Question

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¯¯¯ = 12 hours. Suppose that we know that the study time follows a Normal distribution with standard deviation 7 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.

Answer 1

olution :

Given that,

Point estimate = sample mean = = 12

Population standard deviation =    = 7

Sample size = 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   ( Using z table )

Margin of error = E = Z/2* ( /n)

= 2.576* (7 / 427)

= 0.87

At 99% confidence interval IS

- E < < + E

12- 0.87 < < 12+0.87

11.13< < 12.87

(11.13 TO 12.87 )