Page 302 - Six Sigma Demystified

282 Six SigMa DemystifieD

Difference = µ – µ 2
1
Estimate for difference: –5.500

95% CI for difference: (–6.508,–4.492)
T test of difference = 0 (versus not): T value = –10.90; p value = 0.000; df = 63.
Conclude: Reject null (p < 0.05); means are not equal.

Excel

Calculate p as
Two-sided: TDIST(|t |,ν,2)
0
One-sided Case 1: 1-TDIST(|t |,ν,1) for t > 0; TDIST otherwise
0
0
One-sided Case 2: TDIST(|t |,ν,1) for t > 0; 1-TDIST otherwise
0 0
Two-Sided Example at 5% Significance
=TDIST(ABS(10.90),63,2); p = 0.000; reject null.

Test on Paired Means
Note: This is a special case of the general two-sample mean test. In this case,
each observation in sample 1 has a corresponding observation in sample 2 re-
flecting paired estimates of the same sample unit. For example, if two people
inspected the same item, each of the measurements (from each person) would
have a corresponding paired sample with the measurement for the second per-
son for the same sample unit. The differences between each pair is calculated

first, and then the standard test on mean is applied to the differences, using the
calculated average and standard deviation of the differences.
Null hypothesis H : m – m = 0
0
1
2
Alternate hypothesis H : m – m = 0
1 1 2
Test statistic: t 0
µ
0 t = ( X− )
s/
n
where X and s are calculated using the differences between the paired data.

Reject if t > t α/2,n-1 or t < –t α/2,n-1
0
0
Example: n = 25; average difference = 11.4; S = 2.1
differences