Transcript Week 7, Lecture 3, Estimating the population proportion p

QBM117 - Business Statistics
Estimating the population proportion
Objectives

To give some examples where we need to estimate the
population proportion.

To describe the sampling distribution of the sample
proportion, p̂

To derive the confidence interval estimate of the
population proportion.
Estimating the population proportion
When the random variable of interest is qualitative, the
characteristic we usually consider is the proportion of
successes. For example, we may be interested in
•
the proportion of homes for sale, that have a modern
kitchen.
•
The proportion of voters who will vote for a particular
political candidate
In the topic on probability distributions, we saw that
the number of successes in n trials followed a
binomial distribution.
Now, instead of expressing the number of successes X, we
can easily convert this variable to the proportion of
successes, by dividing by the sample size, n.
Therefore, the point estimator of the population proportion
p, is the sample proportion, p hat,
X
pˆ  where X is the number of successesin n trials
n
The sampling distribution of
p̂
The sample proportion is approximately normal with
mean p and standard deviation  pˆ 
provided that
pq
n
np  5 and nq  5
Therefore the standardised random variable
z
pˆ  p
 pˆ
pˆ  p

pq
n
is approximately standard normally distributed
Finding an interval estimate of the population
proportion, p
The interval estimate of p is found by
pˆ  z / 2
pq
n
however, since we are trying to estimate p, we cannot
determine qp
ˆ and qˆ
therefore we must estimate p and q by p
n
Therefore the confidence interval estimator of p is
found by
pˆ qˆ
pˆ  z / 2
n
as long as npˆ  5 and nqˆ  5
Exercise 1 – Exercise 8.27 p270


Here we want to estimate the population proportion, p.
The sample proportion p̂ is the best estimator of p.
75
pˆ 
 0.3 and qˆ  0.7
250
Now since npˆ  75  5 and nqˆ  175  5
p̂ will be approximately normally distributed.
Therefore the confidence interval estimator of p is
found by
pˆ  z / 2
pˆ qˆ
n
pˆ  0.3 qˆ  0.7 n  250
pˆ qˆ
(0.3)( 0.7)
 pˆ can be estimated by

 0.029
n
250
1    0.99   0.01 Z 0.005  2.575
Therefore the confidence interval is given by
pˆ  z / 2
pˆ qˆ
 0.3  (2.575)(0.029)
n
 0.3  0.075
 99% CI ( p)  0.225 to 0.375
Example 2

In a random sample of 50 homeowners selected from a
large suburban area, 19 said that they had serious problems
with excessive noise from their neighbours.

Find a 99% confidence interval estimate of the percentage
of all homeowners in this suburban area who have such
problems.
Reading for next lecture

S&S Chapter 8 Sections 8.6 - 8.7
Exercises to be completed before next lecture

S&S 8.35