Sample Population Variance Quiz: Sample vs. Population Variance

The mean equals (5 + 10 + 15) ÷ 3 = 10. The deviations are −5, 0, and 5. Squaring each gives 25, 0, and 25. The sum of squared deviations is 50. Dividing by N = 3 gives population variance = 50 ÷ 3 = 16.67.

The inspector has a sample of n = 8 items drawn from a much larger batch. Sample variance applies, and the correct denominator is n − 1 = 8 − 1 = 7. Option A (8) would give the population variance formula applied to sample data, which underestimates the population variance. Option B (9) exceeds the sample size and has no valid justification. Option D (2,000) is the full batch size and would be used only if all 2,000 items were measured.

The mean equals (3 + 7 + 7 + 7) ÷ 4 = 24 ÷ 4 = 6. The deviations are −3, 1, 1, and 1. Squaring each gives 9, 1, 1, and 1. The sum of squared deviations is 12. Dividing by N = 4 gives population variance = 12 ÷ 4 = 3.

The answer is False. Dividing by n − 1 uses fewer degrees of freedom than dividing by n, not more. The degrees of freedom equal n − 1, which is one less than n. One degree of freedom is lost because the sample mean x̄ is estimated from the same data, constraining the deviations so they always sum to zero. This constraint leaves only n − 1 independent deviations available to estimate the spread.

The researcher has data for every member of the cycling team, which constitutes the entire population of interest. Population variance is appropriate, and the correct denominator is N = 12. Option A incorrectly applies the sample formula. Option C divides by n = 12 but uses sample notation, which is inconsistent. Option D applies Bessel's correction to population variance, which is unnecessary when the full population is available.

The mean equals 6 and the sum of squared deviations is 72, as computed for the population variance of the same dataset. For sample variance, divide by n − 1 = 3. Sample variance = 72 ÷ 3 = 24. The sample variance of 24 is larger than the population variance of 18, consistent with the general rule that sample variance always exceeds population variance for the same dataset.

Option A is correct: the only computational difference between the two formulas is the denominator — N for population and n − 1 for sample. Option C is correct: dividing by n − 1 rather than n is specifically called Bessel's correction, and it makes sample variance an unbiased estimator of the population variance. Option B is incorrect: sample variance is always larger than population variance for the same data because n − 1 is smaller than n. Option D is incorrect and reverses the notation: population variance uses the true population mean μ, while sample variance uses the sample mean x̄.

The mean equals (2 + 2 + 8 + 12) ÷ 4 = 24 ÷ 4 = 6. The deviations are −4, −4, 2, and 6. Squaring each gives 16, 16, 4, and 36. The sum of squared deviations is 72. Dividing by N = 4 gives population variance = 72 ÷ 4 = 18.

The answer is True. The researcher has a sample of 200 from a much larger population. Because the data does not include every member of the population, sample variance applies. The sample variance formula divides by n − 1 = 199, applying Bessel's correction to produce an unbiased estimate of the true population variance. Dividing by n = 200 instead would systematically underestimate the population variance.

The mean equals 10 and the sum of squared deviations is 50, as computed for the population variance of the same dataset. For sample variance, divide by n − 1 = 2. Sample variance = 50 ÷ 2 = 25. The sample variance of 25 is larger than the population variance of 16.67 because dividing by a smaller denominator produces a larger result.

Population variance uses the true population mean μ and divides by N, the total number of values in the population. Options A and B both use s² and x̄, which are sample notation. Option D uses N − 1, which is Bessel's correction applied to samples, not populations. Only option C correctly uses population notation and divides by N.

The term n − 1 is called the degrees of freedom. It represents the number of independent pieces of information available after estimating the sample mean x̄ from the data. Because x̄ is estimated from the same dataset, one degree of freedom is used up, leaving n − 1 free deviations that carry independent information about the spread. Dividing by n − 1 rather than n corrects the bias introduced by this constraint.

The mean equals 2 and the sum of squared deviations is 24, as computed for the population variance of the same dataset. For sample variance, divide by n − 1 = 2. Sample variance = 24 ÷ 2 = 12. Comparing to the population variance of 8, the sample variance of 12 is larger because dividing by 2 instead of 3 produces a greater result.

The answer is False. Sample variance is an unbiased estimator of population variance, meaning it is correct on average across many samples. However any single sample variance calculation is subject to sampling variability and will typically differ from the true population variance. Increasing sample size reduces this variability and brings the estimate closer to the true value, but no threshold such as n greater than 30 guarantees an exact estimate.

The mean equals (0 + 0 + 6) ÷ 3 = 2. The deviations are −2, −2, and 4. Squaring each gives 4, 4, and 16. The sum of squared deviations is 24. Dividing by N = 3 gives population variance = 24 ÷ 3 = 8. The single large value of 6 contributes the largest squared deviation and drives the variance upward.

Option A requires population variance because the dataset includes every employee at the company — no one is excluded. Option C requires population variance because all 15 specimens in the petri dish are measured — the petri dish is the defined population. Option B is a random subset of a larger hospital population, so sample variance applies. Option D is a random subset of a much larger city population, so sample variance applies.

The mean equals 10 and the sum of squared deviations is 80, as computed for the population variance of the same dataset. For sample variance, divide by n − 1 = 3. Sample variance = 80 ÷ 3 = 26.67. The sample variance is larger than the population variance of 20 because dividing by 3 instead of 4 produces a greater result.

The mean equals (4 + 8 + 12 + 16) ÷ 4 = 40 ÷ 4 = 10. The deviations are −6, −2, 2, and 6. Squaring each gives 36, 4, 4, and 36. The sum of squared deviations is 80. Dividing by N = 4 gives population variance = 80 ÷ 4 = 20.

The answer is True. The only difference between sample variance and population variance for the same dataset is the denominator: sample variance divides by n − 1 while population variance divides by n. When n is very large, n − 1 is negligibly different from n, so the two values converge. For example with n equals 10,000, dividing by 9,999 versus 10,000 produces results that differ by less than 0.01 percent.

Sample variance uses the sample mean x̄ and divides by n − 1, applying Bessel's correction to produce an unbiased estimate of the population variance. Option C uses correct sample notation but divides by n rather than n − 1, which would underestimate the population variance on average. Options A and B use population notation σ² and μ, and are not sample variance formulas.