Inferences Concerning One Variance - HP 49g+ User Manual

Inferences concerning one variance

The null hypothesis to be tested is , H
α)100%, or significance level α, using a sample of size n, and variance s
The test statistic to be used is a chi-squared test statistic defined as
Depending on the alternative hypothesis chosen, the P-value is calculated as
follows:
2 2
• : σ < σ
H ,
1 o
2 2
• : σ > σ
H ,
1 o
• : σ 2 ≠ σ 2
H ,
1 o
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,
• if P-value < α
Reject H
o
•
Do not reject H
o
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
20, and the sample was drawn from a normal population.
: σ 2 = σ 2
hypothesis, H
o o
2
χ
o
With ν = n - 1 = 25 - 1 = 24 degrees of freedom, we calculate the P-value as,
P-value = P(χ
Since, 0.2587... > 0.05, i.e., P-value > α, we cannot reject the null
: σ 2 =25(= σ
hypothesis, H
o
: σ 2 = σ
o
2
( n − ) 1 s
2
χ =
o
2
σ
0
2 2
P-value = P(χ <χ ) = 1-UTPC(ν,χ
o
2 2
P-value = P(χ >χ ) = UTPC(ν,χ
o
2
P-value =2⋅min[P(χ <χ
o
2
2⋅min[1-UTPC(ν,χ ), UTPC(ν,χ
o
if P-value > α.
2
= 25, α=0.05, n = 25, and s
o
: σ 2 < σ 2
, against H , we first calculate
1 o
2
( n ) 1 s ( 25 ) 1
2
σ 25
0
2
<19.2) = 1-UTPC(24,19.2) = 0.2587...
2
).
o
2
, at a level of confidence (1-
o
2
)
o
2
)
o
2 2 2
), P(χ >χ )] =
o
2
)]
o
To test the
20
189 2 .
Page 18-47
2
.
2
=