A group of 66 randomly selected students have a mean score of 22.4....?
A group of 66 randomly selected students have a mean score of 22.4 with a standard deviation of 2.8 on a placement test. What is the 90% confidence interval for the mean score, μ, of all students taking the test?
ANSWER: 90% Resulting Confidence Interval for 'true mean': = [21.8, 23.0]
SMALL SAMPLE, CONFIDENCE INTERVAL, NORMAL POPULATION DISTRIBUTION
x-bar = Sample mean 22.4
s = Sample standard deviation2.8
n = Number of samples 66
df = degrees of freedom 65
"Look-up" Table 't-critical value'1.7
Look-up Table of t critical values for confidence and prediction intervals. Central two-side area = 90%
with df = 65. Another Look-up method is to utilize Microsoft Excel function:
TINV(probability,degrees_freedom) Returns the inverse of the Student's t-distribution
90% Resulting Confidence Interval for 'true mean':
x-bar +/- ('t critical value') * s/SQRT(n)= 22.4 +/- 1.7 * 2.8/SQRT(66) = [21.8, 23.0]