confidence intervals

Question

A random sample of 35 teenagers averaged 7.3 hours of sleep per night. Assume the population standard deviation is 1.8 hours. Calculate a 98% confidence interval for the mean. 

Answer 1

Okay, so this is a problem designed to use something called the Central Limit Theorem.  

 

In plain words, the central limit theorem is about how random variance will tend to cancel itself out as you take greater and greater size samples.  The people who sleep a lot will tend to cancel the people who sleep very little, leading to a more "average" result as the sample size increases.  If you take a sample of 100 of something, that sample will have only a tenth of the variance (amount of standard deviation) that a sample size of 1 would have.   

 

So, if the population SD of teenagers is 1.8 hours of sleep a night, and if you have 35 teenagers in your sample, then that 1.8 will be divided by the square root of 35.  (Let's assume it was 36 to make working the problem easy, so the square root is 6 and the standard deviation for collective samples of 36 teens is 1.8/6 = .3 hours.  

 

Now, in order to find how many sample SDs you need to use to get 98%, we'll have to check a chart or calculator.  You should memorize that +/- 1 SD is 68.3% (about 2/3), +/- 2 SDs is 95.4% (about 19/20), and +/- 3 SDs is 99.7%.  That means that you know the answer should be between2 and 3 sds.  

 

Off a chart, I find that the number is around 2.33 SDs.  So, we take 2.33 * 0.3 = 0.7 hours.  So we'll have a 98% confidence interval at 7.3 +/- 0.7, or, stated another way, from 6.6 to 8.0 hours for a sample size of 36 teens.

 

Now you can work it for 35, right?