Homework 3 (21 points)
STAT 210
Name(s) _____SOLUTION______
Due: Thursday, September 27th by 4pm
1. Determine the appropriate null and alternative hypotheses for the each of the scenarios given
below. Please use symbol notation.
a. Becker et al. (International Journal of Eating Disorders 2003) conducted a study using a
sample of 50 ethnic Fijian women. The women completed a self-report questionnaire
on dieting and attitudes toward body shape and change. The researchers found that
five of the respondents reported at least weekly episodes of binge eating during the
previous 6 months. (2 points)
Research Question – Is there evidence to conclude that less than 20 percent of
the population of Fijian women engage in at least weekly episodes of binge
eating?
H0: p ≥ 0.20
Ha: p < 0.20
b. A study was designed to determine whether dogs can be trained to identify urine
specimens from individuals with bladder cancer. Dogs were first trained to discriminate
urine samples from patients with bladder cancer or with other conditions. After the
training was completed, the dogs had to pick one of seven new urine specimens. Each
time, only one of the seven urine specimens came from a patient with bladder cancer.
Out of 54 trials, the dogs identified the correct urine specimen 22 times. (2 points)
Research Question – Are dogs able to identify urine specimens from patients
with bladder cancer?
H0: p ≤ 0.14
Ha: p > 0.14
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2. Wild-type fruit flies have red eyes, but a recessive mutation produces white-eyed individuals. A
researcher wants to assess the frequency of heterozygous individuals from among red-eyed fruit
flies. A heterozygous red-eyed fly crossed with a white-eyed mutant will have a mixed progeny.
Of the 100 red-eyed fruit flies crossed with white-eyed mutants, 11 produced a mixed progeny.
a. Construct a 95% confidence interval by hand for the proportion of mixed progeny.
(3 points)
n = 100
pˆ =
11
= 0.11
100
z = 1.96
2(100)(0.11) + 1.962 1.96 1.962 + 4(100)(0.11)(0.89)
±
2 100 + 1.962 
2 100 + 1.962 
25.84
12.85
±
 0.12 ± 0.06  0.06 ≤ p ≤ 0.18
207.68 207.68
b. Interpret the 95% confidence interval constructed in part a. (3 points)
95% confident the true proportion of red-eyed fruit flies crossed with whiteeyed mutants with mixed progeny is between 0.06 and 0.18.
c. Using JMP, verify the 95% confidence interval you constructed in part a. Make sure to
provide your JMP output or a detailed sketch of the output below. (2 points)
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3. Genetic counselors work with pregnant women (usually women at high risk of fetal
abnormalities or those who might not be at high risk but screen positive for abnormalities based
on standard screening tests) and hypothesize that about one-half of all abnormalities are
trisomy 21, one-third trisomy 18, and the remainder are trisomy 13. (Trisomy indicates three
copies of a particular chromosome, e.g. 21, and reflects a particular abnormality associated with
that chromosome.) A random sample of 200 pregnant women who deliver babies with
abnormalities were studied and the results are summarized below.
Abnormality
Trisomy 21 Trisomy 18 Trisomy 13 Total
Observed Frequency
107
70
23
200
Research Question – Does the pattern of abnormalities observed fit the pattern
hypothesized by genetic counselors?
a. Set up the null and alternative hypotheses for this scenario. (2 points)
H0: ptrisomy 21 = 0.50
ptrisomy 18 = 0.33
ptrisomy 13 = 0.17
Ha: Two or more differ
b. Find the expected counts which would be used in the computation of the Chi-square
test statistic. Make sure to show how you computed the expected counts. (2 points)
Abnormality
Trisomy 21
Trisomy 18
Trisomy 13
Total
Expected Count
200(0.50) =
100
200(0.33) =
66
200(0.17) =
34
200
c. Using JMP, find the test statistic and p-value for this test. Paste or provide a detailed
sketch of the appropriate JMP below. (2 points)
Test Statistic = 4.2912
p-value = 0.1170
d. Based on the p-value found in part c, does this study provide evidence the observed
pattern of abnormalities fits the pattern hypothesized by genetic counselors? Explain.
(3 points)
Evidence (p-value = 0.0001) that the observed pattern of abnormalities does
not fit the pattern hypothesized by genetic counselors.
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