Z-Test Statistic
Tests whether a sample mean differs from a hypothesized value.
Formula
Variables
Example
Compare to critical z
How it works
The z-test statistic measures how far a sample mean is from a hypothesized population mean, but it scales that difference by the standard error so the result is in standard-deviation units. This works because, when the population standard deviation is known and the sampling conditions are appropriate, the sampling distribution of the mean is approximately normal with spread \(\sigma/\sqrt{n}\). A large positive or negative z-value means the observed mean would be unlikely if the hypothesized mean were true.
When to use it
Use this when testing a claim about a population mean and the population standard deviation is known, or when the sample size is large enough for a normal approximation to be reliable. Common settings include quality control, manufacturing measurements, standardized test scores, and other measurement-based problems.
Worked example, step by step
A factory claims the fill volume of a drink is 500 mL. A quality inspector knows the population standard deviation is 12 mL and takes a sample of 36 bottles with mean fill volume 495 mL. Step 1: Identify the values: \(\bar{x} = 495\), \(\mu_0 = 500\), \(\sigma = 12\), \(n = 36\). Step 2: Compute the standard error: \(\sigma/\sqrt{n} = 12/\sqrt{36} = 12/6 = 2\). Step 3: Plug into the formula: \(z = (\bar{x} - \mu_0)/(\sigma/\sqrt{n}) = (495 - 500)/2 = -5/2\). Step 4: Simplify: \(z = -2.5\). Final answer: The sample mean is 2.5 standard errors below the hypothesized mean, so the z-test statistic is \(-2.5\).
Common mistakes to avoid
- Using the sample standard deviation \(s\) in place of the known population standard deviation \(\sigma\) without switching to a t-test
- Forgetting to divide by \(\sqrt{n}\), which makes the denominator too large and the z-value too small in magnitude
- Reversing the subtraction order and reporting \((\mu_0 - \bar{x})/(\sigma/\sqrt{n})\) without noticing the sign change
Did You Know?
The z-test assumes a known population standard deviation, a luxury so rare that the t-test usually replaces it.
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