HP 50g User Manual page 601

The confidence interval for the population variance
2
where
n-1, /2
degrees of freedom, exceeds with probabilities /2 and 1-
The one-sided upper confidence limit for
Example 1 – Determine the 95% confidence interval for the population variance
2
based on the results from a sample of size n = 25 that indicates that the
sample variance is s
In Chapter 17 we use the numerical solver to solve the equation
In this program, represents the degrees of freedom (n-1), and
probability of exceeding a certain value of x (
For the present example,
equation presented above results in
On the other hand, the value
values = 24 and
12.4011502175.
The lower and upper limits of the interval will be (Use ALG mode for these
calculations):
2 2
(n-1) S /
2 2
(n-1) S /
n-1,1- /2
Thus, the 95% confidence interval for this example is:
2 2
[(n-1) S /
n-1, /2
2
, and
n-1,1- /2
2
= 12.5.
= 0.05, = 24 and
2
= 0.975. The result is
= (25-1) 12.5/39.3640770266 = 7.62116179676
n-1, /2
= (25-1) 12.5/12.4011502175 = 24.1913044144
7.62116179676 <
2
; (n-1) S /
are the values that a
2
is defined as (n-1) S
2
2
=
n-1, /2
2
=
n-1, /2 24,
2
n-1,1- /2
2
< 24.1913044144.
2
is therefore,
2
].
n-1,1- /2
2
variable, with
/2, respectively.
2
), i.e., Pr[ >
= 0.025. Solving the
2
= 39.3640770266.
24,
is calculated by using the
2
=
24,
= n-1
2 2
/ .
n-1,1-
= UTPC( ,x).
represents the
2
] = .
=
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