274 Some Elementary Statistical Inferencesand define the critical region to beC={(x 1 ,x 2 ):^34 ≤x 1 x 2 }.Findthepower
function of the test.
4.5.4. LetXhave a binomial distribution with the number of trialsn=10and
withpeither 1/4or1/2. The simple hypothesisH 0 :p=^12 is rejected, and the
alternative simple hypothesisH 1 :p=^14 is accepted, if the observed value ofX 1 ,a
random sample of size 1, is less than or equal to 3. Find the significance level and
the power of the test.
4.5.5.LetX 1 ,X 2 be a random sample of sizen= 2 from the distribution having
pdff(x;θ)=(1/θ)e−x/θ, 0 <x<∞, zero elsewhere. We rejectH 0 :θ=2and
acceptH 1 :θ= 1 if the observed values ofX 1 ,X 2 ,sayx 1 ,x 2 ,aresuchthat
f(x 1 ;2)f(x 2 ;2)
f(x 1 ;1)f(x 2 ;1)≤1
2.Here Ω ={θ:θ=1, 2 }. Find the significance level of the test and the power of the
test whenH 0 is false.
4.5.6.Consider the tests Test 1 and Test 2 for the situation discussed in Example
4.5.2. Consider the test that rejectsH 0 ifS≤10. Find the level of significance for
this test and sketch its power curve as in Figure 4.5.1.4.5.7.Consider the situation described in Example 4.5.2. Suppose we have two
tests A and B defined as follows. For Test A,H 0 is rejected ifS≤kA, while for
Test B,H 0 is rejected ifS≤kB. If Test A has a higher level of significance than
Test B, show that Test A has higher power than Test B at each alternative.
4.5.8.Let us say the life of a tire in miles, sayX, is normally distributed with mean
θand standard deviation 5000. Past experience indicates thatθ=30,000. The
manufacturer claims that the tires made by a new process have meanθ> 30 ,000.
It is possible thatθ=35,000. Check his claim by testingH 0 :θ=30,000 against
H 1 :θ> 30 ,000. We observenindependent values ofX,sayx 1 ,...,xn,andwe
rejectH 0 (thus acceptH 1 ) if and only ifx≥c. Determinenandcso that the power
functionγ(θ) of the test has the valuesγ(30,000) = 0.01 andγ(35,000) = 0.98.
4.5.9. LetXhave a Poisson distribution with meanθ. Consider the simple hy-
pothesisH 0 :θ=^12 and the alternative composite hypothesisH 1 :θ<^12 .Thus
Ω={θ:0<θ≤^12 }.LetX 1 ,...,X 12 denote a random sample of size 12 from this
distribution. We rejectH 0 if and only if the observed value ofY=X 1 +···+X 12 ≤2.
Show that the following R code graphs the power function of this test:
theta=seq(.1,.5,.05); gam=ppois(2,theta*12)
plot(gam~theta,pch=" ",xlab=expression(theta),ylab=expression(gamma))
lines(gam~theta)
Run the code. Determine the significance level from the plot.
4.5.10.LetY have a binomial distribution with parametersnandp. We reject
H 0 :p=^12 and acceptH 1 :p>^12 ifY≥c.Findnandcto give a power function
γ(p) which is such thatγ(^12 )=0.10 andγ(^23 )=0.95, approximately.