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326 CHAPTER 9 TESTS OF HYPOTHESES FOR A SINGLE SAMPLE

MIND-EXPANDING EXERCISES

9-91. Suppose that we wish to test H 0 : 0 versus 9-94. When X 1 , X 2 , p , X n is a random sample from a
, where the population is normal with
H 1 : 0 normal distribution and n is large, the sample standard
known . Let 0
, and define the critical region deviation has approximately a normal distribution with
2
so that we will reject H 0 if z 0 z
or if z 0 z
, mean and variance 12n2 . Therefore, a large-sample
where z 0 is the value of the usual test statistic for these test for H 0 : 0 can be based on the statistic
hypotheses.
(a) Show that the probability of type I error for this test
S 0
is . Z 2
(b) Suppose that the true mean is 1 0
. Derive 2 0 12n2
an expression for for the above test.
9-92. Derive an expression for for the test on the Use this result to test H 0 : 10 versus H 1 : 10 for
variance of a normal distribution. Assume that the two- the golf ball overall distance data in Exercise 6-25.
sided alternative is specified. 9-95. Continuation of Exercise 9-94. Using the
9-93. When X 1 , X 2 , p , X n are independent Poisson results of the previous exercise, find an approximately
random variables, each with parameter , and n is large, unbiased estimator of the 95 percentile
1.645 .
the sample mean X has an approximate normal distribu- From the fact that X and S are independent random
tion with mean and variance n . Therefore, variables, find the standard error of . How would you
estimate the standard error?
X 9-96. Continuation of Exercises 9-94 and 9-95.
Z Consider the golf ball overall distance data in Exercise
1 n
6-25. We wish to investigate a claim that the 95 per-
centile of overall distance does not exceed 285 yards.
has approximately a standard normal distribution. Thus
Construct a test statistic that can be used for testing the
we can test H 0 : 0 by replacing in Z by 0 . When X i appropriate hypotheses. Apply this procedure to the data
are Poisson variables, this test is preferable to the large- from Exercise 6-25. What are your conclusions?
sample test of Section 9-2.5, which would use S 1n in
the denominator, because it is designed just for the 9-97. Let X 1 , X 2 , p , X n be a sample from an exponen-
Poisson distribution. Suppose that the number of open cir- tial distribution with parameter . It can be shown that
n
cuits on a semiconductor wafer has a Poisson distribution. 2 i 1 X i has a chi-square distribution with 2n degrees
Test data for 500 wafers indicate a total of 1038 opens. of freedom. Use this fact to devise a test statistic and
Using 0.05, does this suggest that the mean number critical region for H 0 : 0 versus the three usual
of open circuits per wafer exceeds 2.0? alternatives.

IMPORTANT TERMS AND CONCEPTS
In the E-book, click on any Null hypothesis Sample size determina- Test for homogeneity
term or concept below to One- and two-sided tion for hypothesis Test for independence
go to that subject. alternative hypotheses tests Test statistic
Connection between Operating characteristic Significance level of a Type I and type II
hypothesis tests curves test errors
and confidence Power of the test Statistical hypotheses
CD MATERIAL
intervals P-value Statistical versus practi-
Likelihood ratio test
Critical region for a test Reference distribution cal significance
statistic for a test statistic Test for goodness of fit