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Exercises | Section 14.2 (9-16)
9. In an experiment to compare the bond strength of Unexposed 8 Il 12 14 20 43 111
two different adhesives, each adhesive was used in passed 35-56 $3 92 128 150 176 208
five bondings of two surfaces, and the force nec-
essary to separate the surfaces was determined for 13. Reconsider the situation described in Exercise 100
each bonding. For adhesive 1, the resulting values of Chapter 10 and the accompanying MINITAB
were 229, 286, 245, 299, and 250, whereas the output (the Greek letter eta is used to denote a
adhesive 2 observations were 213, 179, 163, 247, median).
and 225. Let i; denote the true average bond Mann-Whitney Conf idence Interval and
strength of adhesive type i. Use the Wilcoxon Test
rank-sum test at level .05 to test Ho: fly = My good N=8 Median = 0.540
versus Hy! fly > Jb. poor N=8 Median = 2.400
~ Point estimate for ETA1 — ETA2 is
10. The article “A Study of Wood Stove Particulate -1.155
Emissions” (J. Air Pollut. Contr. Assoc., 1979: 95.9% CI for ETAL — ETA2 is(—3.160,
724-728) reports the following data on burn time —0.409) W= 41.0
(hours) for samples of oak and pine. Test at level Mest Of ETAL = ERAZ VS ETAT <2TAR is
. : significant at 0.0027
.05 to see whether there is any difference in true
average burn time for the two types of wood. a. Verify that the value of MINITAB’s test statis-
Oak 172 67 155 156 142 123 1.77 48 Cay ot aprepitate teat or nypeTeRes
Pine 98 1.40 1.33 152.73 1.20 vs
using a significance level of 01.
IL. A modification has been made to the process for 44 ‘The Wilcoxon rank-sum statistic can be repre-
producing a certain type of “time-zero” film (film sented as W = R, +Ry+---+Rp, where R; is
that begins to develop as soon as a picture is taken). the rank of X; — Ao among all m + n such differ-
Because the modification involves extra cost, it will ences, When Hy is true, each R; is equally likely to
besincorporated ‘only st sample data‘strongly:indic be one of the first m + n positive integers; that is,
eaten that the modification has decreased imevayer: R; has a discrete uniform distribution on the values
age developing time by more than | s. Assuming 152, 3p2s5 mtn:
that the developing-time distributions differ only a, Determine the mean value of each R; when Ho
with respect to location if at all, use the Wilcoxon GSS RHA THEA SHOW THALTHE RISER WALSOEW.
rank-sum test at level .05 on the accompanying data is m(m + n + 1)/2. [Hint: Use the hint given in
to test the appropriate hypotheses. Exercise 6(a).]
Original b. The variance of each R; is easily determined.
Process 8.6 5.1 4.5 5.4 6.3 6.6 5.7 8.5 However, the R,’s are not independent random.
variables because, for example, if m =n = 10
‘Modified and we are told that Ry = 5, then Rp must
Process 5.5 4.0 3.8 6.0 58 4.9 7.0 5.7 BSE HEEL OmhaR 19 sitegels) BEEWEEE 1
12. The article “Measuring the Exposure of Infants to and 20. However, if a and b are any two
Tobacco Smoke” (New Engl. J. Med., 1984: distinct positive integers between 1 and
1075~1078) reports on a study in which various m+n inclusive, it follows _ that
measurements were taken both from a random P(R; = aandR; = b) = 1/[(m-+n)(m+n—1)]
sample of infants who had been exposed to house- since two integers are being sampled without
hold smoke and from a sample of unexposed replacement from among 1, 2, ... , m+n.
infants. The accompanying data consists of obser- Use this fact to show that Cov(R;,Rj) =
vations on urinary concentration of cotinine, a —(m+n+1)/12 and then show that the vari-
major metabolite of nicotine (the values constitute ance of W is mn(m +n + 1)/12.
a subset of the original data and were read from a c. A central limit theorem for a sum of non-inde-
plot that appeared in the article). Does the data pendent variables can be used to show that
suggest that true average cotinine level is higher in when m > 8 and n > 8, W has approximately
exposed infants than in unexposed infants by more a normal distribution with mean and variance
than 25? Carry out a test at significance level .05. given by the results of (a) and (b). Use this to
--- Trang 784 ---
14.3 Distribution-Free Confidence Intervals 771
propose a large-sample standardized rank-sum level .01 to decide whether true average length
test statistic and then describe the rejection differs for the two types of vitamin C intake.
region that has approximate significance level Compute also an approximate P-value. [Hint:
« for testing Ho against each of the three See Exercise 14.]
commonly encountered alternative hypotheses. Orange Juice 82 94 96 9.7 100 145
[Note: When there are ties in the observed 152 16.1 176 215
values, a correction for the variance derived ee
in (b) should be used in standardizing W; please Ascorbic Acid 4.2 5.2 58 64 7.0 7.3
consult a book on nonparametric statistics for 10.1 11.2 11.3 115
the result.] 16. Test the hypotheses suggested in Exercise 15
15. The accompanying data resulted from an experi- using the following data:
ment to compare the effects of vitamin C in orange Orange Juice 82 95 95 9.7 100 145
juice and in synthetic ascorbic acid on the length 152 161 176 215
of odontoblasts in guinea pigs over a 6-week a ;
period (“The Growth of the Odontoblasts of the AScoRGIEACI ne ion Be ft TO 73
Incisor Tooth as a Criterion of the Vitamin C ° ° ° .
Intake of the Guinea Pig,” J. Nutrit., 1947: [Hint: See Exercise 14.]
491-504). Use the Wilcoxon rank-sum test at
Distribution-Free Confidence Intervals
The method we have used so far to construct a confidence interval (CI) can be
described as follows: Start with a random variable (Z, T, ~, F, or the like) that
depends on the parameter of interest and a probability statement involving the
variable, manipulate the inequalities of the statement to isolate the parameter
between random endpoints, and finally substitute computed values for random
variables. Another general method for obtaining CIs takes advantage of a relation-
ship between test procedures and CIs. A 100(1 — «)% CI for a parameter @ can be
obtained from a level « test for Ho: 0 = 09 versus H,: 6 # 0. This method will
be used to derive intervals associated with the Wilcoxon signed-rank test and the
Wilcoxon rank-sum test.
Before using the method to derive new intervals, reconsider the f test and the
t interval. Suppose a random sample of n = 25 observations from a normal
population yields summary statistics t= 100, s = 20. Then a 90% CI for yu is
( t a z ) (93.16, 106.84) (14.2)
X — 105,24 = 1X + 05,24 > T=] = (93.16, 106. “
V25 Vv 25.