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some probability p independently of the other t0:0,-andisalvefor'@.]
ovary. Test the fit of this model using 7?.
16. Ina genetics experiment, investigators looked at
x = Number of 300 chromosomes of a particular type and
Ovaries Developed 0 12 counted the number of sister-chromatid
: : exchanges on each (“On the Nature of Sister-
Onserved Count wu Ie 58 Chromatid Exchanges in 5-Bromodeoxyuridine-
Sh waib till i = Substituted Chromosomes,” Genetics, 1979:
M4, The saticle Srecding Beploey ik the Ree eyed 1251-1264). A Poisson model was hypothesized
Vireo/and. Associated Foliage Cleaning: Birds for the distribution of the number of exchanges.
(Ecol, Monogr, 1271: 129-152), presents the Test the fit of a Poisson distribution to the data by
aepompanying data on ite varile® —lhenume first estimating 7 and then combining the counts
ber of hops before the first flight and preceded by hers Bani Oublc meesil
a flight. The author then proposed and fit a geo- . . .
metric probability distribution [p@) = P(X =x) x=Number |0 1 2 3 4 5 6 789
= p*'+qforx =1,2,...,whereqg =1—p|to of Exchanges
the data, The total sample size was n = 130. Observed 6 24 42 59 62 44 41 14 6 2
Counts
x 1234567891011 12
Number 4 3120965421121 17. An article in Annals of Mathematical Statistics
of Times x reports the following data on the number of
Glseeved borers in each of 120 groups of borers. Does
the Poisson pmf provide a plausible model for
Fie Beetinod ds (phigh one IpghBagl the distribution of the number of borers in a
aie Ukelthoodas (og) laa) group? (Hint: Add the frequencies for 7, 8, ...,
p>" -q", Show that the mle of p is given by 12 to establish a single category “ > 7.”]
p=(Xxi—n)/Tx;, and compute p for the Stablisia sing gory eh
given data.
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13.2 Goodness-of-Fit Tests for Composite Hypotheses 743
[Hint: Write the likelihood as a function of 0; and
Number GO 12 34S 67 89 10 M1 12 0, take the natural log, then compute 0/00; and
ohBorend 8/002, equate them to 0, and solve for 01,03.)
Frequency | 24 16 16 18 1596534 3 0 1
20. The article “Compatibility of Outer and Fusible
18. The article “A Probabilistic Analysis of Dissolved Interlining Fabrics in Tailored Garments (Textile
Oxygen—Biochemical Oxygen Demand Relation- Res. J., 1997: 137-142) gave the following
ship in Steams” (J. Water Resources Control observations on bending rigidity (WN - m) for
Fed., 1969: 73-90) reports data on the rate of medium-quality fabric specimens, from which
oxygenation in streams at 20°C ina certain region. the accompanying MINITAB output was
The sample mean and standard deviation were obtained:
computed as. = .173 and s = .066, respectively. ate by 4 Be tex OF FS HHS
Based on the accompanying frequency idistnibu- 46.9 68.3 308 116.7 39.5 73.8 80.6 203
tion, can it be concluded that oxygenation rate is a 25.8 30.9 39.2 368 46.6 15.6 323
normally distributed variable? Use the chi-
squared test with 2 = .05. Normal Probability Plot
Rate (per day) Frequency 2?
Below .100 12 96 °
2 280 ®
-100-below .150 20 5 é
150-below .200 23 & a
-200-below .250 IS a”
.250 or more 13 08 *
oo 01
19, Each headlight on an automobile undergoing an sa
annual vehicle inspection can be focused either too 20 70 120
high (H), too low (L), or properly (WV). Checking bending
the two headlights simultaneously (and not distin- Average: 97.4217 Wiest for Normality
he : : : Std Dev. 25.8101 R 0.9116
guishing between left and right) results in the six Nof data: 23 pvalue(approx): <0.0100
possible outcomes HH, LL, NN, HL, HN, and LN.
If the probabilities (population proportions) for the Would you use a one-sample ¢ confidence inter-
single headlight focus direction are P(H) = 01, val to estimate true average bending rigidity?
P(L) = 02, and P(N) = 1 ~ 0 ~ 03 and the two Explain your reasoning.
headlights are focused independently of each 94. ‘The article from which the data in Exercise 20 was
other, the probabilities of the six outcomes for a obtained also gave the accompanying data on the
randomly selected car are the following: composite mass/outer fabric mass ratio for high-
m=8 p= py = (1-0; — quality fabric specimens.
= 2010s ps 2811 — 8) — 0») 1.15 140 1.34 1.29 1.36 1.26 1.22
Po =203(1— 0, — 63) 140 129 141 132 1.34 1.26 1.36
2 136 130 128 145 1.29 1.28 1.38
Use the accompanying data to test the null 1.55 146 1.32
Hypothesis MINITAB gave r = .9852 as the value of the
Ho : pt = (01, 02), ---,P6 = %6(01, 02) Ryan— Joiner test statistic and reported that P-
value > 10. Would you use the one-sample
where the 7;(0;, 02)’s are given previously. t test to test hypotheses about the value of the
. true average ratio? Why or why not?
Outcome HH LL NN HL HN LN
Frequency 49 26 14 20 53 38 22. The article “Nonbloated Bumed Clay Aggregate
Concrete” (J. Mater., 1972: 555-563) reports the
following data on 7 day flexural strength of
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744 = cuarrer 13 Goodness-of-Fit Tests and Categorical Data Analysis
fomronted bumed clay aggregate concrete samples 7. 24 level .10 to decide whether flexural strength is a
Psi: normally distributed variable.
257 327) 317, 3300) 340 340 343.
374-377-386 383. 393. 407) 407
434 427 440 407 450 440 456
460 456 476 480 490 497 526
546 700
Two-Way Contingency Tables
In the previous two sections, we discussed inferential problems in which the count