Transcript PP 8_3

Understandable Statistics
Eighth Edition
By Brase and Brase
Prepared by: Lynn Smith
Gloucester County College
Edited by: Jeff, Yann, Julie, and Olivia
Chapter 8: Estimation
Section 8.3
Estimating p in the Binomial
Distribution
Focus Points
• Compute the maximal margin of error for
proportions using a given level of
confidence.
• Compute confidence intervals for p and
interpret the results.
• Interpret poll results.
Statistics Quote
“There are two kinds of statistics, the kind
you look up and the kind you make
up.” —Rex Stout
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 > 5 and nq > 5, the binomial
distribution can be approximated by the
normal distribution.
A Point Estimate for p, the
Population Proportion of
Successes
r
pˆ ( read as " p hat " ) 
n
Point Estimate for q
(Population Proportion of
Failures)
qˆ ( read as " q hat " )  1  pˆ
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.
r 370
pˆ  
 0.74
n 500
We estimate that there is a 74% chance that any
given flight will depart on time.
Margin of error for “p hat” as a
Point Estimate for p
pˆ  p
A c Confidence Interval for p
for Large Samples
(np > 5 and nq > 5)
pˆ  E  p
 pˆ  E
r
where pˆ 
and E  z
n
c
pˆ (1  pˆ )
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?
npˆ  370
nqˆ  130
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.
npˆ and
nqˆ are
both  5
so the use of the normal distribution is justified.
Out of 500 departures, 370
departed on time. Find a 99%
confidence interval.
r 370
pˆ  
 0.74
n 500
.74(.26)
E  2.58
 0.0506
500
99% confidence interval for the
proportion of airplanes that
depart on time:
E = 0.0506
Confidence interval is:
ˆ E  p p
ˆE
p
.74  0.0506  p  .74  0.0506
0.6894  p  0.7906
99% confidence interval for
the proportion of airplanes that
depart on time
Confidence interval is
0.6894 < p < 0.7906
We are 99% confident that between 69%
and 79% of the planes depart on time.
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 margin of error is the maximal
error of estimate E for a
confidence interval.
Usually, a 95% confidence interval
is assumed.
Interpretation of Poll Results
The proportion responding in a certain way is
p̂
General interpretation of poll
results
• p-hat: the sample estimate of the
population proportion
• E, margin of error: maximal error of a 95%
confidence interval for p
A 95% confidence interval for
population proportion p is:
pˆ  m arg in of error  p  pˆ  m arg in 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%
The poll indicates (with 95%
confidence):
between 80.5% and 87.5% of the
population prefer the new
cereal.
Calculator Instructions
CONFIDENCE INTERVALS FOR THE PROBABILITY OF
SUCCESS p IN A BINOMIAL DISTRIBUTION
To find a confidence interval for a proportion, press the Stat
key and use option A:1-PropZlnt under TESTS. Notice
that the normal distribution will be used.
Example
• The public television station BPBS wants
to find the percent of its viewing population
who give donations to the station. 300
randomly selected viewers were surveyed,
and it was found that 123 made
contributions to the station. Find a 95%
confidence interval for the probability that
a viewer of BPBS selected at random
contributes to the station.
Example
• The letter x is used to
count the number of
successes (the letter r
is used in the text).
• Enter 123 for x and
300 for n. Use 0.95
for the C-level.
Example
• Highlight Calculate
and press Enter.
• The result is the
interval from 0.35 to
0.47.
Section 8.3, Problem 8
Law Enforcement: Escaped Convicts Case studies showed that out
of 10,351 convicts who escaped from u.s. prisons, only 7867 were
recaptured (The Book of Odds, by Shook and Shook, Signet).
(a)
Let p represent the proportion of all escaped convicts who will
eventually be recaptured. Find a point estimate for p.
(b) Find a 99% confidence interval for p. Give a brief statement of the
meaning of the confidence interval.
(c) Is use of the normal approximation to the binomial justified in this
problem? Explain.
Solution
Section 8.3, Problem 17
Lifestyle: Smoking In a survey of 1000 large corporations, 250 said
that, given a choice between a job candidate who smokes and an
equally qualified nonsmoker, the nonsmoker would get the job
(USA Today).
(a) Let p represent the proportion of all corporations preferring a
nonsmoking candidate. Find a point estimate for p.
(b) Find a 0.95 confidence interval for p.
(c) As a new writer, how would you report the survey results regarding
the proportion of corporations that would hire the equally qualified
nonsmoker? What is the margin of error based on a 95%
confidence interval?
Solution