Ask your own question, for FREE!
Mathematics 18 Online
OpenStudy (anonymous):

I WILL GIVE MEDAL AND FAN! A random sample of 125 students is chosen from a population of 4,000 students. If the mean IQ in the sample is 100 with a standard deviation of 8, what is the 98% confidence interval for the students' mean IQ score?

OpenStudy (anonymous):

100−120 98.33−101.67 92−108 108.66−111.34

OpenStudy (anonymous):

@kirbykirby

OpenStudy (kirbykirby):

You will need to use: \[\large \left(\bar{x} -z\frac{s}{\sqrt{n}},\bar{x}+z\frac{s}{\sqrt{n}}\right)\] where \(\bar{x}\) is the mean, \(s\) is the standard deviation and \(z\) is the critical z value. We can use a \(z\)-value since your sample is large (\(\ge 30\)). Since a confidence interval of 98% means there is 1% on each side of the distribution. SO you will need to look a the standard normal distribution at a 0.99, which is about 2.33

OpenStudy (kirbykirby):

All you need to do is substitute the values you are given and noticing that \(n\) is the number of students in the sample (not in the population).

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!