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Choose the right test for a hypothesis concerning a mean

CFA® Level I · Quantitative Methods · Estimation and Hypothesis Testing · QM.7.c

The idea

The choice follows from what is known and how large the sample is. With a known population variance and a normal population, a z-test applies. With an unknown variance — the usual case — a t-test applies, using the sample standard deviation and n - 1 degrees of freedom. For large samples the two converge, because the t distribution approaches the normal. Comparing the means of two independent samples uses a two-sample t-test; comparing paired observations, such as the same portfolios before and after a change, uses a paired test on the differences.

Formulas

Test statistic for a hypothesis about a mean

t = (x̄ − µ₀) ÷ (s ÷ √n)

x̄
the sample mean
µ₀
the mean the null hypothesis asserts
s
the sample standard deviation
n
the sample size

The t-distribution when the population standard deviation is unknown, which is almost always. The denominator is the standard error, so the statistic is the distance from the hypothesis measured in standard errors, with n − 1 degrees of freedom.

A practice question

From the question bank, with the reason each wrong option is wrong.

A sample of 20 observations is drawn from a normal population whose standard deviation is unknown. Which test applies to its mean?

  1. A z-testWhy not: That requires the population standard deviation to be known. Here it is estimated from 20 observations, and the t distribution's fatter tails price in that extra uncertainty.
  2. A t-test — correct
  3. A chi-square testWhy not: That is for inference about a variance, not a mean.

An estimated standard deviation adds uncertainty, which the t distribution's fatter tails account for. A z-test would require the population standard deviation to be known.

A harder one

Why do t-tests and z-tests give nearly identical results for large samples?

  1. Because large samples are always normally distributedWhy not: The data keep the population's shape however many observations are drawn. This is about the distribution of the test statistic, not the data.
  2. Because the sample standard deviation equals the population value at large nWhy not: It approaches it but never equals it, and the convergence of the two tests does not require equality — only that the remaining uncertainty stops mattering.
  3. Because the t distribution converges on the normal as degrees of freedom rise — correct

With many degrees of freedom the extra uncertainty from estimating the standard deviation becomes negligible and the t distribution's tails thin toward the normal's.

The hardest tier

An analyst compares the same set of portfolios before and after a strategy change. Which test is appropriate, and why?

  1. A paired test on the differences, because the two samples are not independent — correct
  2. A two-sample t-test, because there are two sets of observationsWhy not: Two sets, but not independent ones — each portfolio appears in both. Treating them as independent leaves the portfolio-to-portfolio variation in the noise.
  3. A chi-square test, because the data are relatedWhy not: Chi-square tests variances and categorical counts. Relatedness between observations calls for pairing, not a different family of test.

Each portfolio appears in both samples, so the observations are linked. Testing the differences removes the portfolio-to-portfolio variation and is far more powerful than treating the samples as independent.

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