A study of one thousand teens found that the number of hours they spend on social networking sites each week is normally distributed with a mean of 15 hours. The population standard deviation is 2 hours. What is the 95% confidence interval for the mean?

14.88−15 hours

14.88−15.12 hours

15−15.12 hours

14.76−14.24 hours

Question

A study of one thousand teens found that the number of hours they spend on social networking sites each week is normally distributed with a mean of 15 hours. The population standard deviation is 2 hours. What is the 95% confidence interval for the mean? 

14.88−15 hours 

 14.88−15.12 hours 

 15−15.12 hours 

 14.76−14.24 hours

anonymous · Answer

there is really only one answer here that makes sense.

anonymous · Answer

Given 
ConfidenceTarget = 0.95  Then Z = 1.96
StdDev                  = 2
Mean                    = 15
SampleSize           = 1000

Mean +/- ((confidenceTarget/2 + 1/2)*StdDev) / SQRT(sampleSize)


The standard Z for a 95% confidence target is 1.96 (there is a formula to calculate this, but being 95% you can memorize them (.90 = 1.645, .95=1.96,.99=2.576) because they come up a lot.

First Multiply this by your standard deviation 1.96 * 2 = 3.92

Then Take the square root of the sample size SQRT(1000) = (31.623)

Divide these two 3.92/31.623 = 0.12396

Now add and subtract them from the mean and round off

15 +/- 0.12396

cannaaa · Answer

ohhh ok, and the equation that you're using is the margin of error for population mean right? so the answer would be B

anonymous · Answer

correct

cannaaa · Answer

ok, thank you!!