Question 1042994: 1- Find the necessary sample size.
Scores on a certain test are normally distributed with a variance of 68. A researcher wishes to estimate the mean score achieved by all adults on the test. Find the sample size needed to assure with 95 percent confidence that the sample mean will not differ from the population mean by more than 4 units.
2- A college statistics professor has office hours from 9:00 A.M. to 10:30 A.M. daily. A sample of waiting times to see the professor (in minutes) is 10, 12, 20, 15, 17, 10, 30, 28, 35, 28, 19, 27, 25, 22, 33, 37, 14, 21, 20, 23. Assume alpha equals 7.84 find the 95.44% confidence interval for the population mean.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 1- Find the necessary sample size.
Scores on a certain test are normally distributed with a variance of 68. A researcher wishes to estimate the mean score achieved by all adults on the test. Find the sample size needed to assure with 95 percent confidence that the sample mean will not differ from the population mean by more than 4 units.
--------
n = [z*s/E]^2 = [1.96*68/4]^2 = 1110.2 = 1110 when rounded down
-----------------------
2- A college statistics professor has office hours from 9:00 A.M. to 10:30 A.M. daily. A sample of waiting times to see the professor (in minutes) is 10, 12, 20, 15, 17, 10, 30, 28, 35, 28, 19, 27, 25, 22, 33, 37, 14, 21, 20, 23. Assume the
standard deviation equals 7.84 find the 95.44% confidence interval for the population mean.
(1-0.9544)/2 = 0.0228
z = -invNorm(0.0228) = 2
Sample mean = 22.3
ME = 2*7.84/sqrt(20) = 3.06
-----
95.44% CI:: 22.3-3.06 < u < 22.3+3.06
--------------------
Cheers,
Stan H.
------------
|
|
|
