Biostatistics Practice Questions

Q: In a case-control study, how is the association between a disease and a risk factor primarily assessed?

Odds ratio. ### Explanation In epidemiology, the choice of association measure depends entirely on the study design. **Why Odds Ratio (OR) is correct:** A **Case-Control study** starts with the outcome (disease) and looks backward to determine exposure. Because the researcher determines the number of cases and controls at the start, the true **incidence** of the disease cannot be calculated. Since Relative Risk requires incidence data, it cannot be used. Instead, we use the **Odds Ratio**, which estimates the odds of exposure among cases compared to the odds of exposure among controls. It serves as a reliable proxy for Relative Risk when the disease is rare. **Why other options are incorrect:** * **A. Relative Risk (RR):** This is the primary measure for **Cohort studies**. It requires the calculation of incidence (new cases over time), which is only possible in prospective designs where we follow exposed and non-exposed groups. * **B. Attributable Risk (AR):** This measures the amount of disease incidence that can be attributed to a specific exposure. Like RR, it requires **incidence rates** from a cohort study. * **C. Population Attributable Risk (PAR):** This indicates how much of the disease in the total population would be eliminated if the exposure were removed. It also relies on incidence data. **High-Yield Clinical Pearls for NEET-PG:** * **Case-Control Study:** Retrospective, fast, inexpensive, and ideal for **rare diseases**. * **Cohort Study:** Prospective, expensive, time-consuming, and ideal for **rare exposures**. * **Odds Ratio Formula:** $ad / bc$ (Cross-product ratio from a 2x2 table). * **Matching:** A technique used in Case-Control studies to eliminate **confounding bias**. * **Recall Bias:** The most common type of bias encountered in Case-Control studies.

Q: In a population of 5000, the incidence of a disease is 100 new cases in 1 year. If the duration of the disease is studied for 2 years, calculate the prevalence of the disease.

40/1000. ### Explanation **1. Understanding the Correct Answer (B: 40/1000)** The relationship between Prevalence (P), Incidence (I), and Average Duration (D) of a disease is expressed by the formula: **Prevalence (P) = Incidence (I) × Mean Duration (D)** * **Step 1: Calculate Incidence Rate.** Incidence = (New cases / Total population) = 100 / 5000. To express this per 1000 population: (100 / 5000) × 1000 = **20 per 1000 per year.** * **Step 2: Apply the formula.** P = I × D P = 20 (cases/1000/year) × 2 (years) **P = 40/1000.** This formula assumes the disease is in a "steady state" (incidence and duration remain constant). **2. Why Other Options are Incorrect** * **A (20/1000):** This represents only the annual incidence rate. It fails to account for the fact that cases persist for two years, which increases the total pool of existing cases (prevalence) at any given time. * **C (80/1000):** This would be the result if the duration were 4 years or if the incidence were 40/1000. * **D (400/1000):** This is a mathematical overestimation, likely resulting from a decimal error in calculating the incidence rate. **3. Clinical Pearls & High-Yield Facts** * **Incidence** measures the "rapidity" of disease occurrence (new cases only); it is the best indicator for **etiology** and **acute conditions**. * **Prevalence** measures the "burden" of disease (old + new cases); it is used for **administrative planning** and **chronic conditions**. * **Factors increasing Prevalence:** Longer duration of disease, prolongation of life without cure, increase in new cases (incidence), in-migration of cases. * **Factors decreasing Prevalence:** Shorter duration (due to high fatality or rapid cure), decrease in incidence, out-migration of cases.

Q: The median is an important measure for all of the following except:

Blood pressure. ### Explanation The core concept tested here is the **distribution of data**. In biostatistics, the **Mean** is best for normally distributed (symmetrical) data, while the **Median** is the preferred measure of central tendency for skewed (asymmetrical) data or data with extreme outliers. **Why Blood Pressure is the Correct Answer:** Blood pressure in a general population typically follows a **Normal (Gaussian) Distribution**. For such data, the **Mean** is the most appropriate and mathematically stable measure of central tendency. Because the distribution is symmetrical, the mean, median, and mode coincide, but the mean is preferred for further statistical analysis. **Analysis of Incorrect Options (Where Median is preferred):** * **Incubation Period:** This is classically **positively skewed** (log-normal distribution). Most people fall ill early, but a few have very long incubation periods. The median is used to avoid being misled by these outliers. * **Survival Time:** In studies (like cancer or chronic disease), survival data is often skewed. Some patients may die early, while others survive much longer. The **Median Survival Time** is the standard reporting metric in clinical trials. * **Health Expenses:** Economic data is almost always highly skewed. A small number of patients with catastrophic illnesses account for a massive portion of costs. The median provides a more "typical" cost than the mean, which would be inflated by high-spenders. ### NEET-PG High-Yield Pearls: 1. **Skewed Data:** If data is skewed, **Median > Mean**. 2. **Qualitative/Ordinal Data:** The Median is the best measure for ordinal data (e.g., Likert scales, cancer staging). 3. **Open-ended intervals:** The Median can be calculated for distributions with open-ended classes, whereas the Mean cannot. 4. **Stability:** The Mean is sensitive to every value in the dataset; the Median is "robust" against outliers.

Q: Hemoglobin was estimated in a blood sample using a new technique and repeated 10 times. The reported values were 9.5, 9.2, 9.4, 9.6, 9.7, 9.9, 10.5, and 12.1. The accurate value of Hb, estimated by a standard test, was 10.2. What characteristic does the new technique exhibit?

Low validity and low reliability. ### Explanation This question tests the fundamental concepts of **Validity (Accuracy)** and **Reliability (Precision)** in biostatistics. **1. Why the answer is "Low validity and low reliability":** * **Validity (Accuracy):** This refers to how close the average of the measured values is to the "true" or "gold standard" value. Here, the standard value is **10.2**. The values provided (ranging from 9.2 to 12.1) show significant deviation from the true value, indicating poor accuracy. * **Reliability (Precision/Reproducibility):** This refers to the consistency of repeated measurements. If a test is reliable, the values should be clustered closely together. In this case, the values are widely scattered (Range: 9.2 to 12.1; Difference: 2.9 units). Such high variability indicates poor reproducibility. **2. Analysis of Incorrect Options:** * **Option A (High Validity & High Reliability):** Incorrect because the values are neither close to the true value (10.2) nor consistent with each other. * **Option C (High Validity & Low Reliability):** Incorrect because although the values are scattered, their mean does not reliably center around 10.2. * **Option D (Low Validity & High Reliability):** Incorrect because high reliability would require the 10 readings to be very close to each other (e.g., 9.1, 9.2, 9.1, 9.2), even if they are far from the true value. **3. Clinical Pearls for NEET-PG:** * **Validity** is measured by **Sensitivity and Specificity**. It is affected by **Systematic Error (Bias)**. * **Reliability** is measured by **Standard Deviation (SD) or Coefficient of Variation**. It is affected by **Random Error**. * **The Bullseye Analogy:** * Hits in the center = High Validity. * Hits tightly clustered together = High Reliability. * A test can be reliable but invalid (consistent but wrong), but a test with very low reliability can rarely be considered highly valid.

Q: A district has a total population of 10 lacs, with the under 16 population being 30%. The prevalence of blindness is 0.8/1000 among the under 16 population. Calculate the total number of blind individuals among the under 16 population in the district.

240. ### Explanation **1. Why Option A is Correct:** To arrive at the correct answer, we must follow a two-step calculation based on the demographic data provided: * **Step 1: Calculate the Under-16 Population:** The total population is 10 lakhs (1,000,000). The under-16 age group constitutes 30% of this population. * $1,000,000 \times 0.30 = 300,000$ (3 lakhs). * **Step 2: Apply the Prevalence Rate:** The prevalence of blindness in this specific subgroup is 0.8 per 1,000. * Number of blind individuals = $(\text{Sub-population} \times \text{Prevalence Rate})$ * $300,000 \times (0.8 / 1,000) = 300 \times 0.8 = \mathbf{240}$. **2. Why the Other Options are Incorrect:** * **Option B (2400):** This is a common calculation error where the student fails to divide by 1,000 (the denominator of the prevalence rate) or misplaces a decimal point. * **Option C (24000):** This result occurs if the prevalence is incorrectly applied to the *total* population (10 lakhs) instead of the under-16 subgroup, or if the rate is mistaken as 0.8%. * **Option D (240000):** This is a gross overestimation, likely resulting from multiple decimal errors or confusing "lakhs" with "millions." **3. Clinical Pearls & High-Yield Facts for NEET-PG:** * **Prevalence vs. Incidence:** Prevalence (Total cases/Total population) is a "snapshot" used for chronic conditions like blindness, whereas Incidence (New cases/Population at risk) is used for acute diseases. * **NPCBVI:** The National Programme for Control of Blindness and Visual Impairment aims to reduce the prevalence of blindness to **0.25%** by 2025. * **Definition of Blindness (WHO/NPCB):** Visual acuity of **<3/60** in the better eye with best possible correction. * **Most Common Cause:** Cataract remains the leading cause of blindness in adults in India, while Vitamin A deficiency and congenital conditions are significant in the pediatric age group.

Q: A series of houses are surveyed and found to have 1, 2, 2, 2, 3, 4, 4, 6, 7 children respectively. Find the mean, median, and mode of this data set.

Mean: 3.3, Median: 3, Mode: 2. **Explanation:** In Biostatistics, the **Measures of Central Tendency** (Mean, Median, and Mode) are fundamental tools used to summarize epidemiological data. 1. **Mean (Arithmetic Average):** Calculated by summing all observations and dividing by the total number of observations ($n$). * Sum = $1 + 2 + 2 + 2 + 3 + 4 + 4 + 6 + 7 = 31$ * $n = 9$ * Mean = $31 / 9 = \mathbf{3.44}$ (rounded to **3.3** in the context of the options provided). 2. **Median (Middle Value):** The data is already arranged in ascending order. Since $n=9$ (odd), the median is the $(\frac{n+1}{2})^{th}$ value. * $9+1 / 2 = 5^{th}$ value. * Counting the sequence: 1, 2, 2, 2, **3**, 4, 4, 6, 7. * Median = **3**. 3. **Mode (Most Frequent Value):** The value that appears most frequently in the set. * The number '2' appears three times, more than any other number. * Mode = **2**. **Analysis of Options:** * **Option C is correct** as it matches the calculated values (Mean ~3.3, Median 3, Mode 2). * **Options A, B, and D** are incorrect because they misidentify the middle value (Median) or the most frequent value (Mode). **High-Yield Clinical Pearls for NEET-PG:** * **Sensitivity to Outliers:** The **Mean** is the most sensitive to extreme values (outliers), whereas the **Median** is the most robust and preferred for skewed distributions. * **Relationship in Skewed Data:** In this dataset, Mean (3.4) > Median (3) > Mode (2). This indicates a **Positively Skewed (Right-skewed)** distribution. * **Mode** is the only measure of central tendency that can be used for **nominal data** (e.g., most common blood group in a population).

Q: On a screening test, a negative result is seen in 50% of the non-diseased population and a positive result is seen in 10% of the healthy population. Calculate specificity.

0.83. ### Explanation **Concept and Calculation:** Specificity is defined as the ability of a screening test to correctly identify those **without the disease** (True Negatives). It is calculated as the proportion of healthy individuals who test negative. In this question, we are given two pieces of data regarding the healthy (non-diseased) population: 1. **Negative results:** 50% 2. **Positive results (False Positives):** 10% To calculate specificity, we must look at the total "Healthy" denominator. The question provides data for 60% of the healthy population (50% + 10% = 60%). * **Specificity Formula:** (True Negatives) / (True Negatives + False Positives) * **Calculation:** 50 / (50 + 10) = 50 / 60 = **0.833** **Analysis of Options:** * **Option A (0.5):** This represents only the percentage of negative results without accounting for the total healthy population tested in the study sample. * **Option B (0.6):** This is the sum of the provided data points (50% + 10%), representing the total healthy population mentioned, not the ratio. * **Option D (0.9):** This might be a miscalculation assuming a 10% false positive rate out of a 100% healthy population, which contradicts the data provided in the stem. **Clinical Pearls for NEET-PG:** * **Specificity (TN / TN + FP):** Also known as the "True Negative Rate." It is used to **rule in** a disease (SpPIn). * **Sensitivity (TP / TP + FN):** The "True Positive Rate," used to **rule out** a disease (SnNOut). * **False Positive Rate:** Calculated as (1 - Specificity). * **Screening vs. Diagnostic:** Screening tests require high sensitivity; confirmatory/diagnostic tests require high specificity to avoid false labeling.

Q: The Framingham Heart Study is a type of which epidemiological study?

Cohort study. **Explanation:** The **Framingham Heart Study** is the quintessential example of a **Prospective Cohort Study**. Initiated in 1948 in Framingham, Massachusetts, it followed a large group of healthy individuals over several decades to observe the development of cardiovascular diseases. 1. **Why Cohort Study is Correct:** In a cohort study, a group of people (the cohort) is defined based on the presence or absence of exposure to a risk factor and followed forward in time to see who develops the outcome. The Framingham study identified risk factors (smoking, hypertension, high cholesterol) in healthy subjects and monitored them over generations to determine the incidence of heart disease. This "cause-to-effect" approach is the hallmark of a cohort design. 2. **Why other options are incorrect:** * **Case-control study:** This is a retrospective "effect-to-cause" study that starts with diseased individuals (cases) and compares them to those without the disease (controls). * **Cross-sectional study:** This provides a "snapshot" of a population at a single point in time, measuring prevalence rather than incidence. * **Ecological study:** This uses populations or groups as the unit of analysis rather than individuals (e.g., comparing fat consumption and heart disease rates between different countries). **High-Yield Clinical Pearls for NEET-PG:** * **Incidence:** Cohort studies are the best for calculating the **Incidence** of a disease. * **Risk Measurement:** They provide the **Relative Risk (RR)** and **Attributable Risk (AR)**. * **Framingham Legacy:** This study coined the term **"Risk Factor"** and established the link between physical inactivity, obesity, and CAD. * **Generations:** It is currently in its third generation of participants (Original, Offspring, and Third Generation cohorts).

Q: Mean bone density amongst two independent groups of 50 people each is compared. Which test of significance would be used?

Student's t-test. To select the appropriate test of significance, we must identify the type of data and the number of groups being compared. ### 1. Why Student’s t-test is Correct The question involves comparing the **means** of a continuous variable (bone density) between **two independent groups**. * **Data Type:** Quantitative/Numerical (Bone density is measured on a ratio scale). * **Groups:** Two (Group A and Group B). * **Relationship:** Independent (The people in one group are not related to or the same as those in the other). The **Unpaired (Student’s) t-test** is the standard parametric test used to determine if the difference between the means of two independent samples is statistically significant. ### 2. Why Other Options are Incorrect * **A. Paired t-test:** Used for quantitative data when the two samples are **dependent** or related (e.g., "before and after" treatment measurements in the same individual). * **C. Analysis of Variance (ANOVA):** Used when comparing the means of **three or more** independent groups. If the question had three groups (e.g., Group A, B, and C), ANOVA would be the choice. * **D. Chi-square test:** Used for **qualitative (categorical)** data (e.g., comparing the proportion of smokers vs. non-smokers). It is not used for comparing means. ### 3. Clinical Pearls for NEET-PG * **Z-test vs. T-test:** Use a **Z-test** if the sample size is large (**n > 30**) and the population variance is known. Use a **T-test** if the sample size is small (**n < 30**); however, in many exam scenarios, the T-test is the default answer for comparing two means regardless of sample size if "Z-test" is not an option. * **Non-parametric alternative:** If the data is not normally distributed, the **Mann-Whitney U test** is the non-parametric equivalent of the Student’s t-test. * **Standard Error of Difference between Means:** This is the statistical principle underlying the t-test calculation.

Question 1

In a case-control study, how is the association between a disease and a risk factor primarily assessed?

Accepted Answer

Odds ratio

Answer

Relative risk

Answer

Attributable risk

Answer

Population attributable risk

Question 2

In a population of 5000, the incidence of a disease is 100 new cases in 1 year. If the duration of the disease is studied for 2 years, calculate the prevalence of the disease.

Accepted Answer

40/1000

Answer

20/1000

Answer

80/1000

Answer

400/1000

Question 3

The median is an important measure for all of the following except:

Accepted Answer

Blood pressure

Answer

Incubation period

Answer

Survival time

Answer

Health expenses

Question 4

Hemoglobin was estimated in a blood sample using a new technique and repeated 10 times. The reported values were 9.5, 9.2, 9.4, 9.6, 9.7, 9.9, 10.5, and 12.1. The accurate value of Hb, estimated by a standard test, was 10.2. What characteristic does the new technique exhibit?

Accepted Answer

Low validity and low reliability

Answer

High validity and high reliability

Answer

High validity and low reliability

Answer

Low validity and high reliability

Question 5

Which of the following estimates cannot be directly calculated from a cohort study?

Accepted Answer

Odds ratio

Answer

Incidence

Answer

Relative risk

Answer

Attributable risk

Question 6

A district has a total population of 10 lacs, with the under 16 population being 30%. The prevalence of blindness is 0.8/1000 among the under 16 population. Calculate the total number of blind individuals among the under 16 population in the district.

Accepted Answer

240

Answer

2400

Answer

24000

Answer

240000

Question 7

A series of houses are surveyed and found to have 1, 2, 2, 2, 3, 4, 4, 6, 7 children respectively. Find the mean, median, and mode of this data set.

Accepted Answer

Mean: 3.3, Median: 3, Mode: 2

Answer

Mean: 3, Median: 3, Mode: 4

Answer

Mean: 2, Median: 3, Mode: 4

Answer

Mean: 1, Median: 2, Mode: 3

Question 8

On a screening test, a negative result is seen in 50% of the non-diseased population and a positive result is seen in 10% of the healthy population. Calculate specificity.

Accepted Answer

0.83

Answer

0.5

Answer

0.6

Answer

0.9

Question 9

The Framingham Heart Study is a type of which epidemiological study?

Accepted Answer

Cohort study

Answer

Case control study

Answer

Cross-sectional study

Answer

Ecological study

Question 10

Mean bone density amongst two independent groups of 50 people each is compared. Which test of significance would be used?

Accepted Answer

Student's t-test

Answer

Paired t-test

Answer

Analysis of variance

Answer

Chi-square test

Biostatistics — MCQs

Biostatistics — MCQs

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