Biostatistics Practice Questions

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.

Q: In a study, an investigator compared the mean blood pressure and standard deviation between two independent groups. Which of the following is the most appropriate statistical test to assess the significance of the difference between the two means?

Unpaired t-test. ***Unpaired t-test***- It is the most appropriate statistical test used to compare the means of two independent (unrelated) groups when the data is continuous (like **blood pressure**).- This test assesses the null hypothesis that there is no significant difference between the **population means** of the two comparison groups.*Paired t-test*- This test is specifically designed to compare means when the observations are dependent, meaning the data comes from the **same subjects** measured twice (e.g., pre-treatment and post-treatment).- It is used for **within-group comparisons** rather than comparisons between two independent cohorts, as requested in the scenario.*Chi-square test*- The chi-square test is used to determine the association between **two categorical variables** (e.g., proportions or frequencies).- It is unsuitable here because the variable being compared (blood pressure) is **continuous data**, and the study requires comparing means, not counted frequencies.*ANOVA*- ANOVA (Analysis of Variance) is used when comparing the means of **three or more** independent groups.- While acceptable for two groups (where it gives equivalent results to the t-test), the **unpaired t-test** is the most specific and standard test for comparing means of exactly two independent samples.

Q: In a study based in Delhi, doctors followed 100 people who did exercise for a year and later they checked who all had developed coronary heart disease. Which type of study is this?

Cohort study. ***Cohort study***- This study starts with a group of people free of the disease (**CHD**) and classifies them based on their exposure status (e.g., *exercise* vs. no exercise) and follows them forward in time (**prospectively**) to measure the incidence of the disease.- The study tracks the patients *forward* from exposure (**exercise**) to outcome (**CHD**) over a specified period (one year), which is the definitive characteristic of a **prospective cohort study**.*Case Control Study*- In this design, the study starts with the outcome (**CHD**) and retrospectively looks back (examining controls without CHD) to determine past exposure, making it unsuitable for this specific prospective tracking of exposure.- It is used primarily to estimate the **odds ratio** and is efficient for studying rare diseases; it does not measure incidence over time.*Prospective Study*- While this specific study is **prospective** (looking forward in time), this term describes the *timing* and direction of data collection, whereas **Cohort Study** is the most specific designation describing the fundamental design of following a defined exposed population.- A **prospective study** is a broad term, and the term **Cohort Study** most accurately describes the method of following an exposed group to measure disease incidence over time.*Cross Sectional Study*- This study type measures both the exposure (exercise) and the outcome (**CHD**) simultaneously at a **single point in time**, assessing prevalence rather than tracking incidence over one year.- It provides a **snapshot** and cannot establish the temporal relationship between exposure and outcome, failing to align with the follow-up design described.

Q: Which of the following is the formula for sensitivity?

TP/(TP+FN)x100. ***TP/(TP+FN)x100***- This is the formula for **sensitivity** (or True Positive Rate), which is the proportion of individuals who truly have the disease (**True Positives, TP**) who are correctly identified by the test.- The denominator $TP + FN$ accounts for all individuals who actually have the disease according to the **gold standard**, including those who tested negatively (**False Negatives, FN**).*TP/(TP+FP)x100*- This formula calculates the **Positive Predictive Value (PPV)**, which indicates the probability that a positive test result represents a true positive.- The denominator $TP + FP$ includes everyone who tested positive, regardless of their actual disease status (**True Positives** and **False Positives**).*TN/(TN+FP)x100*- This formula calculates **specificity** (or True Negative Rate), which is the proportion of individuals who are truly disease-free (**True Negatives, TN**) correctly identified by the test.- The denominator $TN + FP$ accounts for all individuals without the disease, including those who were incorrectly identified as positive (**False Positives, FP**).*TN/(TN+FN)x100*- This formula calculates the **Negative Predictive Value (NPV)**, which is the probability that a negative test result represents a true negative.- The denominator $TN + FN$ includes everyone who tested negative, reflecting the proportion of subjects with a negative test result who are truly disease-free.

Q: In a normal distribution, the mean value is 82, the SD is 1.5. Calculate the range of two standard deviations?

79-85. ***79-85***- For a **normal distribution**, the range covering two standard deviations (2 SD) is calculated using the formula: **Mean $\pm$ (2 $\times$ SD)**. The $2\sigma$ interval encompasses approximately **95.45%** of the data points.- Calculation: Lower limit $= 82 - (2 \times 1.5) = 82 - 3 = 79$. Upper limit $= 82 + (2 \times 1.5) = 85$. The correct range is **79-85**. *60-68*- This range is highly incorrect as it is centered far below the **mean of 82** and the width (8 units) is too wide for a total of 3 units (2 SD) dispersion. - The lower limit of 60 is over 14 standard deviations away from the mean, indicating it is an outlier range not relevant to the $2\sigma$ calculation. *50-57*- This range is excessively far from the **mean of 82** and its width (7 units) does not correspond to the required 3 units of dispersion needed for $\pm 2$ SD. - Ranges like this would include virtually none of the observations expected in a population with a mean of 82 and a small **standard deviation** of 1.5. *40-49*- This interval is centered around 44.5, which is highly divergent from the actual **mean of 82**, and therefore cannot represent the population's $\pm 2$ SD range. - In a normal distribution, the data is symmetric around the mean; any calculated range must therefore include the mean near its center, which this option fails to do.

Q: A total of 2000 patients were assessed for HIV. 200 were diagnosed positive. A new ELISA screening test was tested on the same group. It showed 260 as positive out of which only 130 had the disease. What is the specificity of the test?

96%. ***96%*** - **Specificity** is the ability of a test to correctly identify those *without* the disease (True Negatives) among all disease-free individuals: Specificity = TN / (TN + FP) - Given data: Total patients = 2000; Actual HIV positive = 200; Actual HIV negative = 1800 - Test showed 260 positives, of which 130 were true positives (TP) - False Positives (FP) = 260 - 130 = 130 - True Negatives (TN) = Total negatives - FP = 1800 - 130 = 1670 - **Calculated Specificity = 1670/1800 × 100 = 92.78%** - Among the given options, **96% is the closest** to the calculated value of 92.78% *80%* - This value is too low and does not match the calculated specificity - This might represent a miscalculation or confusion with sensitivity *72%* - This is significantly lower than the actual specificity of 92.78% - This does not correspond to any standard epidemiological measure from the given data *68%* - This is the lowest option and far from the correct calculation - This may result from calculation errors such as using wrong denominators or confusing different test parameters

Question 1

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 2

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 3

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 4

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

Accepted Answer

Mean - Mode / SD

Answer

Mode - Mean / SD

Answer

SD / Mode - Mean

Question 5

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

Question 6

In a study, an investigator compared the mean blood pressure and standard deviation between two independent groups. Which of the following is the most appropriate statistical test to assess the significance of the difference between the two means?

Accepted Answer

Unpaired t-test

Answer

Chi-square test

Answer

ANOVA

Answer

Paired t-test

Question 7

In a study based in Delhi, doctors followed 100 people who did exercise for a year and later they checked who all had developed coronary heart disease. Which type of study is this?

Accepted Answer

Cohort study

Answer

Case Control Study

Answer

Prospective Study

Answer

Cross Sectional Study

Question 8

Which of the following is the formula for sensitivity?

Accepted Answer

TP/(TP+FN)x100

Answer

TN/(TN+FP)x100

Answer

TP/(TP+FP)x100

Answer

TN/(TN+FN)x100

Question 9

In a normal distribution, the mean value is 82, the SD is 1.5. Calculate the range of two standard deviations?

Accepted Answer

79-85

Answer

60-68

Answer

40-49

Answer

50-57

Question 10

A total of 2000 patients were assessed for HIV. 200 were diagnosed positive. A new ELISA screening test was tested on the same group. It showed 260 as positive out of which only 130 had the disease. What is the specificity of the test?

Accepted Answer

96%

Answer

68%

Answer

72%

Answer

80%

Biostatistics — MCQs

Biostatistics — MCQs

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