Question
A class survey in a large class for first-year college students asked, "About how many minutes do you study on a typical weeknight?" The mean response of the 463 students was x = 118 minutes.
Suppose that we know that the study time follows a Normal distribution with standard deviation σ = 65 minutes in the population of all first-year students at this university.
Use the survey result to give a 99% confidence interval for the mean study time of all first-year students.
Suppose that we know that the study time follows a Normal distribution with standard deviation σ = 65 minutes in the population of all first-year students at this university.
Use the survey result to give a 99% confidence interval for the mean study time of all first-year students.
Answers
95% = mean ± 2.575 SEm
SEm = SD/√n
I'll let you do the calculations.
SEm = SD/√n
I'll let you do the calculations.
Suppose x has a distribution with μ = 11 and σ = 5.
(a) If a random sample of size n = 45 is drawn, find μx, σ x and P(11 ≤ x ≤ 13). (Round σx to two decimal places and the probability to four decimal places.)
(a) If a random sample of size n = 45 is drawn, find μx, σ x and P(11 ≤ x ≤ 13). (Round σx to two decimal places and the probability to four decimal places.)
It is 99%, sorry for the typo.