Biostatistics Practice Questions

Q: A standard normal distribution has which of the following characteristics?

A mean of 0 and a standard deviation of 1. ### Explanation In Biostatistics, a **Standard Normal Distribution** (also known as the **Z-distribution**) is a specific type of normal distribution used to standardize different sets of data for comparison. **1. Why Option B is Correct:** By definition, a standard normal distribution is a normal distribution that has been "standardized" such that the **Mean ($\mu$) is 0** and the **Standard Deviation ($\sigma$) is 1**. This allows any value ($x$) from a normal distribution to be converted into a **Z-score** using the formula: $Z = (x - \mu) / \sigma$. This transformation ensures that the center of the curve sits at zero and the spread is measured in units of 1. **2. Why the Other Options are Incorrect:** * **Option A:** While the standard deviation is 1, the mean must be 0. A mean of 1 would shift the entire bell curve to the right. * **Option C:** This is not a defining characteristic. In a standard normal distribution, the mean (0) is actually *smaller* than the standard deviation (1). * **Option D:** In a normal distribution, only approximately **68%** of scores fall within one standard deviation ($\pm 1\sigma$) of the mean, not all scores. **3. NEET-PG High-Yield Clinical Pearls:** * **Empirical Rule (68-95-99.7 Rule):** * $\pm 1\sigma$ covers **68.2%** of the area. * $\pm 2\sigma$ covers **95.4%** of the area. * $\pm 3\sigma$ covers **99.7%** of the area. * **Key Properties:** The curve is symmetrical, bell-shaped, and the **Mean = Median = Mode**. * **Z-score:** A Z-score of +1.96 corresponds to the 95% confidence interval boundary in a two-tailed test.

Q: In a study, 30 out of 50 smokers developed lung cancer, and 10 out of 50 non-smokers developed lung cancer. What is the odds ratio?

6. ### Explanation **1. Why the Correct Answer (C) is Right** The **Odds Ratio (OR)** is a measure of association used primarily in case-control studies to quantify the relationship between an exposure and an outcome. It is calculated as the ratio of the odds of exposure in cases to the odds of exposure in controls, or more simply, using a 2x2 contingency table: | | Disease (+) | Disease (-) | Total | | :--- | :---: | :---: | :---: | | **Exposed (Smokers)** | 30 (a) | 20 (b) | 50 | | **Non-exposed (Non-smokers)** | 10 (c) | 40 (d) | 50 | * **a (Exposed cases):** 30 * **b (Exposed non-cases):** 50 - 30 = 20 * **c (Non-exposed cases):** 10 * **d (Non-exposed non-cases):** 50 - 10 = 40 **Formula:** $OR = \frac{a \times d}{b \times c}$ **Calculation:** $OR = \frac{30 \times 40}{20 \times 10} = \frac{1200}{200} = \mathbf{6}$ An OR of 6 indicates that the odds of developing lung cancer are 6 times higher in smokers compared to non-smokers. **2. Why Other Options are Wrong** * **Option A (4):** This is a common distractor if a student incorrectly calculates the ratio of diseased individuals (30/10) or makes a calculation error. * **Option B (2.8):** This value is close to the **Relative Risk (RR)**. $RR = \frac{\text{Incidence in exposed}}{\text{Incidence in non-exposed}} = \frac{30/50}{10/50} = \frac{0.6}{0.2} = 3$. * **Option D (7):** Incorrect calculation; does not correspond to any standard epidemiological measure for this data. **3. NEET-PG Clinical Pearls** * **Odds Ratio** is the only measure of association that can be calculated in **Case-Control studies**. * **Relative Risk (RR)** is calculated in **Cohort studies**. * When a disease is rare, the OR is a good approximation of the RR. * **Attributable Risk (AR):** Indicates the amount of disease that can be prevented by removing the exposure. $AR = \frac{I_e - I_u}{I_e} \times 100$.

Q: Which of the following is true regarding case-control studies?

The odds ratio can be calculated.. In epidemiology, a **Case-Control Study** is an observational, analytical study used to identify the association between an exposure and an outcome. ### Why the Correct Answer is Right **Option B (Odds Ratio):** Since case-control studies start with people who already have the disease (cases), we cannot determine the actual risk of developing the disease. Instead, we calculate the **Odds Ratio (OR)**, which is the ratio of the odds of exposure among cases to the odds of exposure among controls. It serves as an estimate of the relative risk. ### Why Other Options are Wrong * **Option A:** This is partially true but technically described as **retrospective**. While it looks backward from effect (disease) to cause (exposure), the standard epidemiological phrasing is that it is "retrospective" in nature. However, Option B is the definitive statistical hallmark of this study design. * **Option C:** **Incidence cannot be calculated** in case-control studies because the denominator (population at risk) is unknown. Incidence can only be calculated in **Cohort Studies**. * **Option D:** Case-control studies are ideal for **rare diseases** and typically require a **smaller sample size** compared to cohort studies, making them inexpensive and quick to conduct. ### High-Yield NEET-PG Pearls * **Direction:** Backward (Effect $\rightarrow$ Cause). * **Measure of Association:** Odds Ratio ($ad/bc$). * **Best for:** Rare diseases or diseases with long latency periods (e.g., Cancer). * **Main Bias:** **Recall Bias** (cases remember past exposures more vividly than controls). * **Matching:** Done to eliminate the effects of **confounding variables**.

Q: Which study design is used to correlate genetic diseases with consanguinity?

Case-control study. ### Explanation **Why Case-control study is correct:** Genetic diseases, particularly those linked to consanguinity (autosomal recessive disorders), are typically **rare** in the general population. The **Case-control study** is the design of choice for studying rare diseases. In this design, researchers start with "Cases" (children with the genetic disease) and "Controls" (healthy children) and look backward (**retrospective**) to determine the frequency of the exposure (consanguineous marriage among parents). It is time-efficient and cost-effective for conditions with long latency or low prevalence. **Why the other options are incorrect:** * **Cohort study:** This starts with the exposure (consanguinity) and follows the group forward to see if the disease develops. Because these genetic diseases are rare, a massive sample size and decades of follow-up would be required, making it impractical and expensive. * **Cross-sectional study:** This measures prevalence at a single point in time. While it can show an association, it is weak at establishing a causal link for rare genetic conditions compared to the case-control method. * **Case report:** This describes a single patient’s clinical features. While it may suggest an association, it lacks a comparison group and cannot be used to scientifically correlate or prove an epidemiological link. **High-Yield Clinical Pearls for NEET-PG:** * **Study of Choice for Rare Diseases:** Case-control study. * **Study of Choice for Rare Exposures:** Cohort study (e.g., a specific chemical leak). * **Measure of Association:** Case-control uses **Odds Ratio (OR)**; Cohort uses **Relative Risk (RR)**. * **Consanguinity Risk:** Increases the risk of **Autosomal Recessive** conditions due to increased homozygosity (sharing of common alleles from a common ancestor).

Q: What is the most appropriate statistical test to study the relationship between the mean height of two groups of children?

Student's t-test. **Explanation:** The core of this question lies in identifying the **type of data** and the **number of groups** being compared. 1. **Why Student’s t-test is correct:** Height is a **quantitative (numerical/continuous)** variable. When comparing the **means** of a quantitative variable between **two independent groups** (e.g., boys vs. girls or Group A vs. Group B), the Student’s independent t-test is the standard parametric test used. It determines if the observed difference in means is statistically significant or due to chance. 2. **Why other options are incorrect:** * **Linear Regression:** This is used to describe the *strength and direction* of a linear relationship between two continuous variables (e.g., age and height) or to predict the value of one variable based on another. It is not a test of difference between group means. * **Chi-square Test:** This is used for **qualitative (categorical)** data to compare proportions or associations (e.g., comparing the number of "stunted" vs. "normal" children in two groups). * **Test of Proportions (Z-test for proportions):** This is used when comparing percentages or ratios between two groups, not mean values of continuous data. **High-Yield Clinical Pearls for NEET-PG:** * **Two groups, comparing means:** Student’s t-test. * **More than two groups (>2), comparing means:** ANOVA (Analysis of Variance). * **Paired data (e.g., weight before and after a diet in the same person):** Paired t-test. * **Non-parametric alternative to t-test:** Mann-Whitney U test (used if data is not normally distributed). * **Rule of thumb:** If the data is in "Mean ± SD," look for t-test or ANOVA. If the data is in "n (%)" or "Proportions," look for Chi-square.

Q: Which of the following is Pearson's measure of skewness?

Mean - Mode / SD. ### Explanation **Pearson’s Coefficient of Skewness** is a measure used in biostatistics to determine the asymmetry of a probability distribution. In a perfectly symmetrical (Normal) distribution, the Mean, Median, and Mode are equal, resulting in a skewness of zero. #### 1. Why Option B is Correct Karl Pearson’s first coefficient of skewness is defined by the formula: **Skewness = (Mean – Mode) / Standard Deviation (SD)** * **Logic:** It measures how far the Mean is pulled away from the Mode relative to the dispersion (SD) of the data. * **Directionality:** * If **Mean > Mode**, the result is positive (**Positive Skew**; tail to the right). * If **Mean < Mode**, the result is negative (**Negative Skew**; tail to the left). #### 2. Why Other Options are Incorrect * **Option A (Mode - Mean / SD):** This is the inverse of the correct formula and would incorrectly sign the direction of the skew. * **Option C (SD / Mode - Mean):** This is mathematically incorrect; the Standard Deviation serves as the denominator to "standardize" the measure, not the numerator. * **Option D:** This is a duplicate of the correct answer in the prompt, but the fundamental formula remains Mean minus Mode divided by SD. #### 3. High-Yield Clinical Pearls for NEET-PG * **Alternative Formula:** Since the Mode can be unstable in some datasets, Pearson’s second coefficient is often used: **3 (Mean – Median) / SD**. * **Relationship in Skewed Data:** * **Positively Skewed:** Mean > Median > Mode (e.g., income distribution, incubation periods). * **Negatively Skewed:** Mode > Median > Mean (e.g., age at death in developed countries). * **Memory Aid:** In a positive skew, the "Mean" is the "Meanest" (highest value) because it is most affected by extreme outliers in the tail.

Q: In a group, the mean blood glucose level is 105 mg/dL with a standard deviation of 10 mg/dL. Assuming a normal distribution, what is the expected range of blood glucose values for approximately 95% of the population?

85 – 125 mg/dL. ***85 – 125 mg/dL*** - This range is calculated using the **Empirical Rule** ($\text{Mean} \pm 2 \text{ SD}$), which states that approximately 95% of observations in a **normal distribution** fall within two standard deviations of the mean. - Calculation: $105 \text{ mg/dL} \pm (2 \times 10 \text{ mg/dL}) = 105 \pm 20 \text{ mg/dL}$, resulting in the range **85 – 125 mg/dL**. *101 – 110 mg/dL* - This range is too narrow, only covering values $5 \text{ mg/dL}$ above and below the mean, and does not represent the required **95%** coverage for a 10 mg/dL standard deviation. - Using this small range indicates an incorrect application of the **standard deviation** multiplier necessary for determining large confidence intervals. *90 – 125 mg/dL* - While the upper limit ($125 \text{ mg/dL}$) is correct ($\text{Mean} + 2 \text{ SD}$), the lower limit ($90 \text{ mg/dL}$) is incorrect, as it must be symmetrical around the mean in a **normal distribution**. - This asymmetrical range does not accurately represent the **95% confidence interval** defined by $\text{Mean} \pm 2 \text{ SD}$. *95 – 115 mg/dL* - This range is calculated using $\text{Mean} \pm 1 \text{ SD}$ ($105 \pm 10 \text{ mg/dL}$), which only includes approximately **68%** of the data according to the **Empirical Rule**, not 95%. - To capture **95%** of the population data, clinicians and students must use **two standard deviations** from the mean.

Question 1

Which statistical test is used to compare Kaplan-Meier survival curves?

Accepted Answer

Log rank test

Answer

T-test

Answer

Chi-square test

Answer

Wilcoxon rank-sum test

Question 2

A standard normal distribution has which of the following characteristics?

Accepted Answer

A mean of 0 and a standard deviation of 1

Answer

A mean of 1 and a standard deviation of 1

Answer

A mean larger than its standard deviation

Answer

All scores within one standard deviation of the mean

Question 3

What is the best method to remove confounding?

Accepted Answer

Stratified randomization

Answer

Randomization

Answer

Restriction

Answer

Multivariate analysis

Question 4

In a study, 30 out of 50 smokers developed lung cancer, and 10 out of 50 non-smokers developed lung cancer. What is the odds ratio?

Accepted Answer

6

Answer

4

Answer

2.8

Answer

7

Question 5

Which of the following is true regarding case-control studies?

Accepted Answer

The odds ratio can be calculated.

Answer

It proceeds from effect to cause.

Answer

Incidence can be calculated.

Answer

It requires a large number of patients.

Question 6

Which study design is used to correlate genetic diseases with consanguinity?

Accepted Answer

Case-control study

Answer

Cohort study

Answer

Cross-sectional study

Answer

Case report

Question 7

What is the most appropriate statistical test to study the relationship between the mean height of two groups of children?

Accepted Answer

Student's t-test

Answer

Linear regression

Answer

Chi-square test

Answer

Test of proportions

Question 8

If the hemoglobin status of a population has a mean value of 10.3 gm% with a standard deviation of 2 gm%, then 5% of the population will be below what value of Hb?

Accepted Answer

7.35

Answer

6.67

Answer

9

Answer

8.6

Question 9

Which of the following is Pearson's measure of skewness?

Accepted Answer

Mean - Mode / SD

Answer

Mode - Mean / SD

Answer

SD / Mode - Mean

Question 10

In a group, the mean blood glucose level is 105 mg/dL with a standard deviation of 10 mg/dL. Assuming a normal distribution, what is the expected range of blood glucose values for approximately 95% of the population?

Accepted Answer

85 – 125 mg/dL

Answer

101 – 110 mg/dL

Answer

95 – 115 mg/dL

Answer

90 – 125 mg/dL

Biostatistics — MCQs

Biostatistics — MCQs

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