Stat 503
Solutions to Homework #9
Problem 10.4 (p. 400)
At a Midwestern hospital there were a total of 932 births in 20 consecutive weeks. Of these births, 216 (15)
occurred on weekends. Do these data reveal more than chance deviation from random timing of the
births? (Test for goodness of fit, with two categories of birth – weekday and weekend. Use a
nondirectional alternative and let  = 0.05.)
If the timing of births is random, then 2/7 of all births should occur on weekends, and 5/7 on weekdays. Do
births deviate from this random timing?
Let 1 be the probability that a birth occurs on a weekend.
Let 2 be the probability that a birth occurs on a weekday.
H0: 1 = 2/7 = 0.286; 2 = 5/7 = 0.714. The births have random weekend/weekday timing
HA: 1  2/7, 2  5/7 ; The timing of births do not have random weekend/weekday timing.
Use a 2 GOF test. Xs2 = (O-E)2/E has a 21 dist'n under H0. Critical value is 3.84 with  = 0.05.
n = 932 births; observed 216 on weekends and 932-216 = 716 on weekdays
ˆ 1 =216/932 = 0.232 ˆ 2 = 716/932 = 0.768 (Note that ˆ 1 < 1 and ˆ 2 > 2 )
Expected under H0:
E1= n 1 = (932)(2/7) = 266.3
E2 = n 2 = (932)(5/7) = 665.7
X2s =

O  E 2
E

216  266.32  716  665.7 2
266.3
665.7
 9.50  3.80  13.3
Reject H0 since 13.3 > 3.84.
This study provides evidence that fewer births occur on weekends at this hospital than would be predicted by
random timing of the births.
Problem 10.5 (p. 400)
In a breeding experiment, white chickens with small combs were mated and produced 190 offspring, of (15)
the types shown in the accompanying table. Are these data consistent with the Mendelian expected
ratios of 9:3:3:1 for the four types? Use a chi-square test at  = 0.10.
Do the probabilities of the four feather/comb phenotypes fit the Mendelian 9:3:3:1 ratio?
Let 1 be the probability that offspring have white feathers, small comb.
Let 2 … white feathers, large comb. Let 3 … dark feathers, small comb. Let 4 … dark feathers, large comb.
H0: 1 = 9/16, 2 = 3/16, 3 = 3/16, 4 = 1/16 The four phenotypes follow the Mendelian 9:3:3:1 ratio.
HA: the probabilities do not follow the 9:3:3:1 ratio.
Use a chi-square GOF test. X2 = (O-E)2/E has a 23 dist'n under H0. With  = 0.10, the critical value is 6.25.
In 190 offspring, the expected numbers of each phenotype under H0 are
E1 = n 1 = (190) (9/16) = 106.9
E2 = n 2 = (190) (3/16) = 35.6
E3 = n 3 = (190) (3/16) = 35.6
E4 = n 4 = (190) (1/16) = 11.9
The test statistic is
X2 =

O  E 2
2
2
2
2




111  106.9
37  35.6
34  35.6
8  11.9




E
106.9
35.6
 0.157  0.055  0.072  1.278  1.562
35.6
11.9
Since 1.562 < 6.25, do not reject H0.
This study provides no evidence at the .10 significance level of deviation from the expected Mendelian ratios.
[or These data are consistent with the expected Mendelian ratios at .10 significance level.]
Problem 10.7 (p. 400)
In an agronomy experiment, peanuts with shriveled seeds were crossed with normal peanuts. The
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genetic model that the agronomists were considering predicted that the ratio of normal to shriveled
progeny would be 3 : 1. They obtained 95 normal and 54 shriveled progeny. Do these data sup.ort the
hypothesized model?
(a) Conduct a chi-square test with  = 0.05. Use a nondirectional alternative.
(b) State your conclusion from part (a) in the context of this setting.
Do the offspring of this cross follow a 3 normal : 1 shriveled ratio?
Let  be the probability of a normal offspring.
H0:  = 0.75; the ratio of normal to shriveled offspring is 3:1
HA:   0.75; the ratio of normal to shriveled offspring is not 3:1
Use a 2 GOF test. X2s = (O-E)2/E has a 21 distribution. Test at level  = 0.05. Critical value is 3.84. If X2s
> 3.84 then will reject H0.
n = 95 + 54 = 149.
E1 = 149  0.75 = 111.75
E2 = 149  0.25 = 37.25
X2s = (95–111.75)2/111.75 + (54 – 37.25)2/37.25 = 10.04
10.04 > 3.84 so reject H0
This study provides evidence at the 0.05 significance level that the ratio of normal to shriveled offspring is not
3:1.
Problem 10.17 (p. 409)
Most salamanders of the species P. cinereus are red-striped, but some individuals are all red. The allred form is thought to be a mimic of the salamander N. viridescens, which is toxic to birds. In order to
test whether the mimic form actually survives more successfully, 163 striped and 41 red individuals of
P. cinereus were exposed to predation by a natural bird population. After two hours, 65 of the striped
and 23 of the red individuals were still alive. Use a chi-square test to assess the evidence that the
mimic form survives more successfully. Use a directional alternative and let  = 0.05.
colour
total
striped
red
survived
65 (70.3)
23 (17.7)
88
outcome
eaten
98 (92.7)
18 (23.3)
116
total
204
163
41
X^2
3.514
(a)
(b)
(c)
(d)
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State the null hypothesis in words
State the null hypothesis in symbols
Find the value of the test statistic and the p-value
State the conclusion of the test in the context of this setting.
Does the red form survive more successfully?
Let 1 = Pr(survived | striped) ; 2 = Pr(survived | red)
[OR, Let 1 = Pr(eaten | striped) ; 2 = Pr(eaten | red)]
a) and b)
H0: 1 = 2 ; striped and red salamanders are equally likely to survive, OR, outcome and colour are independent
HA: 1 < 2 ; red (mimic) form survives more successfully
[this assumes first definition for 's; if  is in terms of "eaten" then should have Pr(eaten | striped)>Pr(eaten |
red)]
Use a 2 contingency table test. X2s = (O-E)2/E has a 21 dist'n under H0. Directional alternative with  =
0.05; critical value for df = 1 is 2.71. Will reject H0 if X2s > 2.71 AND ˆ 1 < ˆ 2 .
ˆ 1 = 65/163 = 0.399 ; ˆ 2 = 23/41 = 0.561. Observe that ˆ 1 < ˆ 2 so continue.
c)
Observed X2 = 3.5 > 2.71, so reject H0.
0.025 < P < 0.05
d)
This study provides evidence at the .05 significance level that the red salamander is more likely to survive bird
predation than the striped salamander is.
Problem 10.20 (p. 410)
Phenytoin is a standard anticonvulsant drug which unfortunately has many toxic side effects. A study
was undertaken to compare phenytoin with valproate, a new drug, in the treatment of epilepsy.
Patients were randomly allocated to receive either phenytoin or valproate for 12 months. Of 20
patients receiving valproate, 6 were free of seizures for the 12 months, while 6 of 17 patients receiving
phenytoin were seizure-free.
(a) Use a chi-square test to compare the seizure-free response rates for the two drugs. Let HA be nondirectional and  = 0.10.
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drug
seizure-free
outcome
seizures
total
X^2
0.118
valproate total
6 (6.5)
14 (13.5)
20
phenytoin
6 (5.5)
11 (11.5)
17
12
25
37
Do the two drugs differ in their seizure-free response rates?
Let 1 = Pr(seizure-free | phenytoin); Let 2 = Pr(seizure-free | valproate)
H0: 1 = 2 ; the two drugs have the same seizure-free response rate
HA: 1  2 ; the two drugs have different seizure-free response rates
Use a chi-square contingency table test. X2s = (O-E)2/E has a 21 dist'n under H0; with  = 0.10, critical
value is 2.71. Will reject H0 if X2s > 2.71.
Observe X2s = (0.5)2(1/5.5+1/11.5+1/6.5+1/13.5) = 0.12 < 2.71, so do not reject H0.
This study does not provide evidence at the 0.10 significance level that the two drugs have different seizure-free
response.
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(b) Does the test in part (a) provide evidence that valproate and phenytoin are equally effective in
preventing seizures? Discuss.
This test does not provide evidence that the two drugs are equally effective in preventing seizures. It shows only (3)
that the difference in effectiveness, if any, is too small to be detected in a study of this size. It is possible that a
difference does exist, which could be found in a larger study, in which case we would have made a type II error
in this study. A confidence interval for the difference in seizure-free response rates could be constructed to
bound the difference in rates.
[It is not necessary to construct a CI to receive full points, but here it is for interest’s sake.
ˆ 1 = 6/17 = 0.353 ; ˆ 2 = 6/20 = 0.3.
SE ˆ 
1
6  11
17 3
 0.116; SE ˆ
2

6  14
20 3
 0.102;
SE ˆ  ˆ  0.116 2  0.102 2  0.154
1
2
95% CI for 1 - 2 is 0.353-0.3  1.960.154 = [-0.25, 0.35].
The 1 could be as much as 0.35 larger than 2, or as much as 0.25 smaller than 2.]
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