This chapter discusses point estimates, confidence intervals, and sample sizes as they relate to estimating population means. Key points: - A point estimate is a single value derived from a sample used to estimate an unknown population parameter. The sample mean is typically used as the point estimate of the population mean. - A confidence interval provides a range of values that is likely to contain the true population parameter. It is calculated using the sample mean, sample size, standard deviation (or standard error), and desired confidence level. - The sample size required depends on the acceptable margin of error, confidence level, and variability in the population as measured by the standard deviation. Larger sample sizes are needed for smaller margins of error,

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Estimation and
Confidence Intervals
Chapter 9

2. 9-2
Learning Objectives
 LO9-1 Compute and interpret a point estimate of a
population mean.
 LO9-2 Compute and interpret a confidence interval for a
population mean.
 LO9-4 Calculate the required sample size to estimate a
population proportion or population mean.
.

3. 9-3
Point Estimates
 A point estimate
is a single value
(point) derived
from a sample
and used to
estimate a
population value.








p
s
s
X
2
2
LO9-1 Compute and interpret a point
estimate of a population mean.

4. 9-4
Confidence Interval Estimates
 A confidence interval estimate is a range of values
constructed from sample data so that the population
parameter is likely to occur within that range at a
specified probability. The specified probability is
called the level of confidence.
LO9-2 Compute and interpret a confidence
interval for a population mean.
C.I. = point estimate ± margin of error

5. 9-5
Factors Affecting Confidence
Interval Estimates
The width of a confidence interval is determined by:
 The sample size, n
 The variability in the population, usually σ
estimated by s
 The desired level of confidence
LO9-2

6. 9-6
Confidence Intervals for a Mean,
σ Known
sample
in the
ns
observatio
of
number
the
deviation
standard
population
the
level
confidence
particular
a
for
value
-
z
mean
sample




n
σ
z
x
 The width of the interval is determined by the level
of confidence and the size of the standard error of
the mean.
 The standard error is affected by two values:
 Standard deviation
 Number of observations in the sample
LO9-2

7. 9-7
Interval Estimates - Interpretation
For a 95% confidence interval about 95% of similarly
constructed intervals will contain the parameter being
estimated. Also 95% of the sample means for a
specified sample size will lie within 1.96 standard
deviations of the hypothesized population.
LO9-2

8. 9-8
Confidence Interval for a Mean,
σ Known - Example
The American Management Association surveys middle
managers in the retail industry and wants to estimate their
mean annual income. A random sample of 49 managers
reveals a sample mean of $45,420. The standard deviation
of this population is $2,050.
 What is the best point estimate of the population mean?
 What is a reasonable range of values for the population
mean?
 What do these results mean?
LO9-2

9. 9-9
Confidence Interval for a Mean,
σ Known – Example
The American Management Association surveys middle
managers in the retail industry and wants to estimate their
mean annual income. A random sample of 49 managers
reveals a sample mean of $45,420. The standard deviation
of this population is $2,050.
 What is the best point estimate of the population mean?
 Our best estimate of the unknown population mean is
the corresponding sample statistic.
 The sample mean of $45,420 is the point estimate
of the unknown population mean.
LO9-2

10. 9-10
Confidence Interval for a Mean,
σ Known - Example
The American Management Association surveys middle
managers in the retail industry and wants to estimate their
mean annual income. A random sample of 49 managers
reveals a sample mean of $45,420. The standard
deviation of this population is $2,050.
 What is a reasonable range of values for the
population mean?
 Suppose the association decides to use the 95 percent
level of confidence.
LO9-2

11. 9-11
How to Obtain a z-value for a
Given Confidence Level
The 95 percent confidence refers
to the middle 95 percent of the
observations. Therefore, the
remaining 5 percent are equally
divided between the two tails.
Following is a portion of Appendix B.3.
LO9-2

12. 9-12
Confidence Interval for a Mean,
σ Known – Example
The American Management Association surveys middle managers
in the retail industry and wants to estimate their mean annual
income. A random sample of 49 managers reveals a sample mean
of $45,420. The standard deviation of this population is $2,050.
The 95 percent confidence interval estimate is:
LO9-2

13. 9-13
Confidence Interval for a Mean –
Interpretation
The American Management Association
surveys middle managers in the retail
industry and wants to estimate their mean
annual income. A random sample of 49
managers reveals a sample mean of
$45,420. The standard deviation of this
population is $2,050.
What is the interpretation of the confidence
limits $45,846 and $45,994?
If we select many samples of 49 managers,
and for each sample we compute the mean
and then construct a 95 percent confidence
interval, we could expect about 95 percent
of these confidence intervals to contain
the population mean. Conversely, about 5
percent of the intervals would not contain the
population mean annual income, µ.
LO9-2

14. 9-14
Confidence Intervals for a Mean,
σ Unknown
In most sampling situations the population standard deviation (σ) is not
known. Below are some examples where it is unlikely the population
standard deviations would be known.
 The Dean of the Business College wants to estimate the mean number of
hours full-time students work at paying jobs each week. He selects a
sample of 30 students, contacts each student, and asks them how many
hours they worked last week.
 The Dean of Students wants to estimate the distance the typical commuter
student travels to class. She selects a sample of 40 commuter students,
contacts each, and determines the one-way distance from each student’s
home to the center of campus.
 The Director of Student Loans wants to know the mean amount owed on
student loans at the time of his/her graduation. The director selects a
sample of 20 graduating students and contacts each to find the information.
LO9-2

15. 9-15
Using the t-Distribution: Confidence
Intervals for a Mean, σ Unknown
 It is, like the z distribution, a continuous distribution.
 It is, like the z distribution, bell-shaped and symmetrical.
 There is not one t distribution, but rather a family of t
distributions. All t distributions have a mean of 0, but their standard
deviations differ according to the sample size, n.
 The t distribution is more spread out and flatter at the center than
the standard normal distribution As the sample size increases,
however, the t distribution approaches the standard normal
distribution.
LO9-2

16. 9-16
Comparing the z and t Distributions
When n is Small, 95% Confidence
Level
LO9-2

17. 9-17
Using the t-Distribution; Confidence
Intervals for a Mean, σ Unknown
A tire manufacturer wishes to
investigate the tread life of its tires.
A sample of 10 tires driven 50,000
miles revealed a sample mean of
0.32 inch of tread remaining with a
standard deviation of 0.09 inch.
Construct a 95 percent confidence
interval for the population mean.
Would it be reasonable for the
manufacturer to conclude that after
50,000 miles the population mean
amount of tread remaining is 0.30
inches?
LO9-2
Given in the problem:
n =10
x = 0.32
s = 0.09
Compute the confidence interval using
the t-distribution (since s is unknown).
x ±t
s
n

18. 9-18
Using the Student’s t-Distribution
Table
LO9-2

19. 9-19
Confidence Interval Estimates for
the Mean – Example
The manager of the Inlet Square Mall, near Ft.
Myers, Florida, wants to estimate the mean
amount spent per shopping visit by customers. A
sample of 20 customers reveals the following
amounts spent.
Based on a 95% confidence interval, do
customers spend $50 on average? Do they
spend $60 on average?
LO9-2

20. 9-20
Confidence Interval Estimates for
the Mean – Example
LO9-2
Compute the Confidence Interval
using the t-distribution (since s is unknown)
with 19 degrees of freedom.
x ±t
s
n
= x ±t
s
n
= 49.35±t
9.01
20
= 49.35± 2.093
9.01
20
= 49.35± 4.22
The endpoints of the confidence interval are $45.13 and $53.57
Conclude: It is reasonable that the population mean could be $50.
The value of $60 is not in the confidence interval. Hence, we
conclude that the population mean is unlikely to be $60.

21. 9-21
When to Use the z or t Distribution for
Confidence Interval Computation
LO9-2

22. 9-22
When to Use the z or t Distribution
for Confidence Interval Computation
Use Z-distribution,
If the population standard
deviation is known.
Use t-distribution,
If the population standard
deviation is unknown.
LO9-2

23. 9-23
Selecting an Appropriate Sample Size
There are 3 factors that determine the size of a sample,
none of which has any direct relationship to the size of the
population:
 The level of confidence desired
 The margin of error the researcher will tolerate
 The variation in the population being studied
LO9-4 Calculate the required sample size to estimate
a population proportion or population mean.

24. 9-24
Selecting an Appropriate Sample
Size: What if the Population Standard
Deviation is not Known?
 Conduct a pilot study
 Use a comparable study
 Use a range-based approach
LO9-4

25. 9-25
Sample Size for Estimating the
Population Mean
2





 

E
z
n

LO9-4

26. 9-26
Sample Size for Estimating
Population Mean – Example 1
A student in public administration wants to
determine the mean amount members of city
councils in large cities earn per month. She
would like to estimate the mean with a 95%
confidence interval and a margin of error of
less than $100. The student found a report by
the Department of Labor that estimated the
standard deviation to be $1,000. What is the
required sample size?
Given in the problem:
 E, the maximum allowable error, is $100,
 The value of z for a 95 percent level of
confidence is 1.96,
 The estimate of the standard deviation is
$1,000.
n =
z×s
E
æ
è
ç
ö
ø
÷
2
=
(1.96)($1,000)
$100
æ
è
ç
ö
ø
÷
2
= (19.6)2
= 384.16
= 385
LO9-4

27. 9-27
Sample Size for Estimating
Population Mean – Example 2
A consumer group would like to estimate the mean monthly
electricity charge for a single family house in July within $5
using a 99 percent level of confidence. Based on similar
studies, the standard deviation is estimated to be $20.00. How
large of a sample is required?
n =
(2.58)(20)
5
æ
è
ç
ö
ø
÷
2
=107
LO9-4
n =
z×s
E
æ
è
ç
ö
ø
÷
2