Title: A professor at a large university believes that students take an average of 15 credit hours per ...
Post by: barmour44 on Aug 30, 2019
A professor at a large university believes that students take an average of 15 credit hours per term. A random sample of 24 students in her class of 250 students reported the following number of credit hours that they were taking:
Does this sample indicate that students are taking more credit hours than the professor believes? Test an appropriate hypothesis and state your conclusion.
Does this sample indicate that students are taking more credit hours than the professor believes? Test an appropriate hypothesis and state your conclusion.
Title: A professor at a large university believes that students take an average of 15 credit hours per ...
Post by: Kailz1218 on Aug 30, 2019
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Title: A professor at a large university believes that students take an average of 15 credit hours per ...
Post by: barmour44 on Aug 30, 2019
Helps a lot... Now I'm ready for my quiz
Title: A professor at a large university believes that students take an average of 15 credit hours per ...
Post by: DragoxCM16 on Aug 30, 2019
A professor at a large university believes that students take an average of 15 credit hours per term. A random sample of 24 students in her class of 250 students reported the following number of credit hours that they were taking:
Find a 95% confidence interval for the number of credit hours taken by the students in the professor's class. Interpret your interval.
Find a 95% confidence interval for the number of credit hours taken by the students in the professor's class. Interpret your interval.
Title: A professor at a large university believes that students take an average of 15 credit hours per ...
Post by: cooldawg69 on Aug 30, 2019
With the conditions satisfied (from Problem 1), we can find a t-interval for mean credit hours.
We know: n = 24,
= 16.6 , and s = 2.22 . So, SE() = 
= 0.453.
Our confidence interval has the form ± t*n-1 
. We have t*23 = 2.069. Our 95% confidence interval is then 16.6 ± 2.069 = 16.6 ± 0.94, or 15.66 to 17.54.
We are 95% confident that the interval 15.66 to 17.54 contains the true mean number of credit hours that students in the professor's class are taking.
We know: n = 24,
= 16.6 , and s = 2.22 . So, SE() = = 0.453.Our confidence interval has the form
± t*n-1 . We have t*23 = 2.069. Our 95% confidence interval is then 16.6 ± 2.069 = 16.6 ± 0.94, or 15.66 to 17.54.We are 95% confident that the interval 15.66 to 17.54 contains the true mean number of credit hours that students in the professor's class are taking.