The Z-test is used to compare means of two distributions with known variance. One sample Z-tests are useful when a sample is being compared to a population, such as testing the hypothesis that the distribution of the test statistic follows a normal distribution. Two-sample Z-tests are more appropriate for comparing the means of two samples of data.

Requirements for the Z-test:

In cases where the population variance is unknown, or the sample size is less than 30, the Student's t-test  may be more appropriate.

To calculate a Z-test statistic, the following formula can be used:

z = $\frac{x\mathit{-}\mathrm{\μ}}{\mathrm{SE}}$,

z = $\frac{x-\mu }{\frac{\sigma }{\sqrt{n}}}$,

where x is the sample mean, m is the population mean, and SE is the standard error, which can be calculated using the following formula:

SE = $\frac{\mathrm{\σ}}{\sqrt{n}}$,

where s is the population standard deviation and n is the sample size.

For each significance level, α, the Z-test has a critical value. For example, the significance level α = 0.01, has a critical value of 2.326. If the Z-test statistic is greater than this critical value, this may provide evidence for rejecting the null hypothesis.

Example

A company produces metal discs with a mean weight of 120 g and standard deviation of 30 g. Suppose that a sample of size 50 has the mean weight of 118 g. Assuming a significance level of $p < 0.05$, is the company correct in accepting the null hypothesis that the sample does not have different weights on average than the population of metal discs?   To start, you can state the null and alternative hypotheses:   Null Hypothesis: H0:  $x \&equals; \mathrm{\&mu;}$ Alternative Hypothesis: HA: $x\ne \mathrm{\&mu;}$   To determine the z-value, first calculate the standard error. You can do this using the formula from the above section:   $\mathrm{SE}\&equals;\frac{30}{\sqrt{50}}$, $\mathrm{SE}\&equals;4.242$.   Plug in the known values and the standard error to calculate the z-value:   $z\&equals;\frac{x-\mu }{\mathrm{SE}}$, $z\&equals;\frac{118-120}{4.242}$, $z\&equals;-0.471$. Now you can look up the probability: $P\left(z<-0.471\right)$ on a probability table:      $P\left(z<-0.471\right)\&equals;0.3192$.   The result, 0.3192, is greater than P = 0.05, and so you cannot reject the null hypothesis that the sample mean is significantly different than the metal disc population mean.