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# PPS9_Solution_.doc

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York University

Mathematics and Statistics

MATH 2565

Jagmohan Chawla

Winter

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owner Page 149 10/01/2013owner
Page 149 10/01/2013123520813.doc
PPS 9 (Solution)
Problem19.23. Rickets.
a) Independence Assumption: It is reasonable to think that the
randomly selected children are mutually independent in
regards to vitamin D deficiency.
Randomization Condition: The 2,700 children were chosen at
random.
10% Condition: 2,700 children are less than 10% of all English
children.
Success/Failure Condition: np= (2,700)(0.20) = 540 and nq=
(2,700)(0.80) = 2160 are both greater than 10, so the sample
is large enough.
Since the conditions are met, we can use a one-proportion z-
interval to estimate the proportion of the English children
with vitamin D deficiency.
∗ pqˆ (0.2)0.80)
p ± z = (0.2)± 2.326 = (18.2%, 2.8%)
n 2700
b) We are 98% confident that between 18.2% and 21.8% of
English children are deficient in vitamin D.
c) About 98% of random samples of size 2,700 will produce
confidence intervals that contain the true proportion of
English children that are deficient in vitamin D.
24. Pregnancy.
a) Independence Assumption: There is no reason to believe that
one woman’s ability to conceive would affect others.
Randomization Condition: These women are not chosen at
random. Assume that they are representative of all women
under 40 that had previously been unable to conceive.
10% Condition: 207 women is less than 10% of all such
women.
Success/Failure Condition: np= 49 and nqˆ= 158 are both
greater than 10, so the sample is large enough.
Since the conditions are met, we can use a one-proportion z-
interval to estimate the proportion of the births to women at
the clinic.
49 158
p ±z∗ pq= 49 ±1.645 207(207=(18 .8%,28.5%)
n 207 207
b) We are 90% confident that between 18.8% and 28.5% of
women under 40 who are treated at this clinic will give birth.
c) About 90% of random samples of size 207 will produce
confidence intervals that contain the true proportion of
women under 40 who are treated at this clinic that will give
birth. owner Page 150 10/01/2013owner
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d) These data do not refute the clinics claim of a 25%
success rate, since 25% is in the interval.
Problem 20.17: Law school.
a) H 0 The law school acceptance rate for LSATisfaction is 63% (p
= 0.63)
H A The law school acceptance rate for LSATisfaction is greater
than 63% (p > 0.63)
b) Randomization Condition: These 240 students may be
considered representative of the population of law school
applicants.
10% Condition: There are certainly more than 2,400 law school
applicants.
np nq
Success/Failure Condition: = 151.2 and = 88.8 are both
greater than 10, so the sample is large enough.
The conditions have been satisfied, so a Normal model can be
used to model the sampling distribution of the proportion, with
μ = p = ˆ pq (0.63)(0.37)
pˆ 0.63 and σ(p) = = ≈ 0.0312
n 240
We can perform a one-proportion z-test. The observed success
163
rate is p= = 0.6792
240

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