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MTH 1210 Biostatistics
Apr 18’09
Homework #11 (answers)
(
Sampling Distribution of the Sample Proportion
)
Use the handout #
12
and MS Excel program
Three_distributions.xls,
sheet
3
(see the textbook
pgs 148, 149, 183)
Below:
is a sample
proportion for an arbitrary random sample;
p
is a
population
proportion; it is assumed below that the sample size
n
is large
enough and the sampling distribution of
is approximately normally distributed
with the mean
µ
= p ;
0
is the given value of
with respect to which a probability of interest is estimated;
*
is the value of
for a particular random sample under consideration.
σ
is a
standard deviation
of the sampling distribution of the sample proportion.
The sample size
n
is treated as "large enough" if
n > 20
and
if
both
np
and
n(1 – p
)
are greater than 5 .
5.5.1
. Smith et al. (A5) performed a retrospective analysis of data on 782 eligible patients
admitted with myocardial infarction to a 46bed cardiac service faculty. Of these patients, 248 (
32
percent) reported a past myocardial infarction. Use
0.32
as the
population
proportion. Suppose
50
subjects are chosen at random from the population, what is the probability that over
40
percent
would report previous myocardial infarction?
np =
16
> 5
n(1 – p ) =
34
> 5
p =
0.32
σ
= [p(1 – p)/n]
1/2
=
0.065970
σ
2
=
0.004352
n =
50
0
=
0.4
z
0
= (
0
– p)/σ
=
1.212678
P(
≤
0
) =
f(Z ≤ z
0
) =
0.887374
P(
>
0
) =
f(Z > z
0
) =
1 – f(Z ≤ z
0
) =
0.112626
ANS:
0.113
5.5.2
. In the study cited in the problem 5.5.1,
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 Spring '05
 Kzaer
 Statistics, Biostatistics

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