IXL - Find confidence intervals for population proportions (Precalculus practice)

Executives at a television network commissioned a phone survey to find out whether people had heard of the network's new show. The executives surveyed 200 randomly chosen members of the show's target demographic.

Response	People
no	81
yes	119

Find a

90%

confidence interval for

the proportion of the target demographic who haven't heard of the show.

Round your answers to the nearest thousandth.

ref_doc_title.

Learn with an example

question

The president of an audio technology company is alarmed by the declining sales of the company's MP3 players. To better understand market trends, company employees asked 75 randomly chosen teenagers whether they currently own MP3 players.

Owns an mp3 player?	Teenagers
yes	9
no	66

Find a

95%

confidence interval for

the proportion of teenagers who do not own MP3 players.

Round your answers to the nearest thousandth.

key idea

You can use a random sample to calculate a

confidence interval for the population proportion. The population proportion is the proportion of items in a population with a specified characteristic.

A confidence interval may or may not contain the population proportion. A

confidence interval will contain the population proportion for

of random samples whose elements were chosen randomly and independently.

The

confidence interval for the population proportion,

has endpoints

where:

p
is the sample
proportion,
the proportion of items in the sample with the specified characteristic.
q
=
1
–
p
n
is the sample size.
z
is the critical value for the confidence level
c
%
.
It is the
z
-value
for which the interval from
–
z
to
z
is the middle
c
%
of the standard normal distribution.
c
%
–
z
z
Here are some common critical values:
Confidence level,
c
%
Critical value,
z
90%
1.645
95%
1.960
99%
2.576

The formula for the endpoints of a confidence interval can be used when all of the following inequalities hold:

n
≥
30
n
p
≥
5
n
q
≥
5

solution

The population proportion,

is the proportion of teenagers who do not own MP3 players.

To find a

95%

confidence interval for the population proportion, start by listing the given information:

The sample size is
n
=
75
.
In this sample, 66 teenagers answered no.
Owns an mp3 player? Teenagers
yes 9
no 66
The critical value for a
95%
confidence level is
z
=
1.960
.

Find

and

Use these values to find sample proportion,

and its complement,

0.88

–

0.12

Verify that the formula for the endpoints of a confidence interval can be used.

n
=
75
≥
30
n
p
=
75
66
75
=
66
≥
5
n
q
=
75
9
75
=
9
≥
5

Notice that

is the number of teenagers who answered no and

is the number of teenagers who answered yes.

Since all three inequalities are true, the formula for the endpoints of a confidence interval can be used.

Find the endpoints of a confidence interval.

The sample size is

0.88

0.12

and the critical value for a

95%

confidence interval is

1.960

Use these values to find the endpoints of a confidence level for the population proportion.

0.88

1.96

0.88

0.12

0.88

0.073545…

The left endpoint is:

0.88

–

0.073545…

0.806454…

Round to the nearest thousandth

≈

0.806

Round to the nearest thousandth

The right endpoint is:

0.88

0.073545…

0.953545…

Round to the nearest thousandth

≈

0.954

Round to the nearest thousandth

State the confidence interval.

So, a

95%

confidence interval for the proportion of teenagers who do not own MP3 players is:

0.806

0.954

Looking for more? Try this:

$Confetti$

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FF.5Find confidence intervals for population means

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Confidence level, c %	Critical value, z
90%	1.645
95%	1.960
99%	2.576