Transcript Estimating a population proportion

Estimating a population
proportion
ASW, 6.3, 7.6, 8.4
Economics 224 notes for October 20, 2008
Normal approximation to binomial (ASW, 6.3)
• If a probability experiment has n independent trials with p as
the probability of success and 1-p as the probability of failure,
the probabilities of the number of successes, x, have a
binomial probability distribution.
• The probabilities for x, where x = 0, 1, 2, 3, ... , n are given by
the expression
n!
x
( n x)
f ( x) 
p (1  p)
x!(n  x)!
• For small n, it is not too difficult to obtain the values of f(x)
with a calculator or from binomial tables.
• For large n, the calculation is more difficult if a computer
program is not available.
• Fortunately, when n is large, the normal probability
distribution can be used to approximate the binomial
probabilities.
Which normal distribution?
• For the binomial probability distribution, the mean and
standard deviation, respectively, are
  np and  np(1 p)
• If np ≥ 5 and n(1-p) ≥ 5, the normal distribution with the
above mean and standard deviation provides a reasonable
approximation to the binomial probabilities (ASW, 243).
• When calculating these, there is a continuity correction factor
(ASW, 243) that must be used. For example, the probability of
obtaining exactly 4 successes would be the area under the
normal curve between 3.5 and 4.5.
• The larger the value of n, the more closely the normal
distribution approximates the binomial probabilities.
Population proportion p
• When conducting research about a population, researchers
are often more interested in the proportion of a population
with a particular characteristic, rather than the number of
population elements with the characteristic.
–
–
–
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Proportion of population who support the Liberals.
Proportion of manufactured objects that are defect free.
Proportion of employees with extended health care plans.
Percentage of the labour force that is unemployed.
• In each of these situations, the actual number of population
elements with the characteristic will vary with the sample
size. But the aim of obtaining samples is to estimate the
proportion, or percentage, of the population with the
characteristic.
• Let the proportion of a population with a particular
characteristics be represented by p.
Terminology and notation for proportions
• p is the proportion of a population with a particular
characteristic.
• Draw a random sample of size n elements from the
population that contains N elements.
• Let x be the number of sample elements with the
characteristic.
• Define the sample proportion as p where
x
p
n
• That is, p is the proportion of elements of the sample of size
n that have the characteristic.
Sampling distribution of p
• If samples of size n are drawn from a population with
proportion p having a particular characteristic, the sample
proportion p will differ from sample to sample. Some
samples will have a larger proportion of sample elements with
the characteristic and some will have a smaller proportion.
The distribution of when therep is repeated sampling is
p
termed the sampling distribution of .
• If the sample size n is only a small proportion of the
population size N, the sampling distribution of p has a
binomial distribution with a mean of p and a standard
deviation of
p 
p(1  p)
n
• See ASW, 279-280 for these results.
Normal approximation for a proportion
• Recall that a binomial variable x has a mean of μ = np with
variance σ2 = np(1-p).
• For a binomial variable p = x/n, where x is divided by n, it
should make sense that the mean and standard deviation of x
divided by n produce a mean of μ = p and a standard
deviation
p 
p(1  p)
n
for x/n.
• If np ≥ 5 and n(1-p) ≥ 5, the normal distribution provides a
reasonable approximation to the binomial probabilities, so
the distribution of the sample proportion is approximated by
the normal distribution with the above mean and standard
deviation (ASW, 280-281).
• From this, the probability of different levels of sampling error
for the sample proportion can be calculated (ASW, 281-282).
Estimating a population proportion
• Let p be the proportion of a population with a particular
characteristic. If a large random sample of n elements of the
population is drawn from this population, the sample
proportion p  x n is approximated by a normal distribution
with mean and standard deviation, respectively, being
  p and  p 
p(1  p)
n
• Since the population proportion is unknown and is being
estimated, the above standard deviation is also unknown.
However, the sample proportion often is a reasonable
estimate of p, so in practice the mean and standard deviation,
respectively, of the distribution of the sample proportion are
  p and  p 
p(1  p)
n
Margin of error for a proportion
• From the previous slides, it follows that (1 – α)100% of the
random samples are associated with the following margin of
error E when estimating a population proportion:
p(1  p)
E  Z
2
n
• This result holds only if the sample size n is large, that is np ≥ 5
and n(1-p) ≥ 5, so the binomial probabilities are approximated
by areas under the normal distribution.
Interval estimate for a population
proportion p
• When n is large, the (1-α)100% confidence interval for
estimating p, the proportion of a population with a particular
characteristic, is
p  Z
2
p(1  p)
n
where p  x n is the sample proportion and x is the number of
sample elements with the characteristic.
• For this interval estimate, large n means
np  5 and n(1  p)  5
For smaller n, the interval will be wider than given by this
formula.
Example of opinion polling - I
• From the October 6, 2008 example of opinion polls prior to
the November 2003 Saskatchewan provincial election, what is
the margin of error for the Cutler poll?
• What is the interval estimate for the percentage of decided
voters who say they will vote NDP?
• Use the 95% level of confidence in each case.
Percentage of respondents, votes, and number of seats by
party, November 5, 2003 Saskatchewan provincial election
Political Party
CBC Poll, Cutler Poll,
Oct. 20-26 Oct. 29 –
Nov. 5 P
P
Election
Result
P
Number
of Seats
NDP
42%
47%
44.5%
30
Saskatchewan Party
39%
37%
39.4%
28
Liberal
18%
14%
14.2%
0
Other
1%
2%
1.9%
0
Total
100%
100%
100.0%
58
15%
16%
800
773
Undecided
Sample size (n)
Sources: CBC Poll results from Western Opinion Research, “Saskatchewan Election Survey for The
Canadian Broadcasting Corporation,” October 27, 2003. Obtained from web site
http://sask.cbc.ca/regional/servlet/View?filename=poll_one031028, November 7, 2003. Cutler poll
results provided by Fred Cutler and from the Leader-Post, November 7, 2003, p. A5.
Example of opinion polling - II
• For the Cutler poll, n = 773 and the conditions for a large
sample size appear to hold. Using even the smallest value for
the sample proportion reported (other at 2% or 0.02),
np  773 0.02  15.46  5 and n(1  p)  773 0.98  757.54  5
• Given this large n, the sample proportion is approximated by a
normal distribution. At 95% confidence level, the Z value is
1.96 and the margin of error is
E  Z
2
p(1  p)
0.5  0.5
 1.96
 1.96 0.0003234 1.96 0.017984 0.035
n
773
• In this case, a value of 0.5 is used for the estimate of the
sample proportion, since this produces the widest possible
margin of error.
Example of opinion polling - III
• For the Cutler poll, the margin of error is plus or minus 0.035
or 3.5 per cent, with 95% confidence. This means that with a
sample of size n = 773, the estimate of the proportion of the
population who support any political party may be incorrect
by as much as 3.5 percentage points in 95 out of 100 samples.
• Each public opinion poll should provide an estimate of the
margin of error when reporting poll results. The margin of
error is the amount E by which the sample proportion differs
from the population proportion, plus a confidence level.
• For purposes of generating this margin of error that applies to
any characteristic, use p  0.5 and this will provide an upper
bound for the estimated margin of error.
Example of opinion polling - IV
• For the 95% confidence interval for the estimate of the
proportion who support a party, note that the sample of
decided voters is only 84% of the 773 (16% were undecided)
so that the actual sample size was n = 0.84 x 773 = 649.
• For the NDP, the sample proportion is 0.47 and the conditions
for large sample size are met, so the normal distribution can
be used. At 95% confidence, Z = 1.96 and the interval is
p  Z
2
p(1  p)
0.47 0.53
 0.47  1.96
 0.47  1.96 0.0003838 0.47  0.03840
n
649
and the 95% interval estimate for the proportion who support
the NDP is from 0.432 to 0.508. Note that this interval
includes the actual proportion p = 0.445 who supported the
NDP in the election.
Sample size for a proportion
• For confidence level (1-α)100% and margin of error E, the
required sample size is determined by solving the following
expression for n.
E  Z
2
p(1  p)
n
• This gives the formula for sample size
2
 Z  p(1  p)

2

n
E2
Estimating sample size
• In the formula for sample size required for estimating a
proportion, the value of the sample proportion is unknown.
ASW (315) revise the formula to use a planning value p* giving
the formula
2
 Z  p* (1  p* )

2

n
E2
• When using the formula, if you let p* = 0.5, this produces the
maximum possible value for n for any given E and α.
• If you consider it possible that the population proportion
differs considerably from p = 0.5, say p  0.2 or p ≥ 0.8, then
use one of the guidelines in ASW (315).
Example of sample size for a proportion
• What sample size would be required to obtain an estimate of
the proportion of University of Regina students who use
Regina Transit to travel to the University, accurate to within 5
percentage points, with 90% confidence?
• For this question, neither the sample nor population
proportion are known so use a planning proportion of p* =
0.5. E = 0.05 and Z = 1.645. The required sample size is
2
 Z  p* (1  p* )

1.6452  0.5  0.5 0.67651
2

n


 270.6
E2
0.052
0.0025
• A random sample of n = 271 UR students will give at least the
precision necessary, and perhaps even greater precision.
• Assume that sampling method produces a random sample. If
N = 12,000, the sample is 2.3% of N, so the sample size is a
small proportion of the population size.
Notes about sample size for estimating a
population proportion
• Random sample of a population.
• If the sample size is a small proportion of the population size
(less than 5-10% of population), then it does not matter how
large the population is, the required n is independent of
population size.
• This formula is especially useful, since it does not require
knowledge of the population variability. If p* = 0.5 is used in
the above formula, the sample size will be more than
sufficient to achieve the required margin of error with the
specified level of confidence.
• Not too many nonsampling errors such as poorly constructed
questions, nonresponse, refusals, etc.
• For more complex sampling procedures, consult a text on
sampling procedures.
• Monday, Oct. 20 – we will discuss the above slides and then
have some time for review.
• Tuesday, Oct. 21, 3:30 – 4:30 p.m. Optional review period
with your two instructors. CL232.
• Wednesday, Oct. 22, 2:30 – 3:45 is the midterm. You are
permitted to bring a text, photocopies of the tables (normal, t,
binomial), and one extra sheet. Make sure you bring a
calculator. No communication with other individuals inside
or outside of the classroom using electronic devices.
• The midterm covers the topics discussed in class to October
20, that is, the assigned sections of chapters 1-8 of the text
and any additional materials discussed in class.
• We are hoping to have Assignment 3 graded and available to
pick up at the Tuesday review session. Answers will be
posted on UR Courses some time on Tuesday.