## Elementary Statistics (12th Edition)

We know that $α=0.05$, hence we know that $t_{α/2}=2.201$. Also, using a standard deviation calculator, we can find that the standard deviation of the 150 AD data set is $5.02$. Hence the error is: $\frac{2.201\cdot5.02}{\sqrt n}=\frac{2.201\cdot5.02}{\sqrt{12}}=3.1896$ We do the same thing for the 4000 BC data set. We know that $α=0.05$, hence we know that $t_{α/2}=2.201$. Also, using a standard deviation calculator, we can find that the standard deviation of the 4000 BC data set is $4.64$. Hence the error is: $\frac{2.201\cdot4.64}{\sqrt n}=\frac{2.201\cdot5.02}{\sqrt{12}}=3.1896$ Then, using the means as the center of the data set and the errors to determine the minimums and maximums, we can see that the data overlap, meaning that we cannot say that the lengths of skull breadths have changed over the time.