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Statistics · Hypothesis testing

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Hypothesis testing for a normal mean

Testing a claim about the mean μ of a normal population from a sample, using that the sample mean is N(μ, σ²/n) — forming the test statistic z = (x̄ − μ₀)/(σ/√n) and comparing it to the critical z-value.

Statistics · Hypothesis testing

Hypothesis testing for a normal mean

Testing a claim about the mean μ of a normal population from a sample, using that the sample mean is N(μ, σ²/n) — forming the test statistic z = (x̄ − μ₀)/(σ/√n) and comparing it to the critical z-value.

Why it works

If individual values come from N(μ,σ2)N(\mu, \sigma^2), then the mean of a sample of nn of them is itself normal — and less spread out, because averaging cancels out the extremes: Xˉ∼N ⁣(μ, σ2n),standard error σn.\bar{X} \sim N\!\left(\mu,\ \frac{\sigma^2}{n}\right), \qquad \text{standard error } \frac{\sigma}{\sqrt{n}}. The standard deviation of the sample mean, σn\dfrac{\sigma}{\sqrt n}, shrinks as nn grows — bigger samples give means clustered more tightly around μ\mu. This is what makes a sample mean good evidence about μ\mu.

The test. To test a claimed mean μ0\mu_0, set H0:μ=μ0,H_0: \mu = \mu_0, with H1:μ>μ0H_1: \mu > \mu_0, μ<μ0\mu < \mu_0 (one-tailed) or μ≠μ0\mu \ne \mu_0 (two-tailed). Under H0H_0 the sample mean is N ⁣(μ0,σ2n)N\!\left(\mu_0, \tfrac{\sigma^2}{n}\right), so the test statistic measures how many standard errors the observed mean xˉ\bar{x} is from μ0\mu_0: z=xˉ−μ0σ/n.z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}}. Compare zz to the critical value:
  • one-tailed at 5%5\%: reject if z>1.6449z > 1.6449 (upper) or z<−1.6449z < -1.6449 (lower);
  • two-tailed at 5%5\%: reject if ∣z∣>1.96|z| > 1.96;
  • one-tailed at 1%1\%: 2.32632.3263; two-tailed at 1%1\%: 2.57582.5758.
If zz falls in the rejection region, reject H0H_0: the sample mean is too far from μ0\mu_0 to be chance. Finish with a conclusion in context. (Equivalently you can compare the pp-value P(Z>z)P(Z > z) with the significance level.)