HP 48gII - Manual (Page 618)

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 ).