In: Math
(16.19) A class survey in a large class for first-year college students asked, "About how many hours do you study in a typical week?". The mean response of the 427 students was x¯¯¯ = 17 hours. Suppose that we know that the study time follows a Normal distribution with standard deviation 8 hours in the population of all first-year students at this university. What is the 99% confidence interval (±0.001) for the population mean? Confidence interval is from to hours.
solution
Given that,
= 17
= 8
n = 427
At 99% confidence level the z is ,
= 1 - 99% = 1 - 0.99 = 0.01
/ 2 = 0.01 / 2 = 0.005
Z/2 = Z0.005 = 2.576
Margin of error = E = Z/2* ( /n)
= 2.576* (8 / 427)
= 0.997
At 99% confidence interval estimate of the population mean is,
- E < < + E
17-0.997< < 17+0.997
16.003< < 17.997
(16.003,17.997 )
. What is the 99% confidence interval (±0.001) for the population mean
(16.003,17.997 )