Why???

SMALL SAMPLE, CONFIDENCE INTERVAL, NORMAL POPULATION DISTRIBUTION
x-bar = Sample mean 637.5
s = Sample standard deviation59.3
n = Number of samples 35
df = degrees of freedom 34

significant digits2

Confidence Level90
“Look-up” Table ‘t-critical value’1.69
Look-up Table of t critical values for confidence and prediction intervals. Central two-sided area = 90%
with df = 34. 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)= 637.5 +/- 1.69 * 59.3/SQRT(35) = [$620.56, $654.44]

ANSWER: 95% Resulting Confidence Interval for ‘true mean’: = [$617.15, $657.85]

Why???

SMALL SAMPLE, CONFIDENCE INTERVAL, NORMAL POPULATION DISTRIBUTION
x-bar = Sample mean 637.5
s = Sample standard deviation59.3
n = Number of samples 35
df = degrees of freedom 34

significant digits2

Confidence Level95
“Look-up” Table ‘t-critical value’2.03
Look-up Table of t critical values for confidence and prediction intervals. Central two-sided area = 95%
with df = 34. Another Look-up method is to utilize Microsoft Excel function:
TINV(probability,degrees_freedom) Returns the inverse of the Student’s t-distribution

95% Resulting Confidence Interval for ‘true mean’:
x-bar +/- (’t critical value’) * s/SQRT(n)= 637.5 +/- 2.03 * 59.3/SQRT(35) = [$617.15, $657.85]