Compute a 95% confidence interval for the population mean, based on the sample 50, 54, 55, 51, 52, 51, 54, 52, 56, and 53. Solution ANSWER: 95% Resulting Confidence Interval for \'true mean\': = [51.4, 54.2] Why??? SMALL SAMPLE, CONFIDENCE INTERVAL, NORMAL POPULATION DISTRIBUTION x-bar: Sample mean = 52.8 s: Sample standard deviation = 1.9 n: Number of samples = 10 df: degrees of freedom = 9 significant digits = 1 Confidence Level = 95 \"Look-up\" Table (\'t-critical value\') = 2.3 Look-up Table of (\'t critical values\') for confidence and prediction intervals. Central two-sided area = 95% with df = 9. 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.

1. Compute a 95% confidence interval for the population mean, based on the sample 50, 54, 55, 51,
52, 51, 54, 52, 56, and 53.
Solution
ANSWER: 95% Resulting Confidence Interval for 'true mean': = [51.4, 54.2]
Why???
SMALL SAMPLE, CONFIDENCE INTERVAL, NORMAL POPULATION DISTRIBUTION
x-bar: Sample mean = 52.8
s: Sample standard deviation = 1.9
n: Number of samples = 10
df: degrees of freedom = 9
significant digits = 1
Confidence Level = 95
"Look-up" Table ('t-critical value') = 2.3
Look-up Table of ('t critical values') for confidence and prediction intervals. Central two-sided
area = 95% with df = 9. 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