HW #10 problem #3b
Suppose that 49 students
Find a 97% Confidence Interval for the true mean score of this class on Exam 1.
mean = 76.857143
std = 10.61445555
se =1.516351
me = 2.85073988
[m-me,m+me] = [74.00640312, 79.70788288]
please help.
Comments
Alright, the first thing you do is find the z* score. This can be done by doing 1-0.97 to get 0.03.
Then multiply that by 0.5 to get 0.015, which is what you'll look for within the normal probability table. This will give you z*=-2.17, but for the purposes of this problem the negative is unnecessary.
Did you multiply the correct z (star)-value? I did 2.17 times (std/sqrt(49)) and got a different answer compared to your ME. That might be where you went wrong. If putting them into a calculator, I would solve for SE, then multiply the answer you get by 2.17.
Hope this helps!