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Estimation

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The probability density function of a χ random variable is

1
f(x) = x (k/2)−1 −x/2 , x > 0 (2.33)
e
2 k/2 Γ(k/2)
2
where k is the number of degrees of freedom. The mean and variance of the χ distribution are k
and 2k, respectively. Note that the chi-square random variable is nonnegative and that the
probability distribution is skewed to the right. However, as k increases, the distribution becomes
more symmetric. As k → ∞, the limiting form of the chi-square distribution is the normal
2
distribution. The percentage points of the χ distribution are given in special table of the
textbooks. Define χ 2 α,k as the percentage point or value of the chi-square random variable with k
2
degrees of freedom such that the probability that X exceeds this value is α. That is,
∫
∞
2
P(X > χ 2 ) = f(u)du = α.
α,k
χ 2
α,k
Definition 2.3. Confidence Interval on the Variance: If s 2 is the sample
variance from a random sample of n observations from a normal distribution
2
2
with unknown variance σ , then a 100(1 − α)% confidence interval on σ is
(n − 1)s 2 2 (n − 1)s 2
≤ σ ≤ (2.34)
χ 2 χ 2
α/2,n−1 1−α/2,n−1
where χ 2 and χ 2 are the upper and lower 100α/2 percentage points
α/2,n−1 1−α/2,n−1
of the chi-square distribution with n − 1 degrees of freedom, respectively. A
confidence interval for σ has lower and upper limits that are the square roots of
the corresponding limits in Equation (2.34). ✓

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It is also possible to find a 100(1−α)% lower confidence bound or upper confidence bound on σ .
One-Sided Confidence Bounds on the Variance: The 100(1 − α)% lower and upper
2
confidence bounds on σ are
(n − 1)s 2 2 2 (n − 1)s 2
≤ σ and σ ≤ (2.35)
χ 2 χ 2
α,n−1 1−α,n−1
respectively.

Example 2.9. Detergent Filling: an automatic filling machine is used to fill
bottles with liquid detergent. A random sample of 20 bottles results in a sample
2
2
variance of fill volume of s = 0.0153 (fluid gram) . If the variance of fill volume
is too large, an unacceptable proportion of bottles will be under- or overfilled.
We will assume that the fill volume is approximately normally distributed. A 95%
2 (n−1)s 2
upper confidence bound is found from Equation (2.35) as follows: σ ≤ or
χ 2 0.95,19
2 19·0.0153 2
σ ≤ = 0.0287 (fluid gram) .
10.117
This last expression may be converted into a confidence interval on the standard
deviation σ by taking the square root of both sides, resulting in σ ≤ 0.17
Practical Interpretation: Therefore, at the 95% level of confidence, the data
indicate that the process standard deviation could be as large as 0.17 fluid gram.
The process engineer or manager now needs to determine if a standard deviation this
large could lead to an operational problem with under-or over filled bottlers. ,

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