One Sample Z Test is used to test if the sample studied follows a normal distribution, assuming that the standard deviation and mean are known. The mean is equal to the test value based on the sample drawn.

Cases where standard deviations are known are rare; a more common method of test in that case is the One Sample T Test.

Requirements for using One Sample Z Test:

Here, the goal is to test the hypothesis that the population where the sample is drawn from follows a normal distribution with the mean equal to the value that is set.

The population studied is assumed to follow a normal distribution.

The standard deviation of the population is already known.

The formula is:

where $X$ is the sample, ${\mathrm{\mu }}_0$ is the test value of mean, $\mathrm{\sigma }$ is The known standard deviation, and $N$ is the sample size.

When the sample size, $N$, is sufficiently large, $Z$ is approximately $\mathrm{Normal}\left(0\,1\right)$.

Determine the null hypothesis:

Null Hypothesis: ${\mathrm{\mu }}_0=64.5$ (the actual mean).

Substitute the information into the formula:

$z=\frac{\left(65-64.5\right)}{\left(\frac{15}{\sqrt{100}}\right)}$ = 0.333333

Compute the p-value:

$p-\mathrm{value}=\mathrm{Probability}\left(\right|Z|>0.333333)=\mathrm{Probability}\left(Z<-0.333333\right)+\mathrm{Probability}\left(Z>0.333333\right)$= 0.738883 $Z˜\mathrm{Normal}\left(0,1\right)$.

Draw the conclusion:

This statistical test does not provide enough evidence to conclude that the null hypothesis is false, so we fail to reject the null hypothesis.