0.693}) = 0.3$

Explanation: When you encounter logistic regression problems, you're working with the logit function that models the log-odds of an outcome. The key insight is that logistic regression coefficients represent changes in log-odds, and you need to convert between probabilities and odds to solve these problems properly. To find the probability for smokers, you first convert the baseline probability to odds, then apply the coefficient. For non-smokers with probability 0.1, the odds are

\frac{0.1}{1-0.1} = \frac{0.1}{0.9}

. The log-odds (logit) is

\ln(0.1/0.9)

. Since the smoking coefficient is 0.693, the log-odds for smokers becomes

\ln(0.1/0.9) + 0.693

. Converting back to probability using the inverse logit function gives

\frac{1}{1 + e^{-(\ln(0.1/0.9) + 0.693)}} = 0.182

, making A correct. B incorrectly treats the coefficient as an additive probability change, but logistic regression works on the log-odds scale, not the probability scale directly. C multiplies the baseline probability by

e^{0.693} = 2

, which conceptually relates to the odds ratio but ignores that this multiplication should happen in odds space, not probability space. D appears to use some form of odds calculation but applies an incorrect formula that doesn't follow logistic regression principles. Remember: logistic regression coefficients affect log-odds, so always convert probabilities to odds, apply the coefficient, then convert back to probability using the logit function.

Question 5

In a logistic model, the log-odds coefficient for BMI is 0.04 per kg/m². A patient with BMI 25 has predicted log-odds of -1.5. What are the predicted log-odds for a patient with BMI 35?

$-1.5 + 0.04(35 - 25) = -1.1$ (correct answer)
$-1.5 + 0.04(35) = -0.1$
$-1.5 \times e^{0.04 \times 10} = -2.24$
$0.04(35) - 1.5 = -0.1$
$-1.5 + 0.04 \times \frac{35}{25} = -1.44$

Explanation: When you encounter logistic regression questions involving log-odds calculations, remember that coefficients represent the change in log-odds per unit change in the predictor variable. The key is understanding how to apply these coefficients to predict outcomes for new values. The correct approach uses the fundamental principle that log-odds change linearly with predictor variables. Since the BMI coefficient is 0.04 per kg/m², each unit increase in BMI increases the log-odds by 0.04. For a patient moving from BMI 25 to BMI 35 (a 10-unit increase), the log-odds increase by

0.04 \times 10 = 0.4

. Adding this to the baseline log-odds:

-1.5 + 0.4 = -1.1

. Answer A correctly applies this logic. Answer B incorrectly multiplies the coefficient by the new BMI value (35) rather than the change in BMI, ignoring that we already know the log-odds at BMI 25. Answer C makes a conceptual error by involving the exponential function and multiplication, confusing log-odds calculations with odds ratio transformations. Answer D commits the same mistake as B, multiplying by the absolute BMI value instead of the change, and coincidentally arrives at the same incorrect result through different arithmetic. Remember this pattern: in linear relationships like logistic regression's log-odds, when you know the outcome at one predictor value, calculate the new outcome by adding (coefficient × change in predictor). Don't multiply by absolute values or introduce exponential functions unless you're specifically converting between log-odds and odds.

Question 6

In a study of surgical complications, the log-odds equation is: log-odds = -3.5 + 0.6(surgery_type) + 0.05(age), where surgery_type is coded as 0 for laparoscopic and 1 for open surgery. What is the odds ratio comparing open to laparoscopic surgery for patients of the same age?

$e^{0.6} = 1.82$ (correct answer)
$e^{0.6 + 0.05} = e^{0.65} = 1.92$
$e^{-3.5 + 0.6} = e^{-2.9} = 0.055$
$\frac{0.6}{0.05} = 12$
$0.6 + 1 = 1.6$

Explanation: When you encounter a logistic regression equation asking for an odds ratio between two groups, you need to focus on how the coefficient translates to the odds ratio for that variable. In this log-odds equation, the coefficient for surgery_type is 0.6. Since surgery_type is coded as 0 for laparoscopic and 1 for open surgery, this coefficient represents the change in log-odds when moving from laparoscopic (0) to open surgery (1). To convert from log-odds to an odds ratio, you exponentiate the coefficient. Therefore, the odds ratio comparing open to laparoscopic surgery is

e^{0.6} = 1.82

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

e^{0.65} = 1.92

) incorrectly adds both coefficients together, but you only need the surgery_type coefficient since you're comparing surgery types at the same age. Answer C (

e^{-2.9} = 0.055

) mistakenly uses the intercept plus the surgery coefficient, but the intercept represents baseline log-odds, not the comparison between groups. Answer D simply divides the coefficients (

\frac{0.6}{0.05} = 12

), but this arithmetic operation has no meaning in logistic regression - you must exponentiate coefficients to get odds ratios. Remember this key principle: in logistic regression, the odds ratio for any variable equals

e^{\text{coefficient}}

. The intercept and other variables don't factor into calculating a specific odds ratio when comparing groups, assuming other variables are held constant.

Question 7

In a logistic regression for disease diagnosis, a biomarker has a coefficient of 0.25 per unit. The baseline log-odds (biomarker = 0) is -4.0. At what biomarker level do the odds of disease equal 1.0?

Biomarker = 16 units (correct answer)
Biomarker = 4 units
Biomarker = 20 units
Biomarker = 0.25 units
Biomarker = 1 unit

Explanation: When you encounter logistic regression problems involving odds, remember that you're working with the relationship between log-odds (logit) and the predictor variables. The logistic regression equation is: log-odds = intercept + coefficient × predictor. To find where odds equal 1.0, you need to determine when log-odds equals 0 (since ln(1) = 0). Setting up the equation: 0 = -4.0 + 0.25 × biomarker level. Solving for the biomarker level: 0.25 × biomarker = 4.0, so biomarker = 4.0 ÷ 0.25 = 16 units. Let's verify: At biomarker = 16, log-odds = -4.0 + 0.25(16) = -4.0 + 4.0 = 0. Since e⁰ = 1, the odds equal 1.0, confirming answer A is correct. Answer B (4 units) represents the value you get before dividing by the coefficient—this would give log-odds = -4.0 + 0.25(4) = -3.0, making odds = e⁻³ ≈ 0.05. Answer C (20 units) might result from incorrectly adding rather than dividing: -4.0 + 0.25(20) = 1.0, giving odds ≈ 2.7. Answer D (0.25 units) confuses the coefficient value with the biomarker level, yielding log-odds = -4.0 + 0.25(0.25) ≈ -3.94. Remember this key strategy: when odds equal 1.0, log-odds must equal 0. Set your logistic equation equal to zero and solve algebraically. This approach works for any logistic regression odds calculation.

Question 8

A logistic model for predicting readmission has coefficients: intercept = -2.8, length_of_stay = 0.12 per day, comorbidities = 0.4 per condition. A patient with 5 days length of stay and 2 comorbidities has what probability of readmission?

$\frac{1}{1 + e^{-(-2.8 + 0.12 \times 5 + 0.4 \times 2)}} = \frac{1}{1 + e^{1.6}} = 0.202$ (correct answer)
$\frac{1}{1 + e^{-(0.12 \times 5 + 0.4 \times 2)}} = \frac{1}{1 + e^{1.4}} = 0.198$
$e^{-1.6} = 0.202$
$\frac{e^{-1.6}}{1 + e^{-1.6}} = 0.168$
$-2.8 + 0.6 + 0.8 = -1.6$

Explanation: When you encounter logistic regression problems, remember that you're calculating the probability of an event using the logistic function:

P = \frac{1}{1 + e^{-(\text{linear combination})}}

, where the linear combination includes all terms from your model. To find this patient's readmission probability, you need to substitute the given values into the complete logistic model. The linear combination is: intercept + (length_of_stay coefficient × days) + (comorbidities coefficient × conditions) = -2.8 + (0.12 × 5) + (0.4 × 2) = -2.8 + 0.6 + 0.8 = -1.6. Therefore:

P = \frac{1}{1 + e^{-(-1.6)}} = \frac{1}{1 + e^{1.6}} = 0.202

Choice A correctly applies the full logistic model with all components and arrives at the right probability of 0.202. Choice B omits the intercept term (-2.8), a critical error since the intercept represents the baseline log-odds when all predictors equal zero. This gives an incorrect linear combination of 1.4 instead of -1.6. Choice C uses only the exponential term

e^{-1.6}

without the proper logistic transformation. This calculates neither probability nor odds correctly. Choice D appears to use an alternative form of the logistic function but incorrectly applies it, yielding 0.168 instead of the correct 0.202. Study tip: Always double-check that you've included every term in your logistic model, especially the intercept. The intercept is often forgotten but represents crucial baseline information that significantly affects your probability calculations.

Question 9

A logistic regression model predicting treatment dropout has the equation: log-odds = -1.5 + 0.4(anxiety_score). At what anxiety score is the predicted probability of dropout equal to 80%?

Anxiety score = $\frac{\ln(0.8/0.2) + 1.5}{0.4} =$ 7.22 (correct answer)
Anxiety score = $\frac{0.8 + 1.5}{0.4} = 5.75$
Anxiety score = $\frac{\ln(0.8) + 1.5}{0.4} = 3.18$
Anxiety score = $\frac{1.5}{0.4} = 3.75$
Anxiety score = $\frac{0.8}{0.4} = 2.0$

Explanation: When you encounter logistic regression problems asking for a specific probability threshold, you need to work backwards from the probability to find the predictor value using the relationship between odds, log-odds, and probability. Starting with 80% probability, you first convert this to odds: odds = probability/(1-probability) = 0.8/0.2 = 4. Since the logistic regression equation gives you log-odds, you take the natural logarithm: log-odds = ln(4) ≈ 1.386. Now you can solve for the anxiety score using the given equation: log-odds = -1.5 + 0.4(anxiety_score). Substituting: 1.386 = -1.5 + 0.4(anxiety_score). Rearranging: anxiety_score = (1.386 + 1.5)/0.4 = 2.886/0.4 ≈ 7.22. This matches the setup in choice A:

\frac{\ln(0.8/0.2) + 1.5}{0.4}

. Choice B incorrectly uses the probability (0.8) directly instead of converting to log-odds first—you can't substitute probability into a log-odds equation. Choice C takes the natural log of the probability rather than the odds, missing the critical step of converting 0.8 to odds of 4. Choice D ignores the probability requirement entirely and just solves for when the linear predictor equals zero. Study tip: Remember the sequence for logistic regression probability problems: probability → odds → log-odds → solve for predictor. The key conversion is odds = p/(1-p), then take the natural log. Always convert probabilities to log-odds when working with logistic regression equations.

Question 10

In a clinical trial analysis, treatment group (1=active, 0=placebo) has a coefficient of 0.693. The placebo group has a baseline probability of response of 0.25. What is the odds ratio for response comparing active treatment to placebo?

$e^{0.693} = 2.00$ (correct answer)
$\frac{0.25 \times e^{0.693}}{0.25} = e^{0.693} = 2.00$
$\frac{0.25 \times 2}{1 - 0.25 \times 2} = 1.0$
$\frac{1/(1 + e^{-(\ln(0.25/0.75) + 0.693)})}{0.25} = 2.0$
$0.693 \times 2 = 1.386$

Explanation: When you encounter a logistic regression coefficient in a clinical trial, remember that the coefficient directly represents the log odds ratio. The odds ratio is simply the exponential of this coefficient. In logistic regression, the coefficient for a binary predictor (like treatment group) tells you how the log odds change when moving from the reference category (placebo, coded as 0) to the comparison category (active treatment, coded as 1). Since the coefficient is 0.693, the odds ratio is

e^{0.693} = 2.00

. This means patients on active treatment have twice the odds of responding compared to those on placebo. Answer A is correct:

e^{0.693} = 2.00

directly applies the fundamental relationship between logistic regression coefficients and odds ratios. Answer B makes an unnecessary calculation by multiplying and dividing by the baseline probability (0.25), which cancels out to give the same result but shows conceptual confusion about what the coefficient represents. Answer C attempts to calculate something involving probabilities and gets 1.0, but this approach is fundamentally flawed. An odds ratio of 1.0 would indicate no difference between groups, which contradicts the positive coefficient. Answer D presents an overly complex calculation involving the inverse logit transformation and baseline odds, unnecessarily complicating what should be a straightforward conversion. Remember: in logistic regression, coefficients are log odds ratios. To get the odds ratio, simply exponentiate the coefficient. Don't overthink it with probability calculations or complex transformations.

Question 11

In a study of hospital readmission, the log-odds of readmission increases by 1.2 for each additional comorbidity. If the baseline log-odds (no comorbidities) is -3.0, what is the odds ratio comparing patients with 2 comorbidities to patients with no comorbidities?

$e^{(1.2 \times 2)} = e^{2.4} = 11.02$ (correct answer)
$e^{(-3.0 + 2.4)} = e^{-0.6} = 0.55$
$\frac{e^{-0.6}}{e^{-3.0}} = e^{2.4} = 11.02$
$e^{1.2} = 3.32$
$2 \times e^{1.2} = 6.64$

Explanation: When you encounter logistic regression problems involving odds ratios, you're working with the relationship between log-odds and the exponential function. The key insight is that odds ratios compare the odds between two different exposure levels. Here's how to approach this systematically. First, calculate the log-odds for each group:

No comorbidities: log-odds = -3.0
Two comorbidities: log-odds = -3.0 + (1.2 × 2) = -3.0 + 2.4 = -0.6

The odds ratio is the ratio of odds, which equals the exponential of the difference in log-odds. Since log-odds differ by 2.4, the odds ratio is

e^{2.4} = 11.02

. This matches answer A. Let's examine why the other options miss the mark. Answer B (

e^{-0.6} = 0.55

) gives you the actual odds for the 2-comorbidity group, not the odds ratio. Answer C shows the correct calculation but uses an unnecessarily complicated approach—while mathematically equivalent to A, it's the long way around. Answer D (

e^{1.2} = 3.32

) represents the odds ratio for just one additional comorbidity, not two. The critical insight here is that when log-odds increases by a constant amount per unit change in exposure, the odds ratio for any comparison is simply

e^{\text{(coefficient × difference in exposure)}}

. For logistic regression problems, always remember: the coefficient tells you the log-odds change per unit, so multiply by the exposure difference and exponentiate to get your odds ratio.

Question 12

A logistic model for infection risk includes temperature (coefficient = 0.08 per °C above 37°C). If a patient with temperature 39°C has log-odds of infection equal to -0.5, what are the log-odds for a patient with temperature 41°C?

$-0.5 + 0.08 \times (41-39) = -0.34$ (correct answer)
$-0.5 + 0.08 \times (41-37) = -0.18$
$-0.5 + 0.08 \times 41 = 2.78$
$-0.5 \times e^{0.08 \times 2} = -0.59$
$0.08 \times (41-37) = 0.32$

Explanation: When you encounter logistic regression problems, remember that coefficients represent the change in log-odds per unit change in the predictor variable. The key is understanding what "per °C above 37°C" means and applying it correctly. The coefficient 0.08 represents the change in log-odds for each degree above 37°C. To find the log-odds for the 41°C patient, you need to calculate how their temperature differs from the reference patient's temperature, then apply the coefficient to that difference. The 39°C patient has log-odds of -0.5. The 41°C patient is 2°C warmer than the 39°C patient (41-39=2). Since each degree increases log-odds by 0.08, the difference is

0.08 \times 2 = 0.16

. Therefore:

-0.5 + 0.16 = -0.34

. This matches option A. Option B incorrectly calculates the temperature relative to 37°C instead of relative to the reference patient. While 41°C is indeed 4°C above 37°C, this ignores that we already know another patient's log-odds and should use that as our starting point. Option C makes the same error as B but also fails to subtract 37°C from 41°C, incorrectly using the raw temperature value of 41. Option D incorrectly applies an exponential transformation. In logistic regression, coefficients directly affect log-odds in a linear fashion—you don't exponentiate when calculating log-odds differences. Study tip: In logistic regression problems, always identify your reference point and calculate differences from there. Coefficients affect log-odds linearly, so simple addition/subtraction is usually correct.

Question 13

A study of medication adherence uses logistic regression where the coefficient for education level is 0.3 per year of schooling. If a patient with 12 years of education has odds of adherence equal to 2.0, what are the odds for a patient with 16 years of education?

$2.0 \times e^{0.3 \times 4} = 2.0 \times 3.32 = 6.64$ (correct answer)
$2.0 + 0.3 \times 4 = 3.2$
$2.0 \times e^{0.3} = 2.0 \times 1.35 = 2.70$
$2.0 + e^{0.3 \times 4} = 2.0 + 3.32 = 5.32$
$e^{0.3 \times 16} = e^{4.8} = 121$

Explanation: When you encounter logistic regression problems, remember that coefficients represent the change in log-odds per unit increase in the predictor variable. The key insight is understanding how to convert between different patients' odds using the exponential relationship. In logistic regression, if the coefficient is 0.3 per year of education, then for every additional year of schooling, the log-odds increase by 0.3. Since odds = e^(log-odds), a 4-year difference in education (16 - 12 = 4) means the odds ratio is

e^{0.3 \times 4} = e^{1.2} = 3.32

. Therefore, the patient with 16 years of education has odds that are 3.32 times higher than the patient with 12 years:

2.0 \times 3.32 = 6.64

. Choice A correctly applies this exponential relationship. Choice B incorrectly adds the coefficient change directly to the odds, treating this as a linear rather than exponential relationship. Choice C makes the error of only accounting for one additional year of education (

e^{0.3}

) instead of the full 4-year difference. Choice D incorrectly adds the odds ratio to the original odds instead of multiplying, which violates the multiplicative nature of odds ratios. Remember this pattern: in logistic regression, odds change multiplicatively with predictor variables. When comparing two individuals who differ by Δx units on a predictor, multiply the baseline odds by

e^{\beta \times \Delta x}

, where β is the coefficient. Always multiply odds ratios, never add them.

Question 14

A logistic model predicts treatment success with log-odds = -1.2 + 0.8(dose). If the current dose gives a 40% success rate, what dose would be needed to achieve a 70% success rate?

Current dose + $\frac{\ln(70/30) - \ln(40/60)}{0.8} = \text{Current dose} + 1.46$ (correct answer)
Current dose + $\frac{0.7 - 0.4}{0.8} = \text{Current dose} + 0.375$
$\frac{\ln(0.7) - (-1.2)}{0.8} = 1.05$
Current dose $\times \frac{0.7}{0.4} = \text{Current dose} \times 1.75$
$\frac{0.7}{0.8} = 0.875$

Explanation: When you encounter logistic regression problems, remember that the relationship between probability and dose involves log-odds, not direct proportions. The logistic model gives you log-odds = -1.2 + 0.8(dose), where log-odds equals

\ln\left(\frac{p}{1-p}\right)

. To solve this, you need to find the change in dose required to go from 40% to 70% success. First, convert probabilities to odds ratios: 40% success means odds = 40/60, and 70% success means odds = 70/30. The log-odds difference is

\ln(70/30) - \ln(40/60)

. Since the coefficient for dose is 0.8, the required dose change is

\frac{\ln(70/30) - \ln(40/60)}{0.8}

, which equals approximately 1.46. Option A correctly applies this log-odds transformation and accounts for the coefficient, giving the right dose increase of 1.46 units. Option B incorrectly treats this as a linear relationship, simply dividing the probability difference (0.7 - 0.4) by the coefficient. This ignores the logarithmic nature of logistic regression. Option C attempts to solve for an absolute dose value rather than the change needed, and doesn't properly account for the current baseline. Option D uses a simple ratio multiplication, completely missing that logistic models involve logarithmic transformations, not direct proportional relationships. Study tip: In logistic regression problems, always convert probabilities to odds first, then work with log-odds. Never treat probability changes as linear when dealing with logistic models.

Question 15

A logistic model for predicting surgical success includes surgeon experience (coefficient = 0.05 per year) and patient age (coefficient = -0.03 per year). What is the odds ratio comparing a 30-year-old patient with a surgeon having 20 years experience to a 40-year-old patient with a surgeon having 10 years experience?

$e^{0.05 \times (20-10) + (-0.03) \times (30-40)} = e^{0.5 + 0.3} = e^{0.8} = 2.23$ (correct answer)
$e^{0.05 \times 20 + (-0.03) \times 30} = e^{1.0 - 0.9} = e^{0.1} = 1.11$
$e^{0.05 \times 10} = e^{0.5} = 1.65$
$e^{(-0.03) \times 10} = e^{-0.3} = 0.74$
$e^{0.05 \times 20} \times e^{(-0.03) \times 30} = 2.72 \times 0.41 = 1.11$

Explanation: When comparing odds between two groups in logistic regression, you need to calculate the difference in the linear predictor (log-odds) for each variable, then exponentiate to get the odds ratio. The logistic model is: log-odds = intercept + 0.05(surgeon experience) - 0.03(patient age). To compare two scenarios, you calculate the difference in their linear predictors. For the 30-year-old with 20-year surgeon versus the 40-year-old with 10-year surgeon, the difference is: Δ(log-odds) = [0.05 × (20-10)] + [(-0.03) × (30-40)] = 0.05 × 10 + (-0.03) × (-10) = 0.5 + 0.3 = 0.8 The odds ratio is

e^{0.8} = 2.23

, making answer A correct. Answer B calculates the absolute log-odds for just one scenario (

e^{0.05 × 20 + (-0.03) × 30}

) rather than the difference between scenarios. This gives you odds for one group, not a comparison between groups. Answer C only considers the surgeon experience difference (

e^{0.05 × 10}

) while ignoring the patient age difference entirely. This incomplete calculation misses half the comparison. Answer D only accounts for the age difference using the wrong baseline (

e^{(-0.03) × 10}

), ignoring surgeon experience and using an incorrect age comparison. Remember: odds ratios in logistic regression require calculating the difference in linear predictors for all relevant variables, then exponentiating. Always include every variable that differs between your comparison groups—partial calculations lead to wrong conclusions.

Question 16

In a logistic regression, the log-odds of disease occurrence is modeled as -4.0 + 0.1(age). At what age does the predicted probability of disease equal 50%?

Age = 40 years (correct answer)
Age = 45 years
Age = 50 years
Age = 35 years
Age = 4 years

Explanation: This question tests your understanding of logistic regression and the relationship between log-odds and probability. When you see a logistic regression equation, remember that it models log-odds, not probability directly, and you'll need to convert between these measures. The logistic regression equation gives us: log-odds = -4.0 + 0.1(age). A 50% probability corresponds to log-odds of 0, because when P = 0.5, the odds are 1:1, and ln(1) = 0. Setting the equation equal to 0: 0 = -4.0 + 0.1(age). Solving for age: 0.1(age) = 4.0, so age = 40 years. Let's verify why each answer choice works or fails. Choice A (40 years) gives us log-odds = -4.0 + 0.1(40) = 0, which converts to exactly 50% probability using the formula

P = \frac{e^{0}}{1 + e^{0}} = \frac{1}{2} = 0.5

. Choice B (45 years) produces log-odds = 0.5, giving a probability greater than 50%. Choice C (50 years) yields log-odds = 1.0, resulting in about 73% probability. Choice D (35 years) gives log-odds = -0.5, producing a probability less than 50%. Remember this key relationship: in logistic regression, 50% probability always occurs when log-odds equals zero. This creates a simple algebraic equation to solve. Also, the coefficient tells you the direction—positive coefficients mean probability increases with the predictor variable, as we see here with age.

Question 17

A logistic regression model for predicting diabetes includes family history (coefficient = 1.1). The model intercept is -2.3. What is the difference in predicted probabilities between individuals with and without a family history of diabetes?

$\frac{1}{1+e^{-(-1.2)}} - \frac{1}{1+e^{-(-2.3)}} = 0.768 - 0.091 = 0.677$ (correct answer)
$e^{1.1} - 1 = 3.00 - 1 = 2.00$
$\frac{e^{1.1}}{1 + e^{1.1}} = \frac{3.00}{4.00} = 0.75$
$1.1$
$\frac{1.1}{1 + 1.1} = 0.52$

Explanation: When you encounter logistic regression problems asking for probability differences, you need to calculate the actual predicted probabilities using the logistic function, then find their difference. In logistic regression, predicted probability equals

\frac{1}{1+e^{-(\text{intercept} + \text{coefficient} \times \text{variable})}}

. For someone WITH family history, the linear predictor is

-2.3 + 1.1(1) = -1.2

, giving probability

\frac{1}{1+e^{-(-1.2)}} = 0.768

. For someone WITHOUT family history, the linear predictor is

-2.3 + 1.1(0) = -2.3

, giving probability

\frac{1}{1+e^{-(-2.3)}} = 0.091

. The difference is

0.768 - 0.091 = 0.677

, making A correct. Option B calculates the odds ratio minus 1, which gives you the multiplicative change in odds, not the probability difference. Option C appears to calculate some form of probability but uses an incorrect formula that doesn't account for the intercept—this would only be valid if the intercept were zero. Option D simply states the coefficient value, which represents the change in log-odds, not probabilities. The key trap here is confusing different measures: coefficients represent log-odds changes, exponentiating coefficients gives odds ratios, but probability differences require computing actual probabilities via the logistic function. Always remember that in logistic regression, you must transform through the logistic function to get interpretable probabilities, and include all relevant terms (intercept plus coefficients) in your linear predictor.

Question 18

In a logistic model predicting hospital mortality, the coefficient for severity score is 0.15 per point. A patient with a severity score of 20 has a predicted probability of death of 0.6. What is the predicted probability for a patient with a severity score of 25?

$\frac{1}{1 + e^{-(\ln(0.6/0.4) + 0.15 \times 5)}} = 0.75$ (correct answer)
$0.6 + 0.15 \times 5 = 1.35$
$0.6 \times e^{0.15 \times 5} = 1.27$
$\frac{0.6}{1-0.6} \times e^{0.15 \times 5} = 3.17$
$0.6 \times (1 + 0.15 \times 5) = 1.05$

Explanation: When you encounter logistic regression problems, remember that these models predict probabilities using the logistic function, where changes in predictor variables affect the log-odds (logit) linearly, not the probability directly. The correct approach uses the logistic model's fundamental structure. Since the coefficient is 0.15 per point, increasing the severity score by 5 points (from 20 to 25) increases the log-odds by

0.15 \times 5 = 0.75

. First, convert the known probability (0.6) to log-odds:

\ln(0.6/0.4) = \ln(1.5)

. Then add the increase:

\ln(1.5) + 0.75

. Finally, convert back to probability using the logistic function. Answer A correctly implements this:

\frac{1}{1 + e^{-(\ln(0.6/0.4) + 0.15 \times 5)}} = 0.75

. Answer B incorrectly adds the coefficient change directly to the probability (

0.6 + 0.75 = 1.35

), ignoring that logistic regression affects log-odds, not probabilities linearly. This also produces an impossible probability > 1. Answer C multiplies the probability by the exponential of the coefficient change (

0.6 \times e^{0.75}

), which has no basis in logistic regression theory and yields an impossible result > 1. Answer D calculates the odds ratio incorrectly by multiplying initial odds by the exponential term, but then presents this as a probability rather than converting back properly. Study tip: In logistic regression, coefficients affect log-odds linearly. Always convert probability → log-odds → add change → convert back to probability. Never add coefficients directly to probabilities.

Question 19

A logistic regression examining medication adherence yields coefficients for three predictors: Age (per decade): 0.223, Income (per $10K): 0.105, Education (college vs. high school): 0.693. Which predictor has the strongest association with adherence per unit change?

Income, with an odds ratio of 1.11 per $10K increase
Age, with an odds ratio of 1.25 per decade increase
Education, with an odds ratio of 2.0 per category change (correct answer)
Age and Education have equivalent effects on adherence

Explanation: When interpreting logistic regression coefficients, you need to understand that the coefficient represents the log odds ratio, and larger absolute values indicate stronger associations with the outcome. To compare effect sizes meaningfully, convert coefficients to odds ratios using

e^{\text{coefficient}}

. Let's calculate the odds ratios for each predictor. For Age:

e^{0.223} = 1.25

, meaning each decade increase multiplies the odds of adherence by 1.25. For Income:

e^{0.105} = 1.11

, so each $10K increase multiplies odds by 1.11. For Education: $

e^{0.693} = 2.0

$, meaning college education doubles the odds compared to high school. Answer C is correct because education shows the strongest association - an odds ratio of 2.0 represents a 100% increase in odds, which is substantially larger than the other predictors' effects. Answer A incorrectly identifies income as strongest. While the odds ratio calculation (1.11) is accurate, this represents only an 11% increase in odds per $10K, making it the weakest predictor. Answer B miscalculates age's importance. Though the odds ratio (1.25) is correct, this 25% increase is still much smaller than education's 100% increase. Answer D is wrong because the coefficients clearly differ (0.223 vs 0.693), indicating different effect sizes. Remember: when comparing logistic regression predictors, convert coefficients to odds ratios first, then compare their distances from 1.0. The predictor with an odds ratio furthest from 1.0 (whether above or below) has the strongest association per unit change.

Question 20

A logistic regression model predicting treatment response includes gender (coefficient = 0.5, where 1 = female, 0 = male) and age (coefficient = 0.02 per year). If the intercept is -1.8, what is the odds ratio for treatment response comparing a 50-year-old female to a 40-year-old male?

$e^{(0.5 + 0.02 \times 10)} = e^{0.7} = 2.01$ (correct answer)
$e^{0.5} \times e^{0.02 \times 10} = 1.65 \times 1.22 = 2.01$
$e^{0.5} = 1.65$
$e^{0.02 \times 50} \times e^{0.5} = 2.72 \times 1.65 = 4.49$
$\frac{e^{-1.8 + 0.5 + 0.02 \times 50}}{e^{-1.8 + 0.02 \times 40}} = 2.01$

Explanation: When comparing two groups in logistic regression, you need to calculate the difference in their log-odds, then exponentiate to get the odds ratio. The key is identifying what differs between your comparison groups and accounting for all those differences simultaneously. For this comparison, you're contrasting a 50-year-old female versus a 40-year-old male. The logistic regression equation is: log-odds = -1.8 + 0.5(gender) + 0.02(age). The intercept cancels out when calculating odds ratios, so you only need the coefficient differences. The female is coded as 1 (vs. 0 for male), contributing +0.5 to her log-odds. She's also 10 years older (50 vs. 40), contributing +0.02 × 10 = +0.2 to her log-odds. The total difference in log-odds is 0.5 + 0.2 = 0.7, so the odds ratio is

e^{0.7} = 2.01

. This makes A correct. B reaches the same numerical answer but uses an unnecessarily complicated approach by separating the gender and age effects before multiplying. While mathematically equivalent due to exponent rules, it's less efficient. C only accounts for the gender difference (

e^{0.5} = 1.65

) but completely ignores the 10-year age difference between the two individuals. D incorrectly uses the 50-year-old's absolute age rather than the 10-year age difference between groups, leading to

e^{0.02 \times 50} = e^{1.0}

instead of

e^{0.02 \times 10} = e^{0.2}

. Study tip: For logistic regression comparisons, always calculate the difference in predictors between groups, not their absolute values. Focus on what changes, not what stays the same.