Biostatistics Practice Questions

Q: In an investigation to study the effect of smoking on renal cell carcinoma, it is observed that 30 out of 50 patients were smokers as compared to 10 out of 50 control subjects. What is the odd ratio of renal cell carcinoma associated with smoking?

6. ### Explanation **1. Understanding the Correct Answer (C: 6)** The study design described is a **Case-Control Study**, as it starts with known cases (Renal Cell Carcinoma) and compares them to controls to look back at an exposure (smoking). The appropriate measure of association for such studies is the **Odds Ratio (OR)**. To calculate the Odds Ratio, we use a 2x2 contingency table: | | Cases (RCC) | Controls | | :--- | :---: | :---: | | **Exposed (Smokers)** | 30 (a) | 10 (b) | | **Non-Exposed (Non-smokers)** | 20 (c) | 40 (d) | | **Total** | 50 | 50 | * **Formula:** $OR = \frac{ad}{bc}$ (Cross-product ratio) * **Calculation:** $OR = \frac{30 \times 40}{10 \times 20} = \frac{1200}{200} = 6$ * **Interpretation:** Smokers have 6 times higher odds of developing renal cell carcinoma compared to non-smokers. **2. Why Other Options are Incorrect** * **Option A (3):** This is a common error where students simply divide the number of exposed cases by exposed controls ($30/10$). This ignores the non-exposed group. * **Option B (0.33):** This is the inverse of 3 ($1/3$), often resulting from flipping the numerator and denominator. * **Option D (0.16):** This is the inverse of 6 ($1/6$), representing the odds of the control group rather than the cases. **3. High-Yield Clinical Pearls for NEET-PG** * **Odds Ratio (OR):** Used in Case-Control studies. It is an estimate of Relative Risk (RR). * **Relative Risk (RR):** Used in Cohort studies. It cannot be calculated in case-control studies because the incidence cannot be determined. * **Attributable Risk (AR):** Indicates the amount of disease that can be attributed to the exposure. * **Key Rule:** If OR > 1, the exposure is a risk factor. If OR = 1, there is no association. If OR < 1, the exposure is protective.

Q: The mean of 25 plasma volumes is 12.5 litres. The standard deviation is 0.25. Calculate the variance.

0.0625. ### Explanation **1. Why the correct answer is right:** In biostatistics, **Variance** is a measure of the dispersion of data points around the mean. It is mathematically defined as the **square of the Standard Deviation (SD)**. The formula is: $$\text{Variance} = (\text{Standard Deviation})^2$$ Given in the question: * Standard Deviation (SD) = 0.25 * Variance = $(0.25)^2$ * Variance = $0.25 \times 0.25 = \mathbf{0.0625}$ Note that the "Mean" (12.5 litres) and the "Sample size" (25) are provided as distractors; they are not required for calculating variance when the SD is already known. **2. Why the incorrect options are wrong:** * **Option B (0.625):** This is a common calculation error where the decimal point is misplaced. * **Option C (6.25):** This occurs if one squares 2.5 instead of 0.25. * **Option D (625):** This occurs if the decimal point is ignored entirely ($25^2$). **3. Clinical Pearls & High-Yield Facts for NEET-PG:** * **Standard Deviation (SD):** It is the square root of variance. It is preferred over variance in clinical reports because it is expressed in the **same units** as the original data (e.g., litres), whereas variance is expressed in units squared (e.g., litres²). * **Coefficient of Variation (CV):** A measure of relative variation, calculated as $(SD / \text{Mean}) \times 100$. It is unitless and used to compare the variability of two different datasets. * **Standard Error of Mean (SEM):** Calculated as $SD / \sqrt{n}$. It measures how far the sample mean is likely to be from the true population mean. * **Normal Distribution:** In a Gaussian curve, Mean = Median = Mode. Approximately 95% of values lie within Mean ± 2 SD.

Q: In a population of 200 people with a normal distribution, how many people would be included within 1 standard deviation?

136. ### Explanation **1. Understanding the Correct Answer (A):** This question is based on the properties of a **Normal Distribution (Gaussian Distribution)** curve. In a normal distribution, the data is symmetrically distributed around the mean. According to the **Empirical Rule**: * **Mean ± 1 Standard Deviation (SD)** covers approximately **68.2%** of the population. * **Mean ± 2 SD** covers approximately **95.4%** of the population. * **Mean ± 3 SD** covers approximately **99.7%** of the population. To find the number of people within 1 SD in a population of 200: Calculation: $200 \times 68.2\% = 200 \times 0.682 = \mathbf{136.4}$. Rounding to the nearest whole number gives **136**. **2. Analysis of Incorrect Options:** * **Option B (140) and C (150):** These numbers are higher than the 68.2% threshold. They do not correspond to any standard sigma levels in biostatistics. * **Option D (190):** This represents 95% of the population ($200 \times 0.95 = 190$). This would be the approximate number of people included within **2 Standard Deviations** (specifically 1.96 SD), not 1 SD. **3. NEET-PG High-Yield Clinical Pearls:** * **Symmetry:** In a normal distribution, Mean = Median = Mode. * **Skewness:** If the tail is towards the right, it is **Positively Skewed** (Mean > Median > Mode). If the tail is towards the left, it is **Negatively Skewed** (Mode > Median > Mean). * **Standard Normal Curve:** A normal curve with a Mean of 0 and an SD of 1. * **Z-score:** Indicates how many standard deviations an observation is from the mean. For 1 SD, $Z = 1$.

Q: What is the preferred measure of central tendency for ordinal data?

Median. ### Explanation In biostatistics, the choice of central tendency depends entirely on the **scale of measurement** of the data. **Why Median is the Correct Answer:** Ordinal data consists of categories that have a natural rank or order (e.g., stages of cancer, socio-economic status, or Likert scales), but the mathematical distance between these ranks is not uniform or quantifiable. The **Median** is the preferred measure because it identifies the middle value in a ranked distribution. It respects the relative positioning of the data points without requiring the numerical intervals necessary for calculating a mean. **Analysis of Incorrect Options:** * **A. Mean:** Requires **Interval or Ratio (Quantitative)** data. Since ordinal data lacks equal intervals (e.g., the "gap" between Stage I and II cancer isn't necessarily the same as between Stage III and IV), calculating an average is mathematically invalid. * **B. Mode:** While the mode can be used for ordinal data (representing the most frequent category), it is less descriptive than the median because it ignores the inherent ranking of the other data points. It is the preferred measure for **Nominal** data (e.g., blood groups). * **D. Range:** This is a measure of **dispersion (variability)**, not central tendency. It describes the spread between the maximum and minimum values. **High-Yield Clinical Pearls for NEET-PG:** * **Nominal Data:** Use **Mode** (e.g., Gender, Religion). * **Ordinal Data:** Use **Median** (e.g., Pain scales, APGAR score). * **Symmetrical/Normal Distribution:** Mean = Median = Mode. * **Skewed Distribution:** The **Median** is the best measure of central tendency because the Mean is sensitive to outliers (extreme values). * **Most common measure** used in medical statistics is the **Mean**, but the **most robust** is the **Median**.

Q: Which epidemiological study design is represented by the following illustration?

Case control study. ***Case control study*** - This study design works **retrospectively**, starting with individuals who already have the disease (cases) and those without (controls), then looking back to assess **past exposure**. - The illustration shows two groups being compared based on their **outcome status** first, then examining their **exposure history**. *Cohort study* - This design follows participants **prospectively** from exposure to outcome, moving forward in time. - Participants are selected based on their **exposure status** (exposed vs. unexposed), not their disease status. *Ecological study* - This design examines **populations or groups** rather than individuals, comparing disease rates across different communities or regions. - The unit of analysis is the **population level**, not individual cases and controls. *Cross-sectional study* - This design examines both **exposure and outcome simultaneously** at a single point in time. - It provides a **snapshot** of the population without following participants over time or looking backwards.

Q: What is a disadvantage of calculating the mean?

May be influenced by abnormal high or low values. **Explanation:** The **Arithmetic Mean** is the most commonly used measure of central tendency in biostatistics. It is calculated by summing all observations and dividing by the total number of items ($Mean = \Sigma x / n$). **Why Option C is Correct:** The primary disadvantage of the mean is its **sensitivity to extreme values (outliers)**. Because every single value in a dataset is used in the calculation, a single abnormally high or low value will "pull" the mean toward it, making it an unrepresentative measure of the "average." For example, in a study of five patients with recovery times of 3, 4, 5, 6, and 50 days, the mean is 13.6 days—a value that does not accurately reflect the typical recovery time of the group. **Why Incorrect Options are Wrong:** * **Option A & B:** The mean is actually the **easiest** measure of central tendency to calculate mathematically and the most **intuitive** to understand for clinicians and researchers. * **Option D:** This is incorrect as Option C is a well-documented statistical limitation. **High-Yield Clinical Pearls for NEET-PG:** * **Skewed Data:** In a skewed distribution (non-normal), the mean is the most affected measure. In **positively skewed** data, Mean > Median > Mode. In **negatively skewed** data, Mean < Median < Mode. * **Best Measure:** The **Median** is the preferred measure of central tendency for skewed data or data containing outliers (e.g., incubation periods, survival times). * **Normal Distribution:** In a perfectly symmetrical (Gaussian) distribution, the Mean, Median, and Mode are all equal.

Q: Randomization is useful to eliminate which of the following?

Confounding factors. **Explanation:** **Why Confounding Factors is Correct:** Randomization is the "heart" of a Randomized Controlled Trial (RCT). Its primary purpose is to ensure that both known and unknown **confounding factors** are distributed equally between the study and control groups. By allocating participants purely by chance, randomization eliminates **selection bias** and balances prognostic factors, ensuring that any observed difference in outcome is due to the intervention alone and not an extraneous variable. It is the only method that can control for **unknown confounders**. **Why Other Options are Incorrect:** * **A & C (Observer and Patient Bias):** These are types of information/measurement bias. Randomization does not prevent these; they are eliminated through **Blinding** (Masking). Blinding ensures that neither the patient nor the researcher knows who is receiving the treatment, thus preventing subjective prejudice. * **D (Sampling Bias):** This occurs when the study population is not representative of the target population. It is eliminated through **Random Sampling** (e.g., simple random sampling), not Randomization (Random Allocation). **High-Yield Clinical Pearls for NEET-PG:** * **Randomization** = Eliminates Confounding + Selection Bias. * **Blinding** = Eliminates Observer + Patient (Information) Bias. * **Matching** = A technique used in Case-Control studies to eliminate known confounders (but cannot control for unknown ones). * **The "Gold Standard"** for testing a new drug is the Double-Blind Randomized Controlled Trial. * **Confounding by Indication:** A specific type of bias where the factor that dictates the treatment choice is also related to the outcome.

Q: A study was conducted on a group of children to assess the seasonal variation of Sudden Infant Death Syndrome (SIDS) during the summer months (June-July) and a subsequent period (August-September) in a group with similar characteristics from the same area. Which statistical test is used to compare the data from these two groups?

Chi-Square test. ### Explanation **1. Why Chi-Square Test is Correct:** The study aims to compare the occurrence of Sudden Infant Death Syndrome (SIDS) between two distinct time periods (June-July vs. August-September). In biostatistics, SIDS is a **categorical (qualitative)** variable—an infant either experiences SIDS or does not. When comparing the frequencies or proportions of a categorical outcome between two independent groups, the **Chi-Square test** is the standard test of significance. It assesses whether the observed variation in SIDS cases across seasons is due to chance or a statistically significant association. **2. Why Other Options are Incorrect:** * **Paired T-test:** This is used for **quantitative (numerical)** data when comparing means of the same group before and after an intervention (e.g., blood pressure before and after a drug). It is not for categorical outcomes. * **Wilcoxon Rank-Sum/Signed-Rank Test:** These are **non-parametric** tests used for ordinal data or non-normally distributed quantitative data. They are not used for simple frequency comparisons of nominal data like SIDS. * **ANOVA (Analysis of Variance):** This is used to compare the **means of three or more** independent groups for quantitative data. It is not applicable to categorical data or a comparison of only two groups. **3. Clinical Pearls & High-Yield Facts:** * **Rule of Thumb:** If the data is in **proportions, percentages, or 2x2 tables**, think **Chi-Square**. If the data is **means/averages**, think **T-test**. * **Fisher’s Exact Test:** Use this instead of Chi-Square if the sample size is very small (any cell value in the 2x2 table is <5). * **SIDS Risk Factors:** High-yield associations include prone sleeping position (Back to Sleep campaign), maternal smoking, and overheating. Peak incidence typically occurs between 2–4 months of age.

Question 1

Which of the following is FALSE regarding Randomised Control Trials?

Accepted Answer

Randomisation is done while selecting subjects for the study

Answer

Results of attrition are included in the analysis

Answer

Double blinding is the most common form of blinding observed

Answer

Cross-over design helps removing ethical concerns

Question 2

In an investigation to study the effect of smoking on renal cell carcinoma, it is observed that 30 out of 50 patients were smokers as compared to 10 out of 50 control subjects. What is the odd ratio of renal cell carcinoma associated with smoking?

Accepted Answer

6

Answer

3

Answer

0.33

Answer

0.16

Question 3

In a randomized controlled trial, what is the essential purpose of randomization?

Accepted Answer

To eliminate selection bias

Answer

To produce double blinding

Answer

To decrease the follow-up period

Answer

To decrease the sample size

Question 4

The mean of 25 plasma volumes is 12.5 litres. The standard deviation is 0.25. Calculate the variance.

Accepted Answer

0.0625

Answer

0.625

Answer

6.25

Answer

625

Question 5

In a population of 200 people with a normal distribution, how many people would be included within 1 standard deviation?

Accepted Answer

136

Answer

140

Answer

150

Answer

190

Question 6

What is the preferred measure of central tendency for ordinal data?

Accepted Answer

Median

Answer

Mean

Answer

Mode

Answer

Range

Question 7

Which epidemiological study design is represented by the following illustration?

Accepted Answer

Case control study

Answer

Cohort study

Answer

Ecological study

Answer

Cross-sectional study

Question 8

What is a disadvantage of calculating the mean?

Accepted Answer

May be influenced by abnormal high or low values

Answer

Difficult to calculate

Answer

Difficult to understand

Answer

None of the above

Question 9

Randomization is useful to eliminate which of the following?

Accepted Answer

Confounding factors

Answer

Observer bias

Answer

Patient bias

Answer

Sampling bias

Question 10

A study was conducted on a group of children to assess the seasonal variation of Sudden Infant Death Syndrome (SIDS) during the summer months (June-July) and a subsequent period (August-September) in a group with similar characteristics from the same area. Which statistical test is used to compare the data from these two groups?

Accepted Answer

Chi-Square test

Answer

Paired T-test

Answer

Wilcoxon Rank test

Answer

ANOVA

Biostatistics — MCQs

Biostatistics — MCQs

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