
Here’s the formula for a CI for p:
pz
*
p(1  p)
n
p-hat is our
unbiased
Estimate of p.
Z* is called the
critical value. I’ll
teach you how to
calculate that
next.
This is the
standard deviation
of p-hat. Notice
all the hats – we
don’t really ever
know p, so we use
p-hat to estimate
it.




You will be told the confidence level (i.e. 90%).
Draw a normal curve. Label the confidence level in the
MIDDLE.
Notice there is a portion in each tail that is unshaded.
 The value of one of those tails is found by subtracting from
1 and then dividing by 2.
 Look that value up in the BODY of Table A. Make it
positive.
i.e. 90% confidence  1 - .90 = 0.10. Divide by 2  0.05.
Look up in Table A. Z* is 1.64 or 1.65.


Find z* for 94% confidence.
Here are some common z* that you might want
to memorize:
90% confidence  z* = 1.645
 95% confidence  z* = 1.96
 99% confidence  z* = 2.575


Let’s look back at our formula.
pz

p (1  p)
n
The margin of error gets smaller when:



*
The sample size increases
The confidence level decreases
This is true for all CIs.

Every inference procedure will have three
conditions to check. Today, we are studying
constructing a CI for p.
SRS
 Normality

 Do you remember the normality check for proportions?
n p  10 and n(1  p )  10

Independence
 This is called the 10% rule in your book.
 Population ≥ 10n
State: What parameter do you want to estimate, and at what
confidence level?
Plan: Identify the appropriate inference method. Check conditions.
Do: If the conditions are met, perform calculations.
Conclude: Interpret your interval in the context of the problem.

One-Sample z Interval for a Population Proportion
Calculate and interpret a 90% confidence interval for the proportion of red beads in
a container. Your teacher claims 50% of the beads are red. You take a sample
of 251 and 107 are red. Use your interval to comment on this claim.
 sample proportion = 107/251 = 0.426
 Conditions?
 For a 90% confidence level, z* = 1.645
statistic ± (critical value) • (standard deviation of the statistic)
We are 90% confident that the
pˆ (1  pˆ )
pˆ  z *
interval from 0.375 to 0.477
n
captures the actual proportion of
(0.426)(1  0.426) red beads in the container.
 0.426  1.645
251
Since this interval gives a range
 0.426  0.051
of plausible values for p and since
 (0.375, 0.477)
0.5 is not contained in the
interval, we have reason to doubt
the claim.

The margin of error in a confidence interval
only accounts for sampling variability.

Other sources of bias can still make our results
inaccurate. Examples:
 Response bias (did some teens answer the question
untruthfully?)
 Nonresponse
 Wording




STAT, TESTS, A:”1-PropZInt”
X = Number of successes in the sample
N = Sample Size
C-Level = Confidence Level

The Gallup Youth Survey asked a SRS of 439
US teens aged 13 to 17 whether they thought
young people should wait to have sex until
marriage. Of the sample, 246 said “yes.”
Construct and interpret a 95% confidence
interval for the proportion of all US teens who
would say “yes” if asked this question.

Choosing the Sample Size
In planning a study, we may want to choose a sample size that allows
us to estimate a population proportion within a given margin of
error.
The margin of error (ME) in the confidence interval for p is
pˆ (1  pˆ )
ME  z *
n
 z* is the standard Normal critical value for the level of confidence we want.
Because the margin of error involves the sample proportion pˆ , we have to
guess the latter value
 when choosing n. There are two ways to do this :
• Use a guess for pˆ based on past experience or a pilot study
• Use pˆ  0.5 as the guess. ME is largest when pˆ  0.5
Sample Size for Desired Margin of Error
 To determine the sample size n that will yield a level C confidence interval
for a population proportion p with a maximum margin of error ME, solve

the following inequality for n: pˆ (1 pˆ )
z*
 ME
n
where pˆ is a guessed value for the sample proportion. The margin of error
will always be less than or equal to ME if you take the guess pˆ to be 0.5.

Example: Customer Satisfaction
Read the example on page 493. Determine the sample size needed to
estimate p within 0.03 with 95% confidence.
 The critical value for 95% confidence is z* = 1.96.
 Since the company president wants a margin of error of no more than
0.03, we need to solve the equation
pˆ (1  pˆ )
1.96
 0.03
n
Multiply both sides by
square root n and divide
both sides by 0.03.

Square both sides.

Substitute 0.5 for the
sample proportion to
find the largest ME
possible.

1.96
pˆ (1  pˆ )  n
0.03
1.96 2

 pˆ (1  pˆ )  n
0.03 
1.96 2

 (0.5)(1  0.5)  n
0.03 
1067.111  n
We round up to 1068
respondents to ensure
the margin of error is no
more than 0.03 at 95%
confidence.