Estimating a Population Proportion
Center
• Mean = p
Spread
• standard deviation of
p̂ is
p(1− p)
n
• must have population at least 10 times as large as the
sample.
• The standard error of
p̂ is pˆ (1n− pˆ ) because we
don’t actually know p.
Shape
• If np>10 and n(1-p) >10, the distribution of p is
approximately normal.
Confidence Interval for a Population Proportion
1) State what you are estimating and the population.
2) Conditions:
• SRS
• Normality:
npˆ >10 and n(1− pˆ ) >10
• Independence: Each observation is independent.
When sampling without replacement, the population
is at least 10 times as large as the sample.
3)
pˆ (1− pˆ )
pˆ ± z *
4) Interpretation
n
Sample Size for a Desired Margin of Error
Given we are looking for a margin of error of
would set the expression to be ≤ m .
z*
p*(1− p*)
≤m
n
≤ m . We
Example:
An automobile manufacturer would like to know what proportion
of its customers are not satisfied with the service provided by the
local dealer. The customer relations department will survey a
random sample of customers and compute a 99% confidence
interval for the proportion who are not satisfied.
a) Past studies suggest that 20% of customers will not be
satisfied. What will the sample size have to be in order to
have a margin of error of 0.015?
2.576
.2(.8)
= .015
n
n = 4718.77 ~ 4719
b) When the SRS is actually contacted, 10% of the sample say
they are not satisfied. Construct a 99% confidence interval
using the inference steps.
We are estimating the proportion of dissatisfied customers at a
local dealership. Our population is all customers at the dealership.
We are using an SRS of customers In order to check normality:
np ≥ 10 and nq ≥ 10 . In order to check independence, we need
to check that the population is at least 10 times as large as the
sample.
The interval is:
We are 99% confident that the true proportion is between this
interval.