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Estimation and Confidence Intervals 313

12. The U.S. Dairy Industry wants to estimate the mean yearly milk consumption. A sample of
16 people reveals the mean yearly consumption to be 60 gallons with a standard deviation
of 20 gallons.
a. What is the value of the population mean? What is the best estimate of this value?
b. Explain why we need to use the t distribution. What assumption do you need to make?
c. For a 90 percent confidence interval, what is the value of t?
d. Develop the 90 percent confidence interval for the population mean.
e. Would it be reasonable to conclude that the population mean is 63 gallons?
13. Merrill Lynch Securities and Health Care Retirement Inc. are two large employers in down-
town Toledo, Ohio. They are considering jointly offering child care for their employees. As a
part of the feasibility study, they wish to estimate the mean weekly child-care cost of their
employees. A sample of 10 employees who use child care reveals the following amounts
spent last week.

$107 $92 $97 $95 $105 $101 $91 $99 $95 $104

Develop a 90 percent confidence interval for the population mean. Interpret the result.
14. The Greater Pittsburgh Area Chamber of Commerce wants to estimate the mean time
workers who are employed in the downtown area spend getting to work. A sample of
15 workers reveals the following number of minutes spent traveling.

29 38 38 33 38 21 45 34
40 37 37 42 30 29 35

Develop a 98 percent confidence interval for the population mean. Interpret the result.

9.4 A Confidence Interval for a Proportion

The material presented so far in this chapter uses
the ratio scale of measurement. That is, we use
such variables as incomes, weights, distances, and
ages. We now want to consider situations such as
the following:
LO5 Compute a • The career services director at Southern Tech-
confidence interval for nical Institute reports that 80 percent of its
a population proportion. graduates enter the job market in a position
related to their field of study.
• A company representative claims that 45 per-
cent of Burger King sales are made at the
drive-through window.
• A survey of homes in the Chicago area indi-
cated that 85 percent of the new construc-
tion had central air conditioning.
• A recent survey of married men between the
ages of 35 and 50 found that 63 percent felt
that both partners should earn a living.
These examples illustrate the nominal scale
of measurement. When we measure with a nom-
inal scale, an observation is classified into one of two or more mutually exclusive
groups. For example, a graduate of Southern Tech either entered the job market in
a position related to his or her field of study or not. A particular Burger King customer
either made a purchase at the drive-through window or did not make a purchase at
the drive-through window. There are only two possibilities, and the outcome must be
classified into one of the two groups.
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314 Chapter 9

PROPORTION The fraction, ratio, or percent indicating the part of the sample or the
population having a particular trait of interest.

As an example of a proportion, a recent survey indicated that 92 out of 100

surveyed favored the continued use of daylight savings time in the summer. The
Statistics in Action sample proportion is 92/100, or .92, or 92 percent. If we let p represent the sam-
ple proportion, X the number of “successes,” and n the number of items sampled,
Many survey results
we can determine a sample proportion as follows.
reported in newspa-
pers, in news maga-
zines, and on TV use
confidence intervals. X
SAMPLE PROPORTION p⫽ [9–3]
For example, a recent n
survey of 800 TV
viewers in Toledo,
Ohio, found 44 per-
The population proportion is identified by ␲. Therefore, ␲ refers to the percent
cent watched the
of successes in the population. Recall from Chapter 6 that ␲ is the proportion of
evening news on the
“successes” in a binomial distribution. This continues our practice of using Greek
local CBS affiliate.
letters to identify population parameters and Roman letters to identify sample
The article went on to
statistics.
indicate the margin
To develop a confidence interval for a proportion, we need to meet the follow-
of error was 3.4 per-
ing assumptions.
cent. The margin of
error is actually the 1. The binomial conditions, discussed in Chapter 6, have been met. Briefly, these
amount that is added conditions are:
and subtracted from a. The sample data is the result of counts.
the point estimate to b. There are only two possible outcomes. (We usually label one of the outcomes
find the endpoints of a “success” and the other a “failure.”)
a confidence interval. c. The probability of a success remains the same from one trial to the next.
From formula (9–4) d. The trials are independent. This means the outcome on one trial does not
and the 95 percent affect the outcome on another.
level of confidence: 2. The values n␲ and n(1 ⫺ ␲) should both be greater than or equal to 5. This
condition allows us to invoke the central limit theorem and employ the standard
p (1 ⫺ p)
z normal distribution, that is, z, to complete a confidence interval.
C n

.44(1 ⫺ .44)
Developing a point estimate for a population proportion and a confidence inter-
⫽ 1.96 val for a population proportion is similar to doing so for a mean. To illustrate, John
C 800
Gail is running for Congress from the third district of Nebraska. From a random
⫽ 0.034 sample of 100 voters in the district, 60 indicate they plan to vote for him in the
upcoming election. The sample proportion is .60, but the population proportion is
unknown. That is, we do not know what proportion of voters in the population will
vote for Mr. Gail. The sample value, .60, is the best estimate we have of the
unknown population parameter. So we let p, which is .60, be an estimate of ␲,
which is not known.
To develop a confidence interval for a population proportion, we use:

CONFIDENCE INTERVAL FOR A p(1 ⫺ p)

p⫾z [9–4]
POPULATION PROPORTION C n

An example will help to explain the details of determining a confidence interval

and the result.
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Estimation and Confidence Intervals 315

Example The union representing the Bottle Blowers of America (BBA) is considering a pro-
posal to merge with the Teamsters Union. According to BBA union bylaws, at least
three-fourths of the union membership must approve any merger. A random sample
of 2,000 current BBA members reveals 1,600 plan to vote for the merger proposal.
What is the estimate of the population proportion? Develop a 95 percent confidence
interval for the population proportion. Basing your decision on this sample informa-
tion, can you conclude that the necessary proportion of BBA members favor the
merger? Why?

Solution First, calculate the sample proportion from formula (9–3). It is .80, found by
X 1,600
p⫽ ⫽ ⫽ .80
n 2,000
Thus, we estimate that 80 percent of the population favor the merger proposal. We
determine the 95 percent confidence interval using formula (9–4). The z value cor-
responding to the 95 percent level of confidence is 1.96.

p(1 ⫺ p) .80(1 ⫺ .80)

p⫾z ⫽ .80 ⫾ 1.96 ⫽ .80 ⫾ .018
C n C 2,000
The endpoints of the confidence interval are .782 and .818. The lower endpoint is
greater than .75. Hence, we conclude that the merger proposal will likely pass because
the interval estimate includes values greater than 75 percent of the union membership.

To review the interpretation of the confidence interval: If the poll was conducted
100 times with 100 different samples, the confidence intervals constructed from
95 of the samples would contain the true population proportion. In addition, the inter-
pretation of a confidence interval can be very useful in decision making and play a
very important role especially on election night. For example, Cliff Obermeyer is run-
ning for Congress from the 6th District of New Jersey. Suppose 500 voters are con-
tacted upon leaving the polls and 275 indicate they voted for Mr. Obermeyer. We
will assume that the exit poll of 500 voters is a random sample of those voting
in the 6th District. That means that 55 percent of those in the sample voted for
Mr. Obermeyer. Based on formula (9–3):
X 275
p⫽ ⫽ ⫽ .55
n 500
Now, to be assured of election, he must earn more than 50 percent of the votes
in the population of those voting. At this point, we know a point estimate, which is
.55, of the population of voters that will vote for him. But we do not know the per-
cent in the population that will ultimately vote for the candidate. So the question is:
Could we take a sample of 500 voters from a population where 50 percent or less of
the voters support Mr. Obermeyer and find that 55 percent of the sample support him?
To put it another way, could the sampling error, which is p ⫺ ␲ ⫽ .55 ⫺ .50 ⫽ .05 be
due to chance, or is the population of voters who support Mr. Obermeyer greater
than .50. If we develop a confidence interval for the sample proportion and find that
.50 is not in the interval, then we conclude that the proportion of voters supporting
Mr. Obermeyer is greater than .50. What does that mean? Well, it means he should
be elected! What if .50 is in the interval? Then we conclude that it is possible that
50 percent or less of the voters support his candidacy and we cannot conclude that
he will be elected based on the sample information. In this case, using the 95 per-
cent significance level and formula (9–4):

p(1 ⫺ p) .55(1 ⫺ .55)

p⫾z ⫽ .55 ⫾ 1.96 ⫽ .55 ⫾ .044
C n C 500
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316 Chapter 9

So the endpoints of the confidence interval are .55 ⫺ .044 ⫽ .506 and .55 ⫹ .044 ⫽
.594. The value of .50 is not in this interval. So we conclude that probably more than
50 percent of the voters support Mr. Obermeyer and that is enough to get him elected.
Is this procedure ever used? Yes! It is exactly the procedure used by television
networks, news magazines, and polling organizations on election night.

Self-Review 9–3 A market survey was conducted to estimate the proportion of homemakers who would
recognize the brand name of a cleanser based on the shape and the color of the con-
tainer. Of the 1,400 homemakers sampled, 420 were able to identify the brand by name.
(a) Estimate the value of the population proportion.
(b) Develop a 99 percent confidence interval for the population proportion.
(c) Interpret your findings.

Exercises
15. The owner of the West End Kwick Fill Gas Station wishes to determine the proportion of
customers who use a credit card or debit card to pay at the pump. He surveys 100 cus-
tomers and finds that 80 paid at the pump.
a. Estimate the value of the population proportion.
b. Develop a 95 percent confidence interval for the population proportion.
c. Interpret your findings.
16. Ms. Maria Wilson is considering running for mayor of the town of Bear Gulch, Montana.
Before completing the petitions, she decides to conduct a survey of voters in Bear Gulch.
A sample of 400 voters reveals that 300 would support her in the November election.
a. Estimate the value of the population proportion.
b. Develop a 99 percent confidence interval for the population proportion.
c. Interpret your findings.
17. The Fox TV network is considering replacing one of its prime-time crime investigation
shows with a new family-oriented comedy show. Before a final decision is made, net-
work executives commission a sample of 400 viewers. After viewing the comedy, 250
indicated they would watch the new show and suggested it replace the crime investiga-
tion show.
a. Estimate the value of the population proportion.
b. Develop a 99 percent confidence interval for the population proportion.
c. Interpret your findings.
18. Schadek Silkscreen Printing Inc. purchases plastic cups on which to print logos for
sporting events, proms, birthdays, and other special occasions. Zack Schadek, the owner,
received a large shipment this morning. To ensure the quality of the shipment, he selected
a random sample of 300 cups. He found 15 to be defective.
a. What is the estimated proportion defective in the population?
b. Develop a 95 percent confidence interval for the proportion defective.
c. Zack has an agreement with his supplier that he is to return lots that are 10 percent
or more defective. Should he return this lot? Explain your decision.

9.5 Choosing an Appropriate Sample Size

When working with confidence intervals, one important variable is sample size.
However, in practice, sample size is not a variable. It is a decision we make so that
our estimate of a population parameter is a good one. Our decision is based on
LO6 Calculate the three variables:
required sample size to
estimate a population 1. The margin of error the researcher will tolerate.
proportion or population 2. The level of confidence desired, for example, 95 percent.
mean. 3. The variation or dispersion of the population being studied.
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Estimation and Confidence Intervals 317

The first variable is the margin of error. It is designated as E and is the amount
that is added and subtracted to the sample mean (or sample proportion) to deter-
mine the endpoints of the confidence interval. For example, in a study of wages,
we may decide that we want to estimate the population average wage with a mar-
gin of error of plus or minus $1,000. Or in an opinion poll, we may decide that we
want to estimate the population proportion with a margin of error of plus or minus
5 percent. The margin of error is the amount of error we are willing to tolerate in
estimating a population parameter. You may wonder why we do not choose small
margins of error. There is a trade-off between the margin of error and sample size.
A small margin of error will require a larger sample and more money and time to
collect the sample. A larger margin of error will permit a smaller sample and a wider
confidence interval.
The second choice is the level of confidence. In working with confidence inter-
vals, we logically choose relatively high levels of confidence such as 95 percent and
99 percent. To compute the sample size, we need the z-statistic that corresponds
to the chosen level of confidence. The 95 percent level of confidence corresponds
to a z value of 1.96, and a 99 percent level of confidence corresponds to a z value
of 2.58. Notice that larger sample sizes (and more time and money to collect the
sample) correspond with higher levels of confidence. Also, notice that we use a z-
statistic.
The third choice to determine the sample size is the population standard devi-
ation. If the population is widely dispersed, a large sample is required. On the other
hand, if the population is concentrated (homogeneous), the required sample size will
be smaller. Often, we do not know the population standard deviation. Here are three
suggestions for finding a value for the population standard deviation.
1. Conduct a pilot study. This is the most common method. Suppose we want
an estimate of the number of hours per week worked by students enrolled in
the College of Business at the University of Texas. To test the validity of our
questionnaire, we use it on a small sample of students. From this small sam-
ple, we compute the standard deviation of the number of hours worked and
use this value as the population standard deviation.
2. Use a comparable study. Use this approach when there is an estimate of the
standard deviation from another study. Suppose we want to estimate the num-
ber of hours worked per week by refuse workers. Information from certain state
or federal agencies that regularly study the workforce may provide a reliable
value to use for the population standard deviation.
3. Use a range-based approach. To use this approach, we need to know or have
an estimate of the largest and smallest values in the population. Recall from
Chapter 3, the Empirical Rule states that virtually all the observations could be
expected to be within plus or minus 3 standard deviations of the mean, assum-
ing that the distribution follows the normal distribution. Thus, the distance
between the largest and the smallest values is 6 standard deviations. We can
estimate the standard deviation as one-sixth of the range. For example, the
director of operations at University Bank wants to estimate the number of
checks written per month by college students. She believes that the distribu-
tion of the number of checks written follows the normal distribution. The mini-
mum and maximum numbers written per month are 2 and 50, so the range is
48, found by (50 ⫺ 2). Then 8 checks per month, 48/6, would be the value we
would use for the population standard deviation.

Sample Size to Estimate a Population Mean

To estimate a population mean, we can express the interaction among these three
factors and the sample size in the following formula. Notice that this formula is the
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318 Chapter 9

margin of error used to calculate the endpoints of confidence intervals to estimate

a population mean!
␴
E⫽z
1n
Solving this equation for n yields the following result.

n⫽a b
SAMPLE SIZE FOR ESTIMATING z␴ 2
[9–5]
THE POPULATION MEAN E

where:
n is the size of the sample.
z is the standard normal value corresponding to the desired level of confidence.
␴ is the population standard deviation.
E is the maximum allowable error.
The result of this calculation is not always a whole number. When the outcome
is not a whole number, the usual practice is to round up any fractional result to the
next whole number. For example, 201.21 would be rounded up to 202.

Example A student in public administration wants to determine the mean amount members of
city councils in large cities earn per month as remuneration for being a council mem-
ber. The error in estimating the mean is to be less than $100 with a 95 percent level
of confidence. The student found a report by the Department of Labor that reported
a standard deviation of $1,000. What is the required sample size?

Solution The maximum allowable error, E, is $100. The value of z for a 95 percent level of
confidence is 1.96, and the value of the standard deviation is $1,000. Substituting
these values into formula (9–5) gives the required sample size as:
z␴ 2 (1.96)($1,000) 2
n⫽ a b ⫽a b ⫽ (19.6)2 ⫽ 384.16
E $100
The computed value of 384.16 is rounded up to 385. A sample of 385 is required
to meet the specifications. If the student wants to increase the level of confidence,
for example to 99 percent, this will require a larger sample. The z value corre-
sponding to the 99 percent level of confidence is 2.58.
z␴ 2 (2.58)($1,000) 2
n⫽ a b ⫽a b ⫽ (25.8)2 ⫽ 665.64
E $100
We recommend a sample of 666. Observe how much the change in the confidence
level changed the size of the sample. An increase from the 95 percent to the 99 per-
cent level of confidence resulted in an increase of 281 observations or 73 percent
[(666/385)*100]. This would greatly increase the cost of the study, both in terms of
time and money. Hence, the level of confidence should be considered carefully.

Sample Size to Estimate a Population Proportion

To determine the sample size for a proportion, the same three variables need to be
specified:
1. The margin of error.
2. The desired level of confidence.
3. The variation or dispersion of the population being studied.
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Estimation and Confidence Intervals 319

For the binomial distribution, the margin of error is:

␲(1 ⫺ ␲)
E⫽z
C n
Solving this equation for n yields the following equation

z 2
n ⫽ ␲(1 ⫺ ␲)a b
SAMPLE SIZE FOR THE
[9–6]
POPULATION PROPORTION E

where:
n is the size of the sample.
z is the standard normal value corresponding to the desired level of confidence.
␲ is the population proportion.
E is the maximum allowable error.
The choices for the z-statistic and the margin of error, E, are the same as the
choices for estimating the population mean. However, in this case the population
standard deviation for a binomial distribution is represented by ␲(1 ⫺ ␲). To find a
value of the population proportion, we would find a comparable study or conduct
a pilot study. If a reliable value cannot be found, then a value of .50 should be used
for ␲. Note that ␲(1 ⫺ ␲) has the largest value using 0.50 and, therefore, without a
good estimate of the population proportion, overstates the sample size. This differ-
ence will not hurt the estimate of the population proportion.

Example The study in the previous example also estimates the proportion of cities that
have private refuse collectors. The student wants the margin of error to be within
.10 of the population proportion, the desired level of confidence is 90 percent,
and no estimate is available for the population proportion. What is the required
sample size?

Solution The estimate of the population proportion is to be within .10, so E ⫽ .10. The desired
level of confidence is .90, which corresponds to a z value of 1.65. Because no esti-
mate of the population proportion is available, we use .50. The suggested number
of observations is

b ⫽ 68.0625
1.65 2
n ⫽ (.5)(1 ⫺ .5)a
.10
The student needs a random sample of 69 cities.

Self-Review 9–4 The registrar wants to estimate the arithmetic mean grade point average (GPA) of all grad-
uating seniors during the past 10 years. GPAs range between 2.0 and 4.0. The mean GPA
is to be estimated within plus or minus .05 of the population mean. Based on prior expe-
rience, the population standard deviation is 0.279. Use the 99 percent level of confidence.
Will you assist the college registrar in determining how many transcripts to study?
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320 Chapter 9

Exercises
19. A population is estimated to have a standard deviation of 10. We want to estimate the
population mean within 2, with a 95 percent level of confidence. How large a sample is
required?
20. We want to estimate the population mean within 5, with a 99 percent level of confidence.
The population standard deviation is estimated to be 15. How large a sample is required?
21. The estimate of the population proportion is to be within plus or minus .05, with a 95 per-
cent level of confidence. The best estimate of the population proportion is .15. How large
a sample is required?
22. The estimate of the population proportion is to be within plus or minus .10, with a 99 per-
cent level of confidence. The best estimate of the population proportion is .45. How large
a sample is required?
23. A survey is being planned to determine the mean amount of time corporation executives
watch television. A pilot survey indicated that the mean time per week is 12 hours, with
a standard deviation of 3 hours. It is desired to estimate the mean viewing time within
one-quarter hour. The 95 percent level of confidence is to be used. How many execu-
tives should be surveyed?
24. A processor of carrots cuts the green top off each carrot, washes the carrots, and inserts
six to a package. Twenty packages are inserted in a box for shipment. To test the weight
of the boxes, a few were checked. The mean weight was 20.4 pounds; the standard
deviation, 0.5 pounds. How many boxes must the processor sample to be 95 percent
confident that the sample mean does not differ from the population mean by more than
0.2 pounds?
25. Suppose the U.S. president wants an estimate of the proportion of the population who
support his current policy toward revisions in the health care system. The president wants
the estimate to be within .04 of the true proportion. Assume a 95 percent level of confi-
dence. The president’s political advisors estimated the proportion supporting the current
policy to be .60.
a. How large of a sample is required?
b. How large of a sample would be necessary if no estimate were available for the pro-
portion supporting current policy?
26. Past surveys reveal that 30 percent of tourists going to Las Vegas to gamble spend more
than $1,000. The Visitor’s Bureau of Las Vegas wants to update this percentage.
a. The new study is to use the 90 percent confidence level. The estimate is to be within
1 percent of the population proportion. What is the necessary sample size?
b. The Bureau feels the sample size determined above is too large. What can be done to
reduce the sample? Based on your suggestion, recalculate the sample size.

9.6 Finite-Population Correction Factor

The populations we have sampled so far have been very large or infinite. What if the
sampled population is not very large? We need to make some adjustments in the way
we compute the standard error of the sample means and the standard error of the
sample proportions.
LO7 Adjust a A population that has a fixed upper bound is finite. For example, there are
confidence interval for 12,179 students enrolled at Eastern Illinois University, there are 40 employees at
finite populations. Spence Sprockets, Chrysler assembled 917 Jeep Wranglers at the Alexis Avenue
plant yesterday, or there were 65 surgical patients at St. Rose Memorial Hospital in
Sarasota yesterday. A finite population can be rather small; it could be all the stu-
dents registered for this class. It can also be very large, such as all senior citizens
living in Florida.
For a finite population, where the total number of objects or individuals is N
and the number of objects or individuals in the sample is n, we need to adjust
the standard errors in the confidence interval formulas. To put it another way, to
find the confidence interval for the mean, we adjust the standard error of the mean
in formulas (9–1) and (9–2). If we are determining the confidence interval for a
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Estimation and Confidence Intervals 321

proportion, then we need to adjust the standard error of the proportion in for-
mula (9–3).
This adjustment is called the finite-population correction factor. It is often
shortened to FPC and is:
N⫺n
FPC ⫽
CN ⫺ 1
Why is it necessary to apply a factor, and what is its effect? Logically, if the sam-
ple is a substantial percentage of the population, the estimate is more precise. Note
the effect of the term (N ⫺ n)/(N ⫺ 1). Suppose the population is 1,000 and the sam-
ple is 100. Then this ratio is (1,000 ⫺ 100)/(1,000 ⫺ 1), or 900/999. Taking the square
root gives the correction factor, .9492. Multiplying this correction factor by the stan-
dard error reduces the standard error by about 5 percent (1 ⫺ .9492 ⫽ .0508). This
reduction in the size of the standard error yields a smaller range of values in esti-
mating the population mean or the population proportion. If the sample is 200, the
correction factor is .8949, meaning that the standard error has been reduced by more
than 10 percent. Table 9–2 shows the effects of various sample sizes.

TABLE 9–2 Finite-Population Correction Factor for Selected Samples When

the Population Is 1,000

Sample Fraction of Correction

Size Population Factor
10 .010 .9955
25 .025 .9879
50 .050 .9752
100 .100 .9492
200 .200 .8949
500 .500 .7075

So if we wished to develop a confidence interval for the mean from a finite popu-
lation and the population standard deviation was unknown, we would adjust formula
(9–2) as follows:

N⫺n
° ¢
s
X⫾t
1n C N ⫺ 1

We would make a similar adjustment to formula (9–3) in the case of a proportion.

The following example summarizes the steps to find a confidence interval for
the mean.

Example There are 250 families in Scandia, Pennsylvania. A random sample of 40 of these
families revealed the mean annual church contribution was $450 and the standard
deviation of this was $75. Could the population mean be $445 or $425?
1. What is the population mean? What is the best estimate of the population
mean?
2. Develop a 90 percent confidence interval for the population mean. What are the
endpoints of the confidence interval?
3. Interpret the confidence interval.
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322 Chapter 9

Solution First, note the population is finite. That is, there is a limit to the number of people
in Scandia, in this case 250.
1. We do not know the population mean. This is the value we wish to estimate. The
best estimate we have of the population mean is the sample mean, which is $450.
2. The formula to find the confidence interval for a population mean follows.

N⫺n
° ¢
s
X⫾t
1n C N ⫺ 1

In this case, we know X ⫽ 450, s ⫽ 75, N ⫽ 250, and n ⫽ 40. We do not

know the population standard deviation, so we use the t distribution. To find
the appropriate value of t, we use Appendix B.2, and move across the top row
to the column headed 90 percent. The degrees of freedom is df ⫽ n ⫺ 1 ⫽
40 ⫺ 1 ⫽ 39, so we move to the cell where the df row of 39 intersects with
the column headed 90 percent. The value is 1.685. Inserting these values in the
formula:

N⫺n
° ¢
s
X⫾t
1n C N ⫺ 1

250 ⫺ 40
° ¢ ⫽ $450 ⫾ $19.98 1.8434 ⫽ $450 ⫾ $18.35
$75
⫽ $450 ⫾ 1.685
140 C 250 ⫺ 1

The endpoints of the confidence interval are $431.65 and $468.35.

3. It is likely that the population mean is more than $431.65 but less than $468.35.
To put it another way, could the population mean be $445? Yes, but it is not
likely that it is $425. Why is this so? Because the value $445 is within the con-
fidence interval and $425 is not within the confidence interval.

Self-Review 9–5 The same study of church contributions in Scandia revealed that 15 of the 40 families
sampled attend church regularly. Construct the 95 percent confidence interval for the
proportion of families attending church regularly.

Exercises
27. Thirty-six items are randomly selected from a population of 300 items. The sample mean
is 35 and the sample standard deviation 5. Develop a 95 percent confidence interval for
the population mean.
28. Forty-nine items are randomly selected from a population of 500 items. The sample mean
is 40 and the sample standard deviation 9. Develop a 99 percent confidence interval for
the population mean.
29. The attendance at the Savannah Colts minor league baseball game last night was 400. A
random sample of 50 of those in attendance revealed that the mean number of soft drinks
consumed per person was 1.86, with a standard deviation of 0.50. Develop a 99 percent
confidence interval for the mean number of soft drinks consumed per person.
30. There are 300 welders employed at Maine Shipyards Corporation. A sample of 30 welders
revealed that 18 graduated from a registered welding course. Construct the 95 percent
confidence interval for the proportion of all welders who graduated from a registered
welding course.
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Estimation and Confidence Intervals 323

Chapter Summary
I. A point estimate is a single value (statistic) used to estimate a population value (parameter).
II. A confidence interval is a range of values within which the population parameter is
expected to occur.
A. The factors that determine the width of a confidence interval for a mean are:
1. The number of observations in the sample, n.
2. The variability in the population, usually estimated by the sample standard deviation, s.
3. The level of confidence.
a. To determine the confidence limits when the population standard deviation is
known, we use the z distribution. The formula is
␴
X ⫾ z [9–1]
1n
b. To determine the confidence limits when the population standard deviation is
unknown, we use the t distribution. The formula is
s
X ⫾ t [9–2]
1n
III. The major characteristics of the t distribution are:
A. It is a continuous distribution.
B. It is mound-shaped and symmetrical.
C. It is flatter, or more spread out, than the standard normal distribution.
D. There is a family of t distributions, depending on the number of degrees of freedom.
IV. A proportion is a ratio, fraction, or percent that indicates the part of the sample or pop-
ulation that has a particular characteristic.
A. A sample proportion is found by X, the number of successes, divided by n, the number
of observations.
B. We construct a confidence interval for a sample proportion from the following formula.
p(1 ⫺ p)
p⫾z [9–4]
C n
V. We can determine an appropriate sample size for estimating both means and proportions.
A. There are three factors that determine the sample size when we wish to estimate the mean.
1. The margin of error, E.
2. The desired level of confidence.
3. The variation in the population.
4. The formula to determine the sample size for the mean is

n⫽ a b
z␴ 2
[9–5]
E
B. There are three factors that determine the sample size when we wish to estimate a
proportion.
1. The margin of error, E.
2. The desired level of confidence.
3. A value for ␲ to calculate the variation in the population.
4. The formula to determine the sample size for a proportion is
z 2
n ⫽ ␲(1 ⫺ ␲) a b [9–6]
E
N⫺n
VI. For a finite population, the standard error is adjusted by the factor .
CN ⫺ 1

Chapter Exercises
31. A random sample of 85 group leaders, supervisors, and similar personnel at General Motors
revealed that, on average, they spent 6.5 years on the job before being promoted. The
standard deviation of the sample was 1.7 years. Construct a 95 percent confidence
interval.