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Statistics practice problems with answers

Ten exam-style statistics problems with complete worked solutions, covering descriptive statistics, the normal distribution, confidence intervals, hypothesis testing, the binomial distribution, and regression interpretation. Attempt each problem fully before opening the solution — the retrieval attempt is what makes practice testing the most effective study technique in the literature. Free to use, no signup.

Problem 1

For the sample {4,8,6,10,12}\{4, 8, 6, 10, 12\}, calculate the sample mean and the sample standard deviation.
Show worked solution
Mean: xˉ=4+8+6+10+125=405=8\bar{x} = \frac{4+8+6+10+12}{5} = \frac{40}{5} = 8. Squared deviations: (4−8)2+(8−8)2+(6−8)2+(10−8)2+(12−8)2=16+0+4+4+16=40(4-8)^2 + (8-8)^2 + (6-8)^2 + (10-8)^2 + (12-8)^2 = 16 + 0 + 4 + 4 + 16 = 40. Sample variance divides by n−1n - 1: s2=404=10s^2 = \frac{40}{4} = 10, so s=10≈3.16s = \sqrt{10} \approx 3.16.

Problem 2

Exam scores are normally distributed with mean 70 and standard deviation 8. What proportion of students score above 86?
Show worked solution
Standardize: z=86−708=2z = \frac{86 - 70}{8} = 2. From the standard normal table, P(Z>2)≈0.0228P(Z > 2) \approx 0.0228. About 2.3 percent of students score above 86.

Problem 3

A random sample of n=36n = 36 observations has mean xˉ=24\bar{x} = 24 and sample standard deviation s=6s = 6. Construct a 95 percent confidence interval for the population mean.
Show worked solution
The standard error is SE=sn=66=1SE = \frac{s}{\sqrt{n}} = \frac{6}{6} = 1. With n=36n = 36 the normal approximation is reasonable, so the interval is xˉ±1.96×SE=24±1.96\bar{x} \pm 1.96 \times SE = 24 \pm 1.96, giving (22.04, 25.96)(22.04,\ 25.96). Interpretation: the procedure that generated this interval captures the true mean in 95 percent of repeated samples — not "there is a 95 percent probability the mean is in this interval."

Problem 4

Test H0:μ=50H_0: \mu = 50 against H1:μ≠50H_1: \mu \neq 50 at the 5 percent significance level, given xˉ=53\bar{x} = 53, s=9s = 9, n=36n = 36.
Show worked solution
Test statistic: t=xˉ−μ0s/n=53−509/6=31.5=2.0t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{53 - 50}{9/6} = \frac{3}{1.5} = 2.0. With df=35df = 35, the two-sided critical value at the 5 percent level is approximately 2.032.03. Since ∣2.0∣<2.03|2.0| < 2.03, we fail to reject H0H_0 — narrowly. Note what this is not: it is not evidence that μ=50\mu = 50; the data are simply not quite strong enough to rule it out at this significance level.

Problem 5

A study reports a p-value of 0.03. Which statement is correct? (a) There is a 3 percent probability the null hypothesis is true. (b) If the null hypothesis were true, data this extreme or more would occur in about 3 percent of samples. (c) The effect has a 97 percent probability of being real.
Show worked solution
Statement (b). The p-value is a conditional probability computed assuming H0H_0 is true: it measures how surprising the observed data would be under the null. It says nothing direct about the probability that the null is true or that the effect is real — those statements require a prior and a Bayesian framework. Confusing (a) for (b) is among the most common errors in applied statistics.

Problem 6

A factory tests whether a batch of components meets specification. Define Type I and Type II errors in this context, and state which error the significance level α\alpha controls.
Show worked solution
Type I error: rejecting a batch that actually meets specification (a false alarm). Type II error: accepting a batch that is actually defective (a miss). The significance level α\alpha is the probability of a Type I error the test tolerates; the probability of a Type II error is β\beta, and power =1−β= 1 - \beta. Lowering α\alpha without changing anything else raises β\beta — the two errors trade off.

Problem 7

A fair production process produces defective items with probability p=0.2p = 0.2. In a random sample of 10 items, what is the probability of exactly 2 defects?
Show worked solution
Binomial with n=10n = 10, p=0.2p = 0.2: P(X=2)=(102)(0.2)2(0.8)8=45×0.04×0.1678≈0.302P(X = 2) = \binom{10}{2}(0.2)^2(0.8)^8 = 45 \times 0.04 \times 0.1678 \approx 0.302. There is roughly a 30 percent chance of exactly two defective items.

Problem 8

A regression of exam score (yy) on hours studied (xx) gives y^=12+0.8x\hat{y} = 12 + 0.8x. Interpret the slope coefficient, and predict the score for a student who studies 10 hours.
Show worked solution
The slope means each additional hour of study is associated with a 0.8-point higher predicted exam score, on average. Prediction at x=10x = 10: y^=12+0.8×10=20\hat{y} = 12 + 0.8 \times 10 = 20. Two cautions: the relationship is an association, not proven causation, and predictions far outside the observed range of xx (extrapolation) are unreliable.

Problem 9

The correlation between monthly ice cream sales and monthly drowning deaths is r=0.85r = 0.85. Does ice cream cause drowning? Explain the statistical issue.
Show worked solution
No — this is the classic confounding example. A third variable, summer weather, raises both swimming (hence drownings) and ice cream sales. Correlation measures linear association, not causation; a strong rr is fully consistent with no causal link between the correlated variables. Establishing causality requires experimental design or causal-inference methods that control for confounders.

Problem 10

Explain why the standard error of the sample mean is SE=σnSE = \frac{\sigma}{\sqrt{n}}, and compute how the standard error changes when the sample size increases from 25 to 100.
Show worked solution
Averaging reduces noise: independent sampling errors partially cancel, and the variance of the mean is σ2/n\sigma^2 / n, so its standard deviation is σ/n\sigma/\sqrt{n}. Increasing nn from 25 to 100 multiplies n\sqrt{n} by 2, halving the standard error. The square root is the practical headache: each halving of the standard error requires quadrupling the sample size.

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