Statistics Interview Questions

"The mean uses every value, so one extreme value can drag it a long way. The median is just the middle value once you sort, so it barely moves. Say five deliveries took 10, 12, 13, 15 and 90 minutes. The sum is 140, divided by 5 gives a mean of 28 minutes. The median is the third value, 13 minutes. Four of the five deliveries took 15 minutes or less, so 28 describes nobody's real experience. That's why I'd lead with the median for skewed things like delivery times, house prices or time on a page. When the data is roughly symmetric, the two are close and the mean is fine, and it has nicer maths behind it. In a report I often show both, because a big gap between them is itself a signal that the data is skewed."

"First the mean: the eight values add up to 40, so the mean is 5. Next I take each value minus 5: that's minus 3, minus 1, minus 1, minus 1, 0, 0, 2 and 4. Squaring gives 9, 1, 1, 1, 0, 0, 4 and 16, which add up to 32. If these eight numbers are the whole population, I divide by 8, so the variance is 4 and the standard deviation is the square root, 2. If they're a sample and I want to estimate the population, I divide by 7 instead, which gives about 4.57, and a standard deviation of about 2.14. The standard deviation is the useful one to talk about because it's in the same units as the data. It says values typically sit around two units away from the mean of 5."

"When I compute a sample variance, I measure distances from the sample mean, because I don't know the true population mean. But the sample mean is, by construction, the value that makes the sum of squared distances as small as it can be for that sample. So measuring from it always gives a sum that's a bit too small compared with measuring from the true mean. On average, dividing by n underestimates the population variance. Dividing by n minus 1 corrects that exactly, so the sample variance is unbiased. Another way to say it is degrees of freedom: once I know the mean and n minus 1 of the deviations, the last deviation is fixed, because they must add to zero. With large samples the difference hardly matters, but with five or ten points it does."

"My first check is usually the IQR rule because it doesn't assume normal data. I find the first and third quartiles, say Q1 is 20 and Q3 is 40, so the interquartile range is 20. One and a half times that is 30. The lower fence is 20 minus 30, which is minus 10, and the upper fence is 40 plus 30, which is 70. Anything above 70 or below minus 10 gets flagged. For roughly normal data I might also flag values more than about three standard deviations from the mean. But flagging isn't removing. I ask why the value is there. If it's a typo, a test account or a unit mix-up, I fix it or drop it and write that down. If it's real, like one genuinely huge order, it stays, and I use the median or report results with and without it."

"At my last company I reported that the average session length went up by about a third after a redesign, and the team was pleased. A week later I broke it down by user and saw most of the increase came from a small number of sessions that ran for hours. They turned out to be tabs left open overnight, which our tracking counted as active. The mean had been pulled up by a handful of extreme values, and the median had hardly moved. I told my manager and the product lead the same day, sent a short correction with the median and the capped mean, and said plainly that the redesign's effect was small. Since then I always look at the distribution before quoting an average, report the median next to the mean for skewed metrics, and check the top few values before sharing anything."

"The normal distribution is the symmetric bell curve, and it's completely described by two numbers, the mean and the standard deviation. The mean, median and mode are all the same point. The 68-95-99.7 rule says about 68 percent of values fall within one standard deviation of the mean, about 95 percent within two, and about 99.7 percent within three. Say adult heights in a group are normal with mean 170 centimetres and standard deviation 10. Then roughly 68 percent are between 160 and 180, and about 95 percent between 150 and 190. Someone who is 185 has a z-score of 185 minus 170, over 10, which is 1.5, so they're one and a half standard deviations above average. It matters so much because averages of many samples tend to be normal, which is what a lot of tests rely on."

"This is binomial: five independent visitors, each either signs up or doesn't, with the same one-in-ten chance. The probability of exactly two is the number of ways to pick which two, times 0.1 squared, times 0.9 cubed for the three who don't. Five choose two is 10. 0.1 squared is 0.01. 0.9 cubed is 0.729. So it's 10 times 0.01 times 0.729, which is 0.0729, about 7 percent. As a sanity check, the expected number of sign-ups is n times p, which is 0.5, so two sign-ups should be fairly unlikely, and it is. If the question were at least one sign-up, I'd use the complement: 1 minus 0.9 to the fifth, which is about 0.41. The assumption I'd question in real life is independence, like a group of friends all arriving from the same link."

"Calls arriving independently at a steady average rate are a classic Poisson case, with lambda equal to 3 per hour. The probability of exactly k calls is e to the minus 3, times 3 to the k, over k factorial. For zero calls that's just e to the minus 3, which is about 0.050, so roughly a one-in-twenty hour is silent. For one call it's e to the minus 3 times 3, about 0.149. For two it's e to the minus 3 times 9 over 2, about 0.224. Adding those three, two or fewer calls is about 0.42. A neat property is that the mean and variance are both 3. So in real data, if the variance is much bigger than the mean, calls are probably bunched, say around lunch or after an outage, and the plain Poisson model is too optimistic."

"The instinct is to say 99 percent, but that ignores how rare the condition is. I'll use counts. Picture 10,000 people. One in 100 have the condition, so 100 do and 9,900 don't. Of the 100, the test catches 99. Of the 9,900 healthy people, 5 in 100 get a false positive, which is 495 people. So there are 99 plus 495, which is 594 positive results, and only 99 of them are real. 99 over 594 is exactly one in six, about 17 percent. That's Bayes' theorem: the probability of having it given a positive equals the probability of a positive given you have it, times the base rate, divided by the overall probability of a positive. The lesson is that when something is rare, even a good test produces mostly false alarms, which is why a positive screening result is usually followed by a second, different test."

"Simple random sampling gives every unit the same chance of being picked. Stratified sampling splits the population into groups, like regions or plan types, and samples within each, so small groups are sure to show up and the estimate is usually tighter. Cluster sampling picks whole groups, like a few stores, and measures everyone in them. It's cheaper, but less precise, because people in one store tend to be alike. Systematic sampling takes every tenth record from a list, which is fine unless the list has a repeating pattern. Convenience sampling, whoever is easy to reach, is the one I'm wary of. For bias, selection bias is when how you pick people is linked to the answer. Non-response bias is when those who reply differ from those who don't. Survivorship bias is studying only what's left, like asking current customers why people leave. A bigger sample shrinks random error, but it doesn't fix any of these."

"The standard deviation describes the data itself: how far individual values tend to sit from the mean. The standard error describes an estimate: if I repeated the study with a new sample, how much would my sample mean bounce around. For a mean, the standard error is the standard deviation divided by the square root of n. Say order values have a standard deviation of 20 and I sample 100 orders. The standard error of the mean is 20 over 10, which is 2. With 400 orders it's 20 over 20, which is 1. So more data doesn't make customers less varied, the SD stays about 20, but it makes my estimate of the average more precise. When I'm describing a population I report the SD. When I'm saying how sure I am about an average, I report the SE or better, a confidence interval."

"The central limit theorem says that if I take independent samples from almost any distribution with a finite variance and compute the mean, the distribution of that mean gets closer and closer to normal as the sample size grows. It's centred on the true mean, and its spread is the population standard deviation over root n. A single die roll is flat, each face equally likely. But if I average 30 rolls and repeat that many times, the averages pile up in a bell shape around 3.5, with a standard error of about 0.31. This matters because real data like spend or session length is often skewed, yet t-tests and confidence intervals for the mean still work reasonably well with decent sample sizes. The mistake is thinking the data itself becomes normal. It doesn't. Only the average does, and very skewed data needs a larger n."

"I'd hold off, because this number describes the people who answered, not our users. First, who saw the pop-up: only people who were active and opened the feature, so anyone who tried it once and left is missing. That's selection bias. Second, who answered: pop-ups get low response rates, and the people who reply tend to feel strongly, which is voluntary response bias. If only a small share of people shown it responded, 9 in 10 of that small group says little. Third, I'd read the question itself, because 'How much do you love the new feature?' leads the answer. If we want a claim we can defend, I'd send a neutral question to a random sample of all users who had access, including lapsed ones, stratified by segment, and check it against usage data like repeat use."

"A p-value answers one question: if nothing interesting were going on, how surprising would my result be? Take a coin I think might be biased. I flip it 10 times and get 9 heads. I start by assuming the coin is fair. Under that assumption, the chance of 9 or more heads is 11 out of 1,024, and counting the equally extreme tails side too, about 2 in 100. That's the p-value, about 0.02. It's small, so either I saw something rare or the coin isn't fair, and I lean towards the second. What a p-value is not is the chance the coin is fair. It's calculated assuming the coin is fair. It also doesn't tell me how biased the coin is. A tiny effect can give a tiny p-value with enough data, so I always report the size of the effect next to it."

"The null hypothesis is that the true mean weight is 500 grams, and the alternative is that it's different, so it's two-sided. I'll use a 0.05 significance level, chosen before looking. Because I'm estimating the standard deviation from the sample, it's a one-sample t-test. The standard error is 6 over the square root of 36, which is 6 over 6, so 1 gram. The t statistic is 497 minus 500, over 1, which is minus 3. With 35 degrees of freedom, the two-sided cutoff at 0.05 is about 2.03, and 3 is well past it. The p-value comes out around 0.005. So I reject the null: the bags are on average lighter than claimed, by about 3 grams, with a confidence interval of roughly 495 to 499. Then I'd ask whether 3 grams matters for the contract, because significance alone doesn't answer that."

"A type I error is a false alarm: I reject the null hypothesis when it's actually true, like concluding a new feature helps when it doesn't. The chance of that, when the null is true, is alpha, often set at 0.05. A type II error is a miss: there really is an effect, but my test doesn't detect it. Its chance is beta, and power is 1 minus beta, the chance of catching a real effect. For a fixed sample size they pull against each other. If I make alpha stricter, I get fewer false alarms but miss more real effects. The only way to lower both is more data or less noise. Which is worse depends on the cost. Approving a drug that doesn't work is a costly type I error. Missing a serious defect in a safety screen is a costly type II error."

"First I'd check whether 0.3 minutes is actually unusual. I'd pull the weekly average for the last few months and look at how much it normally moves. If it swings by that much most weeks, the honest answer is that this is within normal variation, and I'd say so. If it's outside the usual range, I'd look at the median. If the median didn't move, a few very long calls pushed the mean, and I'd find out what they were. Then I'd break it down by call type, team and channel, because a change in the mix, like more complex billing calls or a batch of new starters, can raise the overall average even if nobody got slower. By the afternoon I'd give my manager a short answer: whether it's real, the likely driver, and what I'm still checking."

"First the standard error: 10 over the square root of 100, which is 1. For 95 percent I multiply by about 1.96, so the margin is about 1.96. The interval is 50 plus or minus 1.96, roughly 48 to 52. Strictly, since the standard deviation comes from the sample I'd use a t value, but with 99 degrees of freedom it's about 1.98, so the answer barely changes. Now the meaning. If I repeated the whole study many times and built an interval each time, about 95 out of 100 of those intervals would contain the true mean. For this one interval, the true mean is either in it or not. I can't attach a 95 percent probability to it in the classical sense. In practice I'd say we're fairly confident the true mean is between about 48 and 52, and that's a lot more useful than a bare p-value."

"The estimate is 200 over 1,000, so 0.2. The standard error for a proportion is the square root of 0.2 times 0.8, over 1,000. 0.2 times 0.8 is 0.16, divided by 1,000 is 0.00016, and the square root of that is about 0.0126. Times 1.96 gives a margin of about 0.025. So the interval runs from about 0.175 to 0.225, or roughly 17.5 to 22.5 percent. The normal approximation is fine here because there are plenty of clicks and non-clicks, 200 and 800. To make it narrower, the main lever is more data, but width shrinks with the square root of n, so to halve the width I need four times as many users, 4,000. I could also accept a lower confidence level, like 90 percent, but that's trading away certainty rather than gaining precision."

"In a previous role, a sales lead asked me how many extra renewals a new onboarding call would bring next quarter, for a planning meeting. From a pilot, my best estimate was about 40 extra renewals, but the confidence interval ran from roughly 10 to 70, because the pilot was small. I knew if I gave only 40 it would go into a plan as a promise. So I put it in his terms: we're fairly sure it helps, somewhere between a small gain and a big one, and 40 is the middle. Then I tied it to his decision. Even the low end covered the cost of the calls, so rolling out was safe, but I suggested planning around 20, not 40. He used that, and the actual number came in near 30. I've kept that approach: range in plain words, then what it means for the decision."

"I start with two questions: what type of outcome do I have, and how many groups am I comparing. If the outcome is a number and I have two groups, like average checkout time on the old page versus the new page, that's a t-test. If the same people are measured twice, before and after training, I use a paired t-test on the differences. If it's a number across three or more groups, like average delivery time from four warehouses, that's one-way ANOVA. If the outcome is a category, I'm working with counts, so chi-square. For example, is the choice of plan, basic or premium, related to which region a customer is in? A chi-square test of independence on that table answers it. I'd also check the assumptions, like independence, rough normality of the means for t-tests, and big enough expected counts for chi-square."

"First the table. A has 30 converted and 20 not, B has 20 converted and 30 not. Each row totals 50, and overall 50 converted and 50 didn't, out of 100. The expected count for each cell is row total times column total over 100, so 50 times 50 over 100, which is 25 in every cell. Every observed count is 5 away from 25. Five squared is 25, over the expected 25 gives 1, and there are four cells, so the chi-square statistic is 4. For a two-by-two table the degrees of freedom are 1 times 1, which is 1. The critical value at 0.05 is 3.84, so 4 just clears it, a p-value of about 0.046. I'd report that A converted better, 60 versus 40 percent, but I'd call the evidence modest, because it's close to the line and the sample is small."

"With four teams there are six possible pairs. If I run six t-tests at 0.05 each, the chance of at least one false positive is much higher than 5 in 100. If the tests were independent, it would be 1 minus 0.95 to the sixth, about 0.26, roughly one in four. ANOVA avoids that by asking one question: is there more variation between the team averages than I'd expect from the variation within teams? That ratio is the F statistic. If F is large, at least one team's mean differs. But ANOVA doesn't tell me which one, so I follow up with a post-hoc test like Tukey's, which compares the pairs while controlling the overall error rate. I'd also check the assumptions: independent observations, roughly normal residuals and similar variances, and if variances differ a lot, use Welch's ANOVA."

"With 12 per group and a heavy right tail, I wouldn't trust a plain t-test. The central limit theorem hasn't had enough data to make the means close to normal, and a couple of very slow users can dominate the result. I have three options. Task times are positive and right-skewed, so a log transform often makes them roughly symmetric, and then a t-test on the logs compares typical times on a ratio scale. Second, a Mann-Whitney U test works on ranks, so the slow users can't dominate. It tests whether values in one group tend to be larger than in the other, which is slightly different from comparing means. Third, a permutation test or a bootstrap interval for the difference in medians, which makes very few assumptions. Whatever I pick, I'd decide before looking at results and show the raw data, because with 12 users a chart says a lot."

"In my final-year project I compared satisfaction ratings on a one-to-five scale across three campus services. My first thought was one-way ANOVA on the average rating. When I plotted the answers, though, most were 4s and 5s, with a small cluster of 1s, so the data was lumpy and bounded, and averaging an ordinal scale was hiding that. I switched to a Kruskal-Wallis test, which works on ranks, and I also compared the share of people giving 1 or 2 across services with a chi-square test, because that's the number the service managers actually cared about. The overall ranking of services stayed the same, but the chi-square showed one service had far more very unhappy users, which the averages had smoothed over. What stuck with me is plotting the raw responses before choosing any test."

"Not from that alone. The two groups weren't randomly assigned, they chose. People who turn on notifications are probably already more engaged, and engagement drives retention on its own, so that's a confounder. There may also be reverse causation: someone who already plans to use the app for months is the kind of person who bothers to set notifications up. To find out whether notifications cause people to stay, the cleanest way is an experiment. Randomly pick half of new users to see a prompt to turn on notifications, and compare retention between the prompted and not-prompted groups, not between users who turned them on and those who didn't. If we can't experiment, I'd at least compare users who looked similar before the choice, same sign-up week, same early activity, and be clear that it's still weaker evidence."

"A Pearson correlation of 0.8 says the two variables have a strong positive straight-line relationship: when one is above its average, the other tends to be above its average too. If I square it I get 0.64, so a simple straight-line fit explains about 64 percent of the variation in one variable from the other. What it doesn't tell me is the slope, so it says nothing about how much one changes per unit of the other. It doesn't say anything about cause. It can also be misleading: one extreme point can produce a high correlation on its own, and a strong curved relationship, like a U shape, can have a correlation near zero. So I always look at a scatter plot first. If the relationship is monotonic but not straight, or the data is ranked, I'd use Spearman's correlation instead."

"It's Simpson's paradox, and it happens when the groups are mixed differently. Say A treated 87 small cases and cured 81, about 93 percent, and 263 large cases and cured 192, about 73 percent. B treated 270 small cases and cured 234, about 87 percent, and 80 large cases and cured 55, about 69 percent. A wins both. But overall A cured 273 of 350, about 78 percent, and B cured 289 of 350, about 83 percent. The reason is that A was given mostly large, hard cases, and B mostly easy ones. Case size affects both which treatment was chosen and the chance of success, so it's a confounder, and the overall number mixes it in. Here I'd trust the split comparison. The general habit I keep is checking whether the mix of segments differs before comparing overall rates."

"Four things drive it: the baseline rate, the smallest lift worth detecting, the significance level and the power. Here the baseline is 0.10 and the lift is one percentage point, so the effect is 0.01. With a 0.05 two-sided level and 80 percent power, a handy rule of thumb is about 16 times the variance divided by the effect squared, per group. For a rate of 0.1 the variance is 0.1 times 0.9, which is 0.09. So it's 16 times 0.09, which is 1.44, divided by 0.0001, giving about 14,400 users per group. A more exact formula lands a little higher, near 15,000, so I'd plan for about 15,000 in each group, around 30,000 in total. The key intuition is the effect is squared in the denominator. Detecting half a point instead of a full point needs four times as many users."

"My first reaction is that one hit out of 20 is exactly what chance would produce. If nothing real changed, each metric still has a 5 in 100 chance of looking significant, so across 20 I'd expect about one false positive. The chance of at least one is 1 minus 0.95 to the 20th, about 0.64, so it's more likely than not. Unless that metric was the one we chose as primary before the test, I wouldn't treat it as a finding. If I need to test many metrics formally, a Bonferroni correction divides 0.05 by 20, so each one needs p below 0.0025. That's strict, so for many metrics I'd rather use a false discovery rate method like Benjamini-Hochberg. The better habit is to name one primary metric up front, treat the rest as guardrails, and treat a surprise like this as a hypothesis for a new test."

"I'd say no to relabelling it, but I'd help with the decision. The 0.05 line was agreed before the test, so moving it after seeing the result means our error rate isn't what we said it was, and next time the line moves again. But 0.06 doesn't mean nothing happened either. I'd show the estimated lift and its confidence interval, which probably runs from slightly negative to clearly positive. That's weak evidence of a positive effect. Then I'd ask what the decision really needs. If the change is cheap, low-risk and easy to roll back, shipping it can be a reasonable business call, as long as the write-up says the test was inconclusive, not that it won. If it's costly or hard to undo, I'd suggest a properly planned follow-up test with enough users, rather than extending this one because it came close."