12 The following sample of n 5 20 observations on dielectric breakdown voltage of a
Example 14.12 The following sample of n 5 20 observations on dielectric breakdown voltage of a
piece of epoxy resin first appeared in Example 4.30.
y i
–1.871 –1.404 –1.127 –.917 –.742 –.587 –.446 –.313 –.186 –.062
We asked Minitab to carry out the Ryan-Joiner test, and the result appears in Figure 14.3. The test statistic value is r 5 .9881 , and Appendix Table A.12 gives .9600 as the critical value that captures lower-tail area .10 under the r sampling distribution curve when n 5 20 and the underlying distribution is actually normal. Since .9881 . .9600 , the null hypothesis of normality cannot be rejected even for
a significance level as large as .10.
Figure 14.3 Minitab output from the Ryan-Joiner test for the data of Example 14.12
■
EXERCISES Section 14.2 (12–23)
12. Consider a large population of families in which each
based on the same number of degrees of freedom as the
family has exactly three children. If the genders of the three
test in part (a)? Explain.
children in any family are independent of one another, the
13. A study of sterility in the fruit fly (“Hybrid Dysgenesis in
number of male children in a randomly selected family will
Drosophila melanogaster: The Biology of Female and Male
have a binomial distribution based on three trials.
Sterility,” Genetics, 1979: 161–174) reports the following data
a. Suppose a random sample of 160 families yields the
on the number of ovaries developed by each female fly in a
following results. Test the relevant hypotheses by pro-
sample of size 1388. One model for unilateral sterility states
ceeding as in Example 14.5.
that each ovary develops with some probability p independ-
Number of 2 ently of the other ovary. Test the fit of this model using x .
Male Children
0 1 2 3 x 5 Number of
Frequency
14 66 64 16 Ovaries Developed
Observed Count
b. Suppose a random sample of families in a nonhuman population resulted in observed frequencies of 15, 20,
14. The article “Feeding Ecology of the Red-Eyed Vireo
12, and 3, respectively. Would the chi-squared test be
and Associated Foliage-Gleaning Birds” (Ecological
CHAPTER 14 Goodness-of-Fit Tests and Categorical Data Analysis
Monographs, 1971: 129–152) presents the accompanying
plausible model for the distribution of the number of borers
data on the variable X5 the number of hops before the first
in a group? [Hint: Add the frequencies for 7, 8, . . . , 12 to
flight and preceded by a flight. The author then proposed and
establish a single category “ 7 .”]
fit a geometric probability distribution [ p(x) 5 P(X 5 x) 5
p x21 q for x 5 1, 2, c , where q 5 1 2 p] to the data.
Number
The total sample size was n 5 130 .
of Borers
18. The article “A Probabilistic Analysis of Dissolved
of Times x
48 31 20 9 6 5 4 2 1 1 2 1 Oxygen–Biochemical Oxygen Demand Relationship in
Observed
Streams” (J. Water Resources Control Fed., 1969: 73–90) reports data on the rate of oxygenation in streams at 20°C
a. The likelihood is (p x 1 21 q) c (p x n 21 q) 5
p a x i 2n q n
in a certain region. The sample mean and standard devia-
. Show that the mle of p is given by pˆ 5
tion were computed as x 5 .173 and s 5 .066 , respec-
(g x i 2 n)g x i , and compute for the given data.
tively. Based on the accompanying frequency distribution,
b. Estimate the expected cell counts using pˆ of part (a)
can it be concluded that oxygenation rate is a normally
[ expected cell counts 5 n (pˆ) x21 qˆ for ], x 5 1, 2, c
distributed variable? Use the chi-squared test with
and test the fit of the model using a x 2 test by combining the
a 5 .05 .
counts for x 5 7, 8, c , and 12 into one cell (x 7) .
Rate (per day)
Frequency
15. A certain type of flashlight is sold with the four batteries included. A random sample of 150 flashlights is obtained,
and the number of defective batteries in each is determined,
.100–below .150
resulting in the following data:
.150–below .200
.200–below .250
Number Defective
0 1 2 3 4 .250 or more
Frequency
26 51 47 16 10 19. Each headlight on an automobile undergoing an annual vehicle inspection can be focused either too high (H ), too
Let X be the number of defective batteries in a randomly
low (L), or properly (N ). Checking the two headlights
selected flashlight. Test the null hypothesis that the distribu-
simultaneously (and not distinguishing between left and
tion of X is Bin(4, u ). That is, with p i 5 P(i defectives) , test
right) results in the six possible outcomes HH, LL, NN,
i bu
0 :p i 5 a (1 2 u)
4 HL, HN, and LN. If the probabilities (population propor-
tions) for the single headlight focus direction are
P(H ) 5 u 1 , P(L) 5 u
2 , and P(N ) 5 1 2 u 1 2u 2 and the
[Hint: To obtain the mle of u , write the likelihood (the func-
two headlights are focused independently of one another,
tion to be maximized) as (1 2 u) v
, where the exponents u
the probabilities of the six outcomes for a randomly
and v are linear functions of the cell counts. Then take the
selected car are the following:
natural log, differentiate with respect to u, equate the result to 0, and solve for .] u ˆ
p 1 5u 1 2 p 2 5u 2 p 3 5 (1 2 u 1 2u 2 ) 2
16. In a genetics experiment, investigators looked at 300 chro-
p
4 5 2u 1 u 2 p 5 5 2u 1 (1 2 u 1 2u 2 )
mosomes of a particular type and counted the number of
sister-chromatid exchanges on each (“On the Nature of
p 6 5 2u 2 (1 2 u 1 2u 2 )
Sister-Chromatid Exchanges in 5-Bromodeoxyuridine- Substituted Chromosomes,” Genetics, 1979: 1251–1264). A
Use the accompanying data to test the null hypothesis
Poisson model was hypothesized for the distribution of the
H 1 ,u
0 :p 1 5p 1 (u 2 ), c, p 6 5p 6 (u 1 ,u 2 )
number of exchanges. Test the fit of a Poisson distribution
to the data by first estimating m and then combining the
where the p i (u 1 ,u 2 ) s are given previously.
counts for x58 and x59 into one cell.
x 5 Number
Frequency
of Exchanges 0 1 2 3 4 5 6 7 8 9
[Hint: Write the likelihood as a function of u 1 and u 2 , take
Observed
the natural log, then compute ' 'u 1 and ' 'u 2 , equate them
Counts
6 24 42 59 62 44 41 14 6 2 to 0, and solve for u ˆ 1 , ˆ u 2 .]
17. An article in Annals of Mathematical Statistics reports
20. The article “Compatibility of Outer and Fusible Interlining
the following data on the number of borers in each of
Fabrics in Tailored Garments (Textile Res. J., 1997:
120 groups of borers. Does the Poisson pmf provide a
137–142) gave the following observations on bending
14.3 Two-Way Contingency Tables
rigidity (mN m) for medium-quality fabric specimens,
from which the accompanying Minitab output was
obtained:
1.28 1.45 1.29 1.28 1.38 1.55 1.46 1.32 24.6 12.7 14.4 30.6 16.1 9.5 31.5 17.2 Minitab gave as r 5 .9852 the value of the Ryan-Joiner test
46.9 68.3 30.8 116.7 39.5 73.8 80.6 20.3 statistic and reported that P-value . .10. Would you use the 25.8 30.9 39.2 36.8 46.6 15.6 32.3 one-sample t test to test hypotheses about the value of the true average ratio? Why or why not?
22. The article “A Method for the Estimation of Alcohol in Fortified Wines Using Hydrometer Baumé and Refractometer Brix” (Amer. J. of Enol. and Vitic., 2006: 486–490) gave duplicate measurements on distilled alcohol content () for a sample of 35 port wines. Here are averages of those duplicate measurements:
Use the Ryan-Joiner test to decide at significance level .05 whether a normal distribution provides a plausible model for alcohol content.
23. The article “Nonbloated Burned Clay Aggregate Concrete” (J. of Materials, 1972: 555–563) reports the following data on 7-day flexural strength of nonbloated burned clay aggre- gate concrete samples (psi):
Would you use a one-sample t confidence interval to estimate
true average bending rigidity? Explain your reasoning.
21. The article from which the data in Exercise 20 was obtained
also gave the accompanying data on the composite mass
Test at level .10 to decide whether flexural strength is a nor-
outer fabric mass ratio for high-quality fabric specimens.
mally distributed variable.