14.4A Two-Sample Problem 599
Hence, by the equivalence of the gamma and chi-square distribution it follows that
2
θ 1
∑n
i= 1
Xi∼χ 22 n
2
θ 2
∑m
i= 1
Yi∼χ 22 m
Hence, it follows from the definition of theF-distribution that
(
2
θ 1
∑n
i= 1
Xi
)
(^2 n
2
θ 2
∑m
i= 1
Yi
)
2 m
∼Fn,m
That is, ifXandYare the two sample means, respectively, then
θ 2 X
θ 1 Y
has anF-distribution withnandmdegrees of freedom
Hence, when the hypothesisθ 1 =θ 2 is true, we see thatX/Yhas anF-distribution with
nandmdegrees of freedom. This suggests the following test of the hypothesis that
θ 1 =θ 2.
Test: H 0 :θ 1 =θ 2 vs. alternativeH 1 :θ 1 =θ 2
Step 1: Choose a significance levelα.
Step 2: Determine the value of the test statisticX/Y— say its value isv.
Step 3: ComputeP{F≤v}whereF∼Fn,m. If this probability is either less
thanα/2 (which occurs whenXis significantly less thanY) or greater
than 1−α/2 (which occurs whenXis significantly greater thanY) then
the hypothesis is rejected.
In other words, thep-value of the test data is given by
p-value=2 min(P{F≤v},1−P{F≤v})
EXAMPLE 14.4a Test the hypothesis, at the 5 percent level of significance, that the lifetimes
of items produced at two given plants have the same exponential life distribution if a sample
of size 10 from the first plant has a total lifetime of 420 hours whereas a sample of 15 from
the second plant has a total lifetime of 510 hours.