HP 49g+, Руководство пользователя

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Page 18-47
Inferences concerning one variance
The null hypothesis to be tested is , H
o
:
σ
2
=
σ
o
2
, at a level of confidence (1-
α
)100%, or significance level
α
, using a sample of size n, and variance s
2
.
The test statistic to be used is a chi-squared test statistic defined as
2
0
2
2
) 1 (
σ
χ
s n
o
−
=
Depending on the alternative hypothesis chosen, the P-value is calculated as
follows:
•
H
1
:
σ 2
<
σ
o
2
, P-value = P(
χ 2
<
χ
o
2
) = 1-UTPC(
ν
,
χ
o
2
)
•
H
1
:
σ 2
>
σ
o
2
, P-value = P(
χ 2
>
χ
o
2
) = UTPC(
ν
,
χ
o
2
)
•
H
1
:
σ 2 ≠ σ
o
2
, P-value =2
⋅
min[P(
χ 2
<
χ
o
2
), P(
χ 2
>
χ
o
2
)] =
2
⋅
min[1-UTPC(
ν
,
χ
o
2
), UTPC(
ν
,
χ
o
2
)]
where the function min[x,y] produces the minimum value of x or y (similarly,
max[x,y] produces the maximum value of x or y). UTPC(
ν
,x) represents the
calculator’s upper-tail probabilities for
ν
= n - 1 degrees of freedom.
The test criteria are the same as in hypothesis testing of means, namely,
•
Reject H
o
if P-value <
α
•
Do not reject H
o
if P-value >
α
.
Please notice that this procedure is valid only if the population from which the
sample was taken is a Normal population.
Example 1 -- Consider the case in which
σ
o
2
= 25,
α
=0.05, n = 25, and s
2
=
20, and the sample was drawn from a normal population. To test the
hypothesis, H
o
:
σ
2
=
σ
o
2
, against H
1
:
σ
2
<
σ
o
2
, we first calculate
2 . 189
25
20 ) 1 25 ( ) 1 (
2
0
2
2
=
⋅ −
=
−
=
σ
χ
s n
o
With
ν
= n - 1 = 25 - 1 = 24 degrees of freedom, we calculate the P-value as,
P-value = P(
χ
2
<
19.2
) = 1-UTPC(24,
19.2
) = 0.2587…
Since, 0.2587… > 0.05, i.e., P-value >
α
, we cannot reject the null
hypothesis, H
o
:
σ 2
=25(=
σ
o
2
).