The Z-Test

Standardizing a sample mean

For one value from $N\left(\mu ,{\sigma }^2\right)$N(μ, σ²), $z=\frac{x-\mu }{\sigma }$z = (x − μ)/σ counts the standard deviations between the value and the mean. A test about a mean uses the sample mean x̄ instead, and the spread of x̄ is not $\sigma$σ but the standard error $\frac{\sigma }{\sqrt{n}}$σ/√n. So the test statistic is $z=\frac{\bar{x}-\mu }{\frac{\sigma }{\sqrt{n}}}$z = (x̄ − μ)/(σ/√n): the number of standard errors between the sample mean and the mean claimed by $H_0$H₀.

Sugar is sold in bags labeled 500 g. The filling machine’s standard deviation is known from long use to be 12 g. An inspector weighs 36 bags and finds a mean of 496 g. The standard error is $\frac{12}{\sqrt{36}}=\frac{12}{6}=2\text{g}$12/√36 = 12/6 = 2 g, so $z=\frac{496-500}{2}=-\frac{4}{2}=-2$z = (496 − 500)/2 = −4/2 = −2. The sample mean is 2 standard errors below the label.

The whole test

Hypotheses: $\mu$μ is the machine’s mean fill. The question is whether it has drifted either way, so $H_0$H₀: $\mu =500$μ = 500 and $H_1$H₁: $\mu \ne 500$μ ≠ 500, a two-tailed test at 5%.

Test statistic: under $H_0$H₀, x̄ ~ $N\left(500,2^2\right)$N(500, 2²), so $z=\frac{496-500}{2}=-2$z = (496 − 500)/2 = −2.

Compare: the two-tailed critical values at 5% are $\pm 1.96$±1.96, and −2 < −1.96, so z is in the critical region. Equally, the p-value is 2 × P(Z < −2) = 2 × 0.0228 = 0.0455, which is below 0.05.

Conclusion, in context: reject $H_0$H₀. There is evidence at the 5% level that the machine’s mean fill is no longer 500 g.

If the inspector’s question had been only whether the bags are underfilled, $H_1$H₁: $\mu <500$μ < 500, the critical value would be −1.645 and the p-value P(Z < −2) = 0.0228. The conclusion is the same.

A two-tailed test at z = −1.5: p = 0.1336, so $H_0$H₀ stands. Drag z to the sugar bags’ −2: the area beyond 2 standard errors on both sides is p = 0.0455, below 0.05, and z lies beyond the critical value −1.96.

When the z-test applies

The test needs $\sigma$σ, the population standard deviation, to be known, as it is for a machine with a long record. It also needs x̄ to be normal: either the population is normal, or n is large enough for the central limit theorem to make x̄ approximately normal anyway.

A large sample sees small differences

Suppose the machine’s mean fill really is 499 g, 1 g under the label, and the sample happens to show exactly that. With 36 bags, the standard error is 2 g and $z=\frac{499-500}{2}=-0.5$z = (499 − 500)/2 = −0.5, with p-value 2 × P(Z < −0.5) = 0.6171. Nothing is detected.

With 900 bags, the standard error is $\frac{12}{\sqrt{900}}=\frac{12}{30}=0.4\text{g}$12/√900 = 12/30 = 0.4 g, and the same 1 g gives $z=\frac{499-500}{0.4}=-2.5$z = (499 − 500)/0.4 = −2.5, with p-value 0.0124. Now $H_0$H₀ is rejected at 5%.

The difference is the same; only the standard error shrank. So a significant result says the difference is unlikely to be chance, not that it is large: a huge sample can detect a difference too small to matter.

Under $H_0$H₀, the mean of 36 bags follows the wide dashed curve, $N\left(500,2^2\right)$N(500, 2²), and the mean of 900 bags the narrow gold curve, $N\left(500,0.4^2\right)$N(500, 0.4²). A mean of 499 is well inside the wide one, but 2.5 standard errors out on the narrow one.

The usual mistakes

Leaving the gap undivided. 496 − 500 = −4 is the gap in grams; z counts standard errors, so it is $-\frac{4}{2}=-2$−4/2 = −2.

Dividing by $\sigma$σ instead of the standard error. $\frac{496-500}{12}=-0.33$(496 − 500)/12 = −0.33 is the z-score of one bag; the test is about the mean of 36 bags.

Dividing $\sigma$σ by n. $\frac{12}{36}=0.33$12/36 = 0.33 is not the standard error; the variance is divided by n, so $\sigma$σ is divided by $\sqrt{n}=6$√n = 6.

Using a one-tailed p-value in a two-tailed test. With $H_1$H₁: $\mu \ne 500$μ ≠ 500, a mean as far above 500 would count just as much, so P(Z < −2) = 0.0228 is doubled to 0.0455.