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CONFIDENCE INTERVAL

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The NWBC found that 16.5% of women-owned businesses did not pro- vide any employee ben- efits. What sample size could be 99% confident that the estimated (sam- ple) proportion is within 6 percentage points of the true population propor- tion? Question Suppose a market re- searcher wants to deter- ...

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Alta - Chapter 8 - Confidence Intervals - Part 2
Study online at https://quizlet.com/_8whnz8

Correct an-
The NWBC found that
swers:$254$254
16.5% of women-owned
businesses did not pro-
EBP = 6%. ±=0.01T . he
vide any employee ben-
z-score is found
efits. What sample size
0.012=0.005=z±2=2.576
could be 99% confident
n=(z±2)2p(Ù
2)(q2)EBP2
that the estimated (sam-
=(2.5762)(0.165)(0.835)0.0
ple) proportion is within
n=253.96
6 percentage points of
the true population propor-
Rounded up to 254 peo-
tion?
ple.
Correct an-
Question
swers:$752$752
Suppose a market re-
Given the information in
searcher wants to deter-
the question, EBP=0.03
mine the current percent-
since 3%=0.03 and z±2=-
age of online shoppers
z0.05=1.645 because the
who are males. How many
confidence level is 90%.
online shoppers should
The values of p2 and q2
the researcher survey in
are unknown, but using
order to be 90% confident
a value of 0.5 for p2 will
that the estimated (sam-
result in the largest pos-
ple) proportion is within 3
sible product of p2q2, and
percentage points of the
thus the largest possible n.


, Alta - Chapter 8 - Confidence Intervals - Part 2
Study online at https://quizlet.com/_8whnz8

If p2=0.5, then q2=1 0.5=
true population proportion Therefore,
of online shoppers who n=z2p2q2EBP2=1.6452(0.5
are males? Round the answer up to
z0.101.282z0.051.645z0.02th 5e1.9
ne6x0tzi0n.t0e1g2e.r32
to6z0
be.0052
5 Use the table of values sure the sample size is
above. Round only at the large enough. The sample
final step. should include 752 online
shoppers.
Question
A recent survey asked
1,379 top executives
about business trends. Answer Explanation
The surveyed showed that Correct answers: 0.77
23% want to strengthen in-
novation to capitalize on The value of q2 is:
new opportunities. What is 1 p2=1 0.23=0.77
the value of q2 as a deci-
mal? Round to the nearest
hundredth.
Correct
an-
swers:1$0.652$0.652 2$0
To form the sample pro-
portion, take the number


, Alta - Chapter 8 - Confidence Intervals - Part 2
Study online at https://quizlet.com/_8whnz8

of successes, and divide
it by n, the number of tri-
Question
als. In this scenario, the
Alice wants to estimate
number of successes is
the percentage of people
the number of people who
who own a mountain bike.
own a mountain bike, 150.
She surveys 230 individu-
The total number of peo-
als and finds that 150 own
ple surveyed was 230.
a mountain bike. What are
p2=xn=150230=0.652
the sample proportions for So, p2=0.652 is the sample
successes, p2, and failures,
proportion, which is the
q2?
point estimate of the pop-
Round your answers to
ulation proportion. Since
three decimal places.
p2+q2=1, we can solve for
q2.q2=1 p2=1 0.652=0.3
Correct
Using p2=0.652, q2=0.348,an-
and n=230, what is the swers:1$0.590$0.590 2$0
95% confidence interval We know n=230, p2=0.652
for the proportion of the and q2=0.348. Since
population who own a the confidence level is
mountain bike? 95%, ±=1 0.95=0.0a5n,d
±2=0.025T . he z-value for
Use the table of common z0.025 is 1.96.
z-scores above. Round EBP=(z±2)(-
p2q2n )=(1.96)((


, Alta - Chapter 8 - Confidence Intervals - Part 2
Study online at https://quizlet.com/_8whnz8

So we can write the
your answers to three dec-
confidence interval as
imal places.
(0.652 0.062,0.652+0.062
The procedure to find the
confidence interval for a
proportion is similar to that
for the population mean,
but the formulas are a bit
different. We are still look-
ing for a range of numbers,
written as: (point estimate
- margin of error, point es-
timate + margin of error),
The Parameters for a Con- with (point estimate - mar-
fidence Interval with a Pro- gin of error) as the lower
portion value and (point estimate
+ margin of error) as the
upper value of the confi-
dence interval.
Since we're looking for
a point estimate for the
true proportion and mar-
gin of error for proportions
here, instead of means,
the point estimate will be
the sample proportion, p2,

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