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 30 hours. The population standard deviation is two hours.

What is the 95% confidence interval for the mean?

Answer 1

D          The confidence limits are 30 plus or minus 1.96(2/square root of 1000) =29.88 and 30.12 

Answer 2

Conf interval = (29.876, 30.124)

Explanation:

Given that 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 30 hours.

Also population std deviation sigma = 2 hours

For 95% confidence interval z critical value is

±1.96

Margin of error = Std error x Z critical

Std error = std dev/sq rt of n where n = sample size = 1000

Hence std error = 2/√1000=0.0632

Margin of error = ±1.96(0.0632)=±0.1240

Confidence interval lower bound = 30-0.124=29.876

Upper bound = 30+0.124 = 30.124

Conf interval = (29.876, 30.124)