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Interview Questions

r power requires larger sample sizes. Before running an A/B test, I'd do a power analysis to determine the minimum sample size needed to detect a meaningful effect — this prevents wasting time on underpowered tests."

Q9: "Explain confidence interval to a stakeholder."​

Answer: "A 95% confidence interval tells us: if we repeated this analysis 100 times with different samples, the true value would fall within our range 95 times out of 100. For example, if customer satisfaction is 78 ± 3 at 95% confidence, we're 95% sure the true satisfaction is between 75 and 81. A narrower interval means more precision — which requires more data."

Q10: "How would you check if data is normally distributed?"​

Answer: "Three methods: (1) Visual — plot a histogram and see if it's bell-shaped, or use a Q-Q plot where normally distributed data falls along a straight line. (2) Statistical test — Shapiro-Wilk test for small samples or Kolmogorov-Smirnov for large samples. (3) Check skewness value — values between -1 and +1 generally indicate approximate normality."

Q11: "What is a confounding variable?"​

Answer: "A confounding variable is a third variable that influences both the independent and dependent variables, creating a misleading association. For example, when studying marketing spend vs sales, seasonality could be a confounder — both spend and sales increase during festivals regardless of campaign effectiveness. I'd address confounders through randomization in experiments, or control for them in regression models."

Q12: "When would you use a non-parametric test?"​

Answer: "When the data violates assumptions of parametric tests — mainly when data isn't normally distributed, the sample size is small, or the data is ordinal rather than continuous. Examples: Mann-Whitney U test instead of t-test, Kruskal-Wallis instead of ANOVA, Spearman correlation instead of Pearson. They're more robust but slightly less powerful."

10. Practice Problems with Solutions​

Problem 1​

Q: A dataset has values: 12, 15, 14, 10, 18, 95. Calculate mean and median. Which would you report and why?

Solution:

Sorted: 10, 12, 14, 15, 18, 95

Mean = (10+12+14+15+18+95)/6 = 164/6 = 27.33
Median = average of 3rd and 4th values = (14+15)/2 = 14.5

Report Median (14.5) because 95 is an outlier that inflates the mean.
The mean of 27.33 doesn't represent any typical data point.

Problem 2​

Q: Heights of 200 men are normally distributed: Mean = 170 cm, SD = 6 cm. How many men are taller than 182 cm?

Solution:

Z = (182 - 170) / 6 = 12/6 = 2.0
Z = 2 means 182 cm is 2 SDs above mean
By 68-95-99.7 rule: 95% fall within ±2σ
→ 5% fall outside → 2.5% above 182 cm
→ 200 × 0.025 = 5 men ✅

Problem 3​

Q: An e-commerce site's A/B test results: Control group conversion = 3.2% (n=4000). Treatment group = 3.8% (n=4000). p-value = 0.07. What's your recommendation?

Solution:

p = 0.07 > 0.05 → Not statistically significant at 95% confidence

However:
- Absolute lift: +0.6 percentage points
- Relative lift: 18.75% improvement
- If each conversion is worth ₹500, potential daily gain = 
  4000 × 0.006 × 500 = ₹12,000/day

Recommendation: The trend is positive. I'd extend the test for 
1-2 more weeks to increase sample size. If the lift holds, it 
would likely achieve significance with more data.

Problem 4​

Q: Two teams' monthly sales (in lakhs):

  • Team A: 10, 12, 11, 13, 10, 12, 11
  • Team B: 5, 20, 8, 15, 3, 22, 6

Which team is more consistent? Prove it.

Solution:

Team A: Mean = 11.29, calculate SD:
  Deviations: -1.29, 0.71, -0.29, 1.71, -1.29, 0.71, -0.29
  Squared:    1.66, 0.50, 0.08, 2.92, 1.66, 0.50, 0.08
  Variance = 7.41/7 = 1.06 → SD ≈ 1.03

Team B: Mean = 11.29, calculate SD:
  Deviations: -6.29, 8.71, -3.29, 3.71, -8.29, 10.71, -5.29
  Squared:    39.56, 75.86, 10.82, 13.76, 68.72, 114.70, 27.98
  Variance = 351.41/7 = 50.20 → SD ≈ 7.09

Team A (SD=1.03) is FAR more consistent than Team B (SD=7.09) ✅
Both teams have the same mean, but Team B's performance 
swings wildly from 3L to 22L.

Problem 5​

Q: A machine fills packets with average weight 500g (SD = 10g). Quality check requires 99.7% of packets to be between 470g and 530g. Is the machine performing within tolerance?

Solution:

Required range: 470 to 530 = 500 ± 30g = μ ± 3σ?
Check: 3 × 10 = 30  → Yes, 500 ± 30 = μ ± 3σ

By 68-95-99.7 rule: 99.7% falls within μ ± 3σ
→ 99.7% of packets will be between 470g and 530g
→ Machine IS performing within tolerance ✅