The Ztest is used to compare means of two distributions with known variance. One sample Ztests 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. Twosample Ztests are more appropriate for comparing the means of two samples of data.
Requirements for the Ztest:
•

The mean and standard deviation of the population distribution are known.

•

The mean of the sample distribution is known.

•

The variance of the sample is assumed to be the same as the population.

•

The population is assumed to be normally distributed

•

The population size is over 30

In cases where the population variance is unknown, or the sample size is less than 30, the Student's ttest
may be more appropriate.
To calculate a Ztest 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 Ztest has a critical value. For example, the significance level α = 0.01, has a critical value of 2.326. If the Ztest 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 $p0.05$, 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: H_{0}: $xequals;\mathrm{mu;}$
Alternative Hypothesis: H_{A}_{: }$x\ne \mathrm{\μ}$
To determine the zvalue, first calculate the standard error. You can do this using the formula from the above section:
$\mathrm{SE}\=\frac{30}{\sqrt{50}}$,
$\mathrm{SE}\=4.242$.
Plug in the known values and the standard error to calculate the zvalue:
$z\=\frac{x\mu}{\mathrm{SE}}$,
$z\=\frac{118120}{4.242}$,
$z\=0.471$.
Now you can look up the probability: $P\left(z<0.471\right)$ on a probability table:
$P\left(z<0.471\right)\=0.3192$.
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.
