In: Statistics and Probability
Among those who take the Graduate Record Examination (GRE), 67 people are randomly selected. This sample group has a mean score of 558 and a standard deviation of 149 on the quantitative portion of the GRE. Find a 99% confidence interval estimate of the population mean.
Solution :
Given that,
Point estimate = sample mean =
= 558
Population standard deviation =
= 149
Sample size = n =67
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 * ( 149/
67)
= 46.8916
At 99% confidence interval estimate of the population mean is,
-
E <
<
+ E
558-46.8916 <
< 558+46.8916
511.1084<
< 604.8916
(511.1084, 604.8916)