Generate a 95% confidence interval for the mean if the standard deviation for the population is known as 4. The sample consists of 30 measurements with a mean 84. Round to the nearest hundreths.

The margin of error is E=z\cdot \frac{\sigma}{\sqrt{n}}

.95 CI corresponds to a z score of 1.96

So, we have E=1.96\cdot \frac{4}{\sqrt{30}}=1.43

The CI is then:

\overline{x}-E<\mu<\overline{x}+E

where \overline{x}=84

Calculate the CI. This tells us that with 95% confidence the mean for the poopulation is between those values.