# 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]

Why?

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

significant digits1

Confidence Level90

"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]

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