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Six Sigma for Electronics Design and Manufacturing
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The sample is large enough to use z tables. From equation 5.6:
95% confidence ( = 0.25) = 20 ± 1.960 · 5/9 = 20 ± 1.09
99.9% confidence ( = 0.0005)= 20 ± 3.290 · 5/9 = 20 ± 1.83
Note that the confidence interval for 99.9% is almost double the one
for 95%.
Example 5.7
For a sample of the following values, 2.6, 2.1, 2.4, 2.5, 2.7, 2.2, 2.3, 2.4,
and 1.9, find the confidence interval of the population average, as-
suming that it is normal, for 90%, 95%, and 99.9% confidence.
For the sample data: n = 9; sample average X = 2.34, and sample
standard deviation s = 0.25. Using the t distribution with t /2,8 and
Equation (5.7):
90% confidence ( = 0.05) = 2.34 ± 1.860 · 0.25/3
= 2.34 ± 0.16 (2.18 – 2.5)
95 % confidence ( = 0.025) = 2.34 ± 2.306 · 0.25/3
= 2.34 ± 0.19 (2.15 – 2.53)
99.9% confidence ( = 0.0005) = 2.34 ± 5.041 · 0.25/3
= 2.34 ± 0.42 (2.76 – 1.92)
In every case, the sample point 1.9 falls outside the lower confi-
dence limit, making it an unusual event. At 99.9% confidence, the
point has a probability of less than 0.005.

5.1.5 Standard deviation for samples and populations
The statistical relationships of the sample and population averages
have been discussed in previous sections. There is a similar distri-
2
bution for the sample variability s , which can be used to learn about
its parametric counterpart, the population variance or . This dis-
2
tribution is called the chi square or . Since the distribution cannot
2
be negative, it is not symmetrical, but is in fact related to the gam-
2
ma distribution. The distribution is shown in Figure 5.4. The
2
probability that that a random sample produces a greater than
some specified value is equal to the area of the curve to the right of
the value. The variable represents the value of above which
2
2
there is the area . The equation for the distribution variable is as
follows:
(n – 1) s
2 2
= (5.8)
2
2