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kionalynum · Apr 15, 2011, 09:46 AM

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.

galactus · Apr 15, 2011, 11:41 AM

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.