Two Sample Z Test is used to test if the difference in the means of two samples collected from two populations, each assumed to be normally distributed, is equal to the difference of the means of the two populations.

To use this test, the standard deviations of the two populations must be known. However, since the cases where the standard deviations are provided are rare, Two Sample T Test is more often used to test for the difference between means.

Requirements for using Two Sample Z Test:

The two populations studied are assumed to be normally distributed.

The two standard deviations of the two populations are known.

The formula is:

where $X$ is the sample drawn from the first population, $Y$ is the sample drawn from the second population, $\mathrm{\beta }$ is the test value of the difference, ${\mathrm{\sigma }}_1$ is the known standard deviation of the first population, ${\mathrm{\sigma }}_2$ is the known standard deviation of the second population, $N_1$ is the sample size of $X$, and $N_2$ is the sample size of $Y$.

When $N_1$ and $N_2$ are sufficiently large, $Z$ is approximately $\mathrm{Normal}\left(0\,1\right)$.

Determine the null hypothesis:

Null Hypothesis: The average difference between the lifespan of dogs and cats is 5 years

Substitute the information into the formula:

$\mathrm{Mean}\left(X\right)=11.6917$,  $\mathrm{Mean}\left(Y\right)=10.2333$,  $\mathrm{\beta }=5$, ${\mathrm{\sigma }}_1=4$,  ${\mathrm{\sigma }}_2=3$,  $N_1=10$,  $N_2=15$

$z=\frac{\left(\left(11.69167-10.23333\right)-5\right)}{\sqrt{\left(\frac{4^2}{12}+\frac{3^2}{15}\right)}}=-2.54715$

Compute the p-value:

$p-\mathrm{value}=\mathrm{Probability}\left(\right|Z|>|-2.54715\left|\right)=\mathrm{Probability}\left(Z<-2.54715\right)+\mathrm{Probability}\left(Z>2.54715\right)=0.0108607$,     $Z˜\mathrm{Normal}\left(0,1\right)$

Draw the conclusion:

This statistical test provides evidence that the null hypothesis is false, so we reject the null hypothesis.