Chapter 7
Estimation
Section 7.3
Estimating p in
the Binomial
Distribution
Review of the
Binomial Distribution
• Completely determined by the number
of trials (n) and the probability of
success (p) in a single trial.
• q=1–p
• If np and nq are both > 5, the binomial
distribution can be approximated by the
normal distribution.
A Point Estimate for p, the
Population Proportion of
Successes
Point Estimate for q
(Population Proportion of
Failures)
For a sample of 500 airplane
departures, 370 departed on
time. Use this information to
estimate the probability that
an airplane from the entire
population departs on time.
We estimate that there is a 74% chance that
any given flight will depart on time.
Error of Estimate for “p hat”
as a Point Estimate for p
Confidence Interval for p for
Large Samples (np and nq > 5)
pˆ  E  p  pˆ  E
r
where pˆ 
n
and
E  zc
pˆ (1 pˆ )
pˆ qˆ
 zc
n
n
zc = critical value for confidence level c
taken from a normal distribution
For a sample of 500 airplane
departures, 370 departed on
time. Find a 99% confidence
interval for the proportion of
airplanes that depart on time.
Is the use of the normal distribution justified?
n  500
pˆ  0.74
For a sample of 500 airplane
departures, 370 departed on
time. Find a 99% confidence
interval for the proportion of
airplanes that depart on time.
Can we use the normal distribution?
For a sample of 500 airplane
departures, 370 departed on
time. Find a 99% confidence
interval for the proportion of
airplanes that depart on time.
So the use of the normal distribution
is justified.
Out of 500 departures, 370
departed on time. Find a
99% confidence interval.
r  370
n  500
r
pˆ 
n
370
pˆ 
500
pˆ  0.74
130
qˆ 
500
qˆ  0.26
99% confidence interval for the proportion
of airplanes that depart on time:
Confidence interval is:
0.74(0.26)
0.74(0.26)
 p  0.74  2.5758
0.74  2.5758
500
500
0.1924
0.1924
 p  0.74  2.5758
0.74  2.5758
500
500
0.74  2.5758 0.003848  p  0.74  2.5758 0.0003848
0.74  2.5758 0.01961  p  0.74  2.5758 0.01961
.74  0.0505  p  .74  0.0505
0.6895  p  0.7905
We can say with 99% confidence that the
population proportion of planes that depart on
time is between 0.6895 and 0.7905.
The point estimate and the
confidence interval do not
depend on the size of the
population.
The sample size, however,
does affect the accuracy of
the statistical estimate.
Margin of Error
the maximal error of estimate E
for a confidence interval
E  zc
pˆ qˆ
n
Interpretation of Poll Results
The proportion responding
in a certain way is
A 95% confidence interval
for population proportion
p is:
pˆ  margin of error  p  pˆ  margin of error
pˆ  poll report
Interpret the following poll
results:
“ A recent survey of 400 households
indicated that 84% of the households
surveyed preferred a new breakfast
cereal to their previous brand. Chances
are 19 out of 20 that if all households had
been surveyed, the results would differ by
no more than 3.5 percentage points in
either direction.”
“Chances are 19 out of 20 …”
19/20 = 0.95
A 95% confidence interval is
being used.
“... 84% of the households
surveyed preferred …”
84% represents the percentage
of households who preferred
the new cereal.
84% represents pˆ .
“... the results would differ by
no more than 3.5 percentage
points in either direction.”
3.5% represents the
margin of error, E.
The confidence interval is:
84% - 3.5% < p < 84% + 3.5%
80.5% < p < 87.5%
We cay say with 95% confidence that
the population proportion of people
that would prefer the new cereal is
between 80.5% and 87.5%.
Sample Size for Estimating p
for the Binomial Distribution
Do you have a preliminary study?
Formula for Minimum Sample
Size for Estimating p for the
Binomial Distribution
If p is an estimate of
the population proportion,
zc 

n  pˆ qˆ  
 E
2
Formula for Minimum Sample
Size for Estimating p for the
Binomial Distribution
If we have no preliminary estimate for p, the
probability is at least c that the point estimate
r/n for p will be in error by less than the
quantity E if n is at least:
1  zc 
n  
4 E
2
The manager of a furniture store
wishes to estimate the proportion
of orders delivered by the
manufacturer in less than three
weeks. She wishes to be 95% sure
that her point estimate is in error
either way by less than 0.05.
Assume no preliminary study is
done to estimate p.
She wishes to be 95% sure ...
z0.95 = 1.9600
... that her point estimate is in
error either way by less than
0.05.
E = 0.05
... no preliminary study is done
to estimate p.
2
1  1.9600 
n 

4  0.05 
1
2
n  39.2 
4
1
n  1536.64 
4
n  384.16
n  385
The minimum required
sample size would be 385
deliveries to construct a 95%
confidence interval for the
proportion of deliveries
completed within 3 weeks with
an error of no more than 0.05
and no preliminary study.
Determine the minimum required
sample size if a preliminary study had
been conducted.
A preliminary estimate of p
indicated that p was
approximately equal to 0.75:
 1.9600 
n  0.75 0.25 
 0.05 
2
n  0.75 0.25 39.2 
2
n  0.75 0.25 1536.64 
n  288.12
n  289
The minimum required sample size would be 289
deliveries to construct a 95% confidence interval for
the proportion of deliveries completed within 3 weeks
with an error of no more than 0.05 and a preliminary
study indicating that the proportion of deliveries that
were completed within 3 weeks was 0.75.