Sec 7.2:
LARGE SAMPLE CONFIDENCE INTERVAL FOR POPULATION PROPORTION P
Suppose I want to find the proportion of all AU students who are cigarette smokers( P ). How do I find
that? How do I construct a confidence interval for that?
We can use sample proportion
P̂
to estimate the population proportion P.
A ( 1 - α) % large sample confidence interval for population proportion P is given by the formula

P̂
Point estimate 
P̂
Z /2
Pˆ ( 1 - pˆ )
n
You need NOT
memorize this
formula
Margin of Error
is the sample proportion:
P̂
= X / n.
n is the sample size. X is called the number of successes out of n trials.
You need to
remember
this
For example, Suppose we randomly select 40 students from AU and find that only 10
students are cigarette smokers. Here n = 40 and X = 10.
P̂
= X / n = 10 / 40 = 1/4 = 0.25.
Using this value, how do we find the confidence interval?
You need not memorize the above formula.
You can use 1-propZ interval command.
Calculator Approach:
We can use
1-PropZInterval Command to find the confidence interval.
The calculator will ask you to enter 3 values. What are they? X value, n value and C-Level value.
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1-propZInterval
X value MUST be
less than n value
X:
n:
Please
remember
these three
things
C Level:
Calculate
►
Both X and n values
must be positive integers
n value should be greater than 30 for the above formula to work.
I have not solved this problem completely. I just set it up for you.
----------------------------------------------------------------------------------------------------------------------------Warm up problem:
Suppose X = 120 and n = 300. Find 99% confidence interval for population proportion.
Solution:
Did you notice that you are asked to construct a confidence interval for population proportion.
In this problem, n
= 300
X
= 120
➔ Did you notice that both X value and n value are positive integers.

➔ Did you also notice that n value is larger than X value.

➔ Finally, did you notice that the sample size n = 300 is LARGE.

So, we are justified in using 1-propZinterval command to solve this problem.
➔
Press STAT, move cursor to TESTS, choose 1-propZint
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Note: If you want 99%
confidence interval, then you
enter 0.99 here.
Highlight Calculate & press Enter
Answer:
99% confidence interval for population proportion ( 32.71% , 47.29% )
Calculate the SAMPLE PROPORTION.
Answer: Sample proportion 𝑃̂
= X / n = 120 / 300 = 0.40
It is 40%.
Sometimes, the homework management system will ask you to explain your answer with a
bell-curve picture. How do we do that? Look at the following question. How do we fill in the
blanks?
In this problem, we got the answer that 99% confidence interval for population proportion (
32.71% , 47.29% )
How do we represent this in the picture?
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99% = This represents the area in the middle of the bell curve.
( This 99% is denoted as ( 1 – α ) % in the bell curve )
The remaining is 100% - 99% = 1%
( This 1% has to be distributed equally to both sides of the bell curve. ( 0.01 / 2 = 0.005 ).
( This 0.005 is denoted as α /2 % in the bell curve ).
32.71% and 47.29% are the lower and upper cutoff limit points on the x-axis of the
bell curve.
You can look at my answer below:
32.71%
47.29%
0.005
0.99
0.005
32.71%
7.29 %
You need to
remember this
little formula
ERROR BOUND: What is it?
It is the other name of ‘MARGIN OF ERROR”. How do we find it?
Margin of error = ( Upper limit - Lower limit ) / 2
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Margin of error = ( 47.29% - 32.71% ) / 2
= 14.58 /2 =7.29 %
What does 7.29 mean?
It tells you with 99% confidence that the difference between the population Proportion P and
the sample proportion of 40% will be NO MORE THAN 7.29%
--------------------------------------------------------------------------------------------------------------------------Now you are ready to do applied problems.
Example-1:
A company wants to place an ad in the business news section of the popular TIME
magazine.
Because of the high cost of advertisement, the company decides to place an ad only when at
least 35 % of subscribers regularly read the business section.
In a random sample of 500 subscribers, 200 state that they are regular readers of the business news.
Construct a 90 % confidence interval and interpret the result.
Solution:
Did you notice that you are asked to construct a confidence interval for population proportion.
In this problem, n
X
= 500
= 200
because we selected a random sample of 500 subscribers.
because the number of subscribers who read the business news is 200.
➔ Did you notice that both X value and n value are positive integers.

➔ Did you also notice that n value is larger than X value. 
➔ Finally, did you notice that the sample size n = 500 is LARGE. 
So, we are justified in using 1-propZinterval command to solve this problem.
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➔
Press STAT, move cursor to TESTS, choose 1-propZint
Note: Both X and n values are
integers.
X value is less than n value
Now enter your X value, n value and C-Level value.
Highlight Calculate & press Enter
This interval is your answer
= 0.4 = Sample proportion
Interpretation:
We are 90% confident that the true population proportion of all TIME magazine subscribers who
regularly read the business news is between 36.396% and 43.604%.
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(b)
Identify the POPULATION in this problem.
The Population is ALL TIME MAGAZINE SUBSCRIBERS.
(c)
What is the Parameter that is being estimated in this problem.
P
=
Proportion of ALL TIME magazine subscribers who regularly read the business
section.
(C)
What is the Point Estimate of the parameter in this problem?
The sample proportion
P̂
= X / n = 200 / 500 = 40%.
40% is the best point estimate of the parameter (Proportion of ALL TIME magazine
subscribers who regularly read the business section).
( D ) Compute & Explain is the Margin of Error in the estimation of the parameter.
Margin of error =
=
=
Upper Limit - Lower Limit
2
You MUST
remember this
formula
0.43604 − 0.36396
2
0.07208
= 0.03604 = 3.604%
2
Interpretation of Margin of Error: otherwise known as ERROR BOUND
We are 95% confident that the difference between the population proportion of all TIME
magazine subscribers who regularly read the business section and the sample proportion of
40% is no more than 3.604%.
The other way of interpreting the Margin of error is:
We are 90% confident that the sample proportion of 40% is within 3.604% of the true
population proportion of all TIME magazine subscribers who regularly read the business
section.
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(e)
Based on your results obtained, would you recommend placing an ad in the magazine or not?
Why? Provide reason.
Yes, I would highly recommend placing an ad in the magazine.
Criteria: The company decides to place an ad only when at least 35 % of subscribers regularly read
the business section.
The interval I obtained in this problem is from 36.396% to 43.604%. My interval is certainly ‘at
least 35% .
( f ) Explain your answer in part-a with a bell-curve picture. How do we do that? Look at the
following question. How do we fill in the blanks?
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In this problem, we got the answer that 90% confident that the true population proportion of all
TIME magazine subscribers who regularly read the business news is between 36.396% and
43.604%.
How do we represent this in the picture?
90% = This represents the area in the middle of the bell curve.
( This 90% is denoted as ( 1 – α ) % in the bell curve )
The remaining is 100% - 90% = 10%
( This 10% has to be distributed equally to both sides of the bell curve. ( 0.10 / 2 = 0.05 ).
( This 0.05 is denoted as α /2 % in the bell curve ).
36.40% and 43.60% are the lower and upper cutoff limit points on the x-axis of the
bell curve.
You can look at my answer below:
36.40%
43.60%
0.05
0.90
0.05
40%
36.40%
43.60%
---------------------------------------------------------------------------------------------------------------------------Augusta University
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Example-2:
A survey collected data on attitudes toward the quality of customer service in retail stores. The
survey found that 28% of Americans feel customer service is better today than it was two years ago.
If 650 adults were included in the sample, develop a 95% confidence interval for the proportion of
adults who feel customer service is better today than it was 2 years ago and interpret the result.
Solution:
Did you notice that you are asked to construct a confidence interval for population proportion.
In this problem, n
= 650
because we selected a random sample of 650 adults.
X = ? It is not given to us directly.
But, the problem says that ‘28% of Americans feel customer service is better today’
X=
28% of 650 = 182
So, X = 182.
➔ Did you notice that both X value and n value are positive integers. 
➔ Did you also notice that n value is larger than X value.

➔ Finally, did you notice that the sample size n = 650 is LARGE. 
So, we are justified in using 1-propZinterval command to solve this problem.
➔
Press STAT, move cursor to TESTS, choose 1-propZint
Note: Both X and n values
are integers.
X value is less than n value
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( 24.548 % , 31.452 % )
Written Interpretation:
We are 95% confident that the true population proportion of all ALL Americans who feel customer
service is better today than it was two years ago is between 24.548 % and 31.452%.
Pictorial Interpretation:
24.55%
31.45%
0.025
0.95
0.025
28%
24.55%
31.45%
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(b)
What is the Parameter that is being estimated in this problem.
P = Proportion of ALL Americans who feel customer service is better today than it was two
years ago.
(C)
What is the Point Estimate of the parameter
The sample proportion 𝑃̂
= 28%.
( D ) Compute & Explain is the Margin of Error in the estimation of the parameter.
Margin of error =
=
Upper Limit - Lower Limit
2
0.31452 − 0.24548
0.06904
=
= 0.03452 = 3.452%
2
2
Interpretation of Margin of Error:
We are 95% confident that the difference between the population proportion of all Americans who
feel customer service is better today than it was two years ago and the sample proportion of 28% is
no more than 3.452%.
( e ) Suppose a company claimed that 30% of its customers feel customer service is better today than
it was two years ago. Based on the results obtained, would you support the company’s claim or not?
Why? Provide reason.
We obtained the true population proportion of all ALL Americans who feel customer service is
better today than it was two years ago is between 24.548 % and 31.452%.
Since this interval of ( 24.548% , 31.452 % ) contain the company’s claimed percentage of 30%,
I would support the company’s claim.
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( f ) Suppose a company claimed that 40% of its customers feel customer service is better today than
it was two years ago. Based on the results obtained, would you support the company’s claim or not?
Why? Provide reason.
We obtained the true population proportion of all ALL Americans who feel customer service is
better today than it was two years ago is between 24.548 % and 31.452%.
Since this interval of ( 24.548% , 31.452 % ) does not contain the company’s claimed
percentage of 40%, I would not support the company’s claim.
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