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```© 2012 McGraw-Hill
Ryerson Limited
© 2009 McGraw-Hill Ryerson Limited
1
Lind
Marchal
Wathen
Waite
© 2012 McGraw-Hill Ryerson Limited
2
Learning Objectives
LO 1 Define a point estimate.
LO 2 Define level of confidence.
LO 3 Construct a confidence interval for the population mean
when the population standard deviation is known.
LO 4 Construct a confidence interval for a population mean
when the population standard deviation is unknown.
LO 5 Construct a confidence interval for a population
proportion.
LO 6 Determine the sample size for attribute and variable
sampling.
© 2012 McGraw-Hill Ryerson Limited
3
LO
1
Point Estimates
© 2012 McGraw-Hill Ryerson Limited
4
Point and Interval Estimates
A point estimate is the statistic (single value), computed
from sample information, that is used to estimate the
population parameter.
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.
LO 1
© 2012 McGraw-Hill Ryerson Limited
5
Types of Estimates
Estimates of the population mean:
The population standard deviation (σ) is known.
The population standard deviation is unknown.
– In this case, we substitute the sample standard
deviation (s) for the population standard deviation (σ)
Estimate of the population proportion.
LO 1
© 2012 McGraw-Hill Ryerson Limited
6
LO
2
Level of Confidence
© 2012 McGraw-Hill Ryerson Limited
7
Point Estimates and Confidence Intervals
The sample mean, x is a point estimate of the population
mean, μ.
p ,a sample proportion, is a point estimate of p, the
population proportion.
s, the sample standard deviation, is a point estimate of σ,
the population standard deviation.
LO 2
© 2012 McGraw-Hill Ryerson Limited
8
Point Estimates and Confidence Intervals
z
0.00 0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08 0.09
:
:
:
:
:
:
:
:
:
:
:
1.7
:
:
0.4573
0.4582
0.4591
0.4599
0.4608
0.4616
:
:
1.8
:
:
0.4656
0.4664
0.4671
0.4678
0.4686
0.4693
:
:
1.9
:
:
0.4726
0.4732
0.4738
0.4744
0.4750
0.4756
:
:
2.0
:
:
0.4783
0.4788
0.4793
0.4798
0.4803
0.4808
:
:
2.1
:
:
0.4830
0.4834
0.4838
0.4842
0.4846
0.4850
:
:
2.2
:
:
0.4868
0.4871
0.4875
0.4878
0.4881
0.4884
:
:
:
:
:
:
:
:
:
:
:
:
:
The area between z = –1.96
and z = +1.96 is 0.95.
LO 3
© 2012 McGraw-Hill Ryerson Limited
9
LO
3
Construct a confidence
interval for the population
mean when the population
standard deviation is known.
© 2012 McGraw-Hill Ryerson Limited
10
Confidence Levels and z-values
LO 3
Confidence
Level
Nearest
Probability
z-Value
80 percent
0.3997
1.28
94 percent
0.4699
1.88
96 percent
0.4798
2.05
© 2012 McGraw-Hill Ryerson Limited
11
Factors Affecting Confidence Interval Estimates
The factors that determine the width of a confidence
interval are:
1. the sample size, n
2. the variability in the population σ, usually estimated by s
3. the desired level of confidence
LO 3
© 2012 McGraw-Hill Ryerson Limited
12
Factors Affecting Confidence Interval Estimates
If the population standard deviation is known or the sample
is 30 or more, we use the z distribution.
x
z

n
LO 3
Sample mean
z-value for a particular confidence level
the population standard deviation
the number of observations in the sample
© 2012 McGraw-Hill Ryerson Limited
13
Example – Population Standard Deviation (σ)
Known
A survey company wants to determine the mean income of
middle level employees in the retail industry. A random
sample of 361 employees reveals a sample mean of
\$54 520. The standard deviation of this population is
\$3060. The company would like answers to the following
questions:
1.What is the population mean? What is a reasonable
value to use as an estimate of the population mean?
2.What is a reasonable range of values for the population
mean?
3.What do these results mean?
LO 3
© 2012 McGraw-Hill Ryerson Limited
14
Solution – Population Standard Deviation (σ)
Known
1. In this case, we do not know the population mean. We
do know the sample mean is \$54 520. Hence, our best
estimate of the unknown population value is the
corresponding sample statistic. Thus, the sample mean
of \$54 520 is a point estimate of the unknown
population mean.
2. Suppose the association decides to use the 95 percent
level of confidence:
xz

n
 \$54520  1.96
\$3060
 \$54520  316
361
The confidence limits are \$54 836 and \$54 204.
The margin of error is ±\$316.
LO 3
© 2012 McGraw-Hill Ryerson Limited
15
Solution – Population Standard Deviation (σ)
Known
Continued
3. If we select many samples of 361 employees, and for
each sample we compute the mean and then construct
a 95% confidence interval, we could expect about 95%
of these confidence intervals to contain the population
mean. Conversely, about 5% of the intervals would not
contain the population mean annual income, µ.
LO 3
© 2012 McGraw-Hill Ryerson Limited
16
Confidence Intervals in Excel
LO 3
© 2012 McGraw-Hill Ryerson Limited
17
You Try It Out!
The mean daily sales are \$3000 for a sample of 40 days at
a convenience store. The standard deviation of the
population is \$400.
a) What is the estimated mean daily sales of the
population? What is this estimate called?
b) What is the 99 percent confidence interval?
c) Interpret your findings.
LO 3
© 2012 McGraw-Hill Ryerson Limited
18
LO
4
Construct a Confidence Interval
for the Population Mean When
the Population Standard
Deviation is Unknown.
© 2012 McGraw-Hill Ryerson Limited
19
Population Standard Deviation (σ) Unknown
In most sampling situations the population standard
deviation (σ) is not known. We can use the sample
standard deviation to estimate the population standard
deviation. But in doing so, we can no longer use formula (81), and because we do not know σ, we cannot use the z
distribution.
However, there is a remedy:
We use the sample standard deviation and replace the z
distribution with the t distribution.
LO 4
© 2012 McGraw-Hill Ryerson Limited
20
Characteristics of the t Distribution
1. It is, like the z distribution, a continuous distribution.
2. It is, like the z distribution, bell-shaped and symmetrical.
3. 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 standard deviation for a t distribution with 5
observations is larger than for a t distribution with 20
observations.
LO 4
© 2012 McGraw-Hill Ryerson Limited
21
Characteristics of the t Distribution
4. The t distribution is more spread out and flatter at the
centre than is the standard normal distribution. As the
sample size increases, however, the t distribution
approaches the standard normal distribution, because
the errors in using s to estimate σ decrease with larger
samples.
LO 4
© 2012 McGraw-Hill Ryerson Limited
22
Characteristics of the t Distribution
The Standard Normal Distribution
and Student’s t Distribution
LO 4
© 2012 McGraw-Hill Ryerson Limited
23
Characteristics of the t distribution
Values of z and t for the 95 Percent Level of Confidence
LO 4
© 2012 McGraw-Hill Ryerson Limited
24
Confidence Interval for Population Standard
Deviation (σ) Unknown
To develop a confidence interval for the population mean
using the t distribution, we adjust the formula as follows:
LO 4
© 2012 McGraw-Hill Ryerson Limited
25
Confidence Interval for Population Standard
Deviation (σ) Unknown
To develop a confidence interval for the population mean
with an unknown population standard deviation:
1.Assume the sampled population is either normal or
approximately normal.
2.Estimate the population standard deviation (σ) with the
sample standard deviation (s).
3.Use the t distribution rather than the z distribution.
LO 4
© 2012 McGraw-Hill Ryerson Limited
26
Confidence Interval for Population Standard
Deviation (σ) Unknown
Assume the
population is
normal.
Is the population
standard
deviation known?
No
Use the t distribution.
Yes
Use the z distribution.
Determining When to Use the z Distribution or the t Distribution
LO 4
© 2012 McGraw-Hill Ryerson Limited
27
Example – t distribution
A light bulb manufacturer wishes to investigate the life of its
bulbs. A sample of 10 bulbs in use since 60 days revealed
a sample mean of 0.71 days of life remaining with a
standard deviation of 0.13 days. Construct a 95%
confidence interval for the population mean. Would it be
reasonable for the manufacturer to conclude that after 60
days the population mean amount of life remaining is 0.70
days?
LO 4
© 2012 McGraw-Hill Ryerson Limited
28
Solution – t distribution
Given in the problem:
n = 10
= x0.71
s = 0.13
Compute the confidence interval using the t distribution,
since σ is unknown.
X t
LO 4
s
n
© 2012 McGraw-Hill Ryerson Limited
29
Solution – t distribution
Df
1
2
3
4
5
6
7
8
9
10
LO 4
Confidence Intervals
80%
90%
95%
98%
Level of significance for One-Tailed Test
0.100
0.050
0.025
0.010
Level of Significance for Two-Tailed Test
0.20
0.10
0.05
0.02
3.078
6.314
12.706
31.821
1.886
2.920
4.303
6.965
1.638
2.353
3.182
4.541
1.533
2.132
2.776
3.747
1.476
2.015
2.571
3.365
1.440
1.943
2.447
3.143
1.415
1.895
2.365
2.998
1.397
1.860
2.306
2.896
1.383
1.833
2.262
2.821
1.372
1.812
2.228
2.764
© 2012 McGraw-Hill Ryerson Limited
Continued
99%
0.005
0.01
63.657
9.925
5.841
4.604
4.032
3.707
3.499
3.355
3.250
3.169
30
Solution – t distribution
Continued
To determine the confidence interval we substitute the
values in formula:
s
0.13
X t
 0.71  2.262
 0.71  0.093
n
10
The endpoints of the confidence interval are 0.613 and
0.803.
It is reasonable to conclude that the population mean is in
this interval. The manufacturer can be reasonably sure (95
percent confident) that the mean remaining life is between
0.613 and 0.803 days. Because the value of 0.70 is in this
interval, it is possible that the mean of the population is
0.70.
LO 4
© 2012 McGraw-Hill Ryerson Limited
31
t distribution In Excel
© 2012 McGraw-Hill Ryerson Limited
32
Example – t distribution
The manager of the Inlet Square Mall wants to estimate the
mean amount spent per shopping visit by customers. A
sample of 20 customers reveals the following amounts
spent in dollars.
\$48.16 \$42.22 \$46.82 \$51.45 \$23.78 \$41.86 \$54.86 \$37.92 \$52.64 48.59
50.82
46.94
61.83
61.69
49.17
61.46
52.68
52.68
58.84
43.88
What is the best estimate of the population mean?
Determine a 95 percent confidence interval.
Interpret the result. Would it be reasonable to conclude that
the population mean is \$50? What about \$60?
LO 4
© 2012 McGraw-Hill Ryerson Limited
33
Solution – Confidence Interval Estimates for
the Mean Using Excel
LO 4
© 2012 McGraw-Hill Ryerson Limited
34
Solution – Confidence Interval Estimates for
the Mean Using Excel
LO 4
© 2012 McGraw-Hill Ryerson Limited
35
You Try It Out!
Ms. Kleman is concerned about absenteeism among her
students. The information below reports the number of days
absent for a sample of 10 students during the last twoweek exam period.
3 2 1 3 3 5 4 0 1 4
a)
b)
c)
d)
e)
LO 4
Determine the mean and the standard deviation of the sample.
What is the population mean? What is the best estimate of that
value?
Develop a 95 percent confidence interval for the population mean.
Explain why the t distribution is used as a part of the confidence
interval.
Is it reasonable to conclude that the typical student does not miss
any days during an exam period?
© 2012 McGraw-Hill Ryerson Limited
36
LO
5
Confidence Interval for a
Proportion
© 2012 McGraw-Hill Ryerson Limited
37
Confidence Interval for a Proportion
A proportion is the fraction, ratio, or percent indicating the
part of the sample or the population having a particular trait
of interest.
1. All binomial conditions are met.
2. The values np and n(1 – p) should both be greater
than or equal to 5.
LO 5
© 2012 McGraw-Hill Ryerson Limited
38
Confidence Interval for a Proportion
To develop a confidence interval for a population
proportion, we change formula (8-1) to:
For sample data:
We can then construct a confidence interval for a
population proportion from the following formula.
LO 5
© 2012 McGraw-Hill Ryerson Limited
39
Example – Confidence Interval for a Proportion
A company decided to take poll on whether employees
should have dress code. Employees will have dress code if
at least three fourths of employees vote in favour of it. A
random sample of 3000 employees reveals 2600 plan to
vote for dress code.
a) What is the estimate of the population proportion?
b) Develop a 95% confidence interval for the population
proportion.
c) Basing your decision on this sample information, can you
conclude that the necessary proportion of employees
favours the dress code?
LO 5
© 2012 McGraw-Hill Ryerson Limited
40
Solution – Confidence Interval for a Proportion
First compute the sample proportion:
X
2,600
p

 0.87
n
3,000
Compute the 95% C.I.
C.I.  p  z
p (1  p )
n
 0.87  1.96
.87(1  .87)
 .87  .012
3,000
 (0.858, 0.882)
We conclude that the dress code proposal will pass
because the interval estimate includes values greater than
75% of the employees.
LO 5
© 2012 McGraw-Hill Ryerson Limited
41
You Try It Out!
A survey was conducted to estimate the proportion of car
drivers who use seatbelts while driving. Of the 1500 drivers
sampled, 520 drivers said they always use a seatbelt while
driving.
a) Estimate the value of the population proportion.
b) Compute the standard error of the proportion.
c) Develop a 99% confidence interval for the population
proportion.
d) Interpret your findings.
LO 5
© 2012 McGraw-Hill Ryerson Limited
42
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.
A population that has a fixed upper bound is finite.
LO 5
© 2012 McGraw-Hill Ryerson Limited
43
Finite Population Correction Factor
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 find the confidence interval for the mean, we adjust the
standard error of the mean.
For the confidence interval for a proportion, we adjust the
standard error of the proportion.
LO 5
© 2012 McGraw-Hill Ryerson Limited
44
Finite Population Correction Factor
This adjustment is called the finite-population correction
factor (FPC).
FPC 
N n
N 1
The usual rule is if the ratio of n/N is less than 0.05, then
the correction factor is ignored.
LO 5
© 2012 McGraw-Hill Ryerson Limited
45
Adjusting the Standard Errors with the FPC
Adjust the standard error of the mean or proportion as
follows:
LO 5
© 2012 McGraw-Hill Ryerson Limited
46
Example – Finite Population Correction
Factor
There are 350 families in one area in Brooks City. A poll of
50 families reveals the mean annual charity contribution is
\$550 with a standard deviation of \$85.
1. Develop a 90 percent confidence interval for the
population mean.
2. Interpret the confidence interval.
LO 5
© 2012 McGraw-Hill Ryerson Limited
47
Solution – Finite Population Correction
Factor
Given in Problem:
N = 350; n = 50 and s = \$85
Since n/N = 50/350 = 0.14, the finite population correction
factor must be used.
The population standard deviation is not known, therefore
use the t distribution.
X t
s
n
N n
\$85
 \$550  t
N 1
50
 \$550  1.677
350  50
350  1
\$85
50
350  50
350  1
 \$550  \$20.16 .93
 \$550  \$18.75
 (\$568.75, \$531.25)
LO 5
© 2012 McGraw-Hill Ryerson Limited
48
Solution – Finite Population Correction
Factor
Continued
It is likely that the population mean is more than \$531.25
but less than \$568.75. The population mean can be \$545
but not \$525, because \$545 is within the confidence
interval and \$525 is not.
LO 5
© 2012 McGraw-Hill Ryerson Limited
49
You Try It Out!
The same study of charity contributions revealed that 20 of
the 50 families sampled donate to charity regularly.
1. Construct the 95 percent confidence interval for the
proportion of families donating to charity regularly.
2. Should the finite-population correction factor be used?
Why or why not?
LO 5
© 2012 McGraw-Hill Ryerson Limited
50
LO
6
Choosing an Appropriate
Sample Size
© 2012 McGraw-Hill Ryerson Limited
51
Choosing An Appropriate Sample Size
There are three factors that determine the size of a sample,
none of which has any direct relationship to the size of the
population.
1. the degree of confidence selected
2. the maximum allowable error
3. the population standard deviation
i. Use a comparable study.
ii. Use a range-based approach.
iii. Conduct a pilot study.
LO 6
© 2012 McGraw-Hill Ryerson Limited
52
Sample Size for a Population Mean
We can express the interaction among these three factors
and the sample size in the following formula.
Ez

n
Solving this equation for n yields the following result.
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.
LO 6
© 2012 McGraw-Hill Ryerson Limited
53
Example – Sample Size for a Population
Mean
An NGO wants to determine the mean number of trees
planted in last month near city. The error in estimating the
mean is to be less than 150 with a 95 percent level of
confidence. An NGO found a report by the Department of
Forestry that estimated the standard deviation to be 1500.
What is the required sample size?
LO 6
© 2012 McGraw-Hill Ryerson Limited
54
Solutions – Sample Size for a Population
Mean
 z  
n

 E 
2
 (1.96)(1, 500) 


150


 (19.6) 2
 384.16
2
 384
LO 6
© 2012 McGraw-Hill Ryerson Limited
55
Sample Size for a Population Mean In Excel
LO 6
© 2012 McGraw-Hill Ryerson Limited
56
Sample Size for a Population Proportion
Again, three items need to be specified:
1. the desired level of confidence
2. the margin of error in the population proportion
3. an estimate of the population proportion
The formula to determine the sample size of a proportion
is:
LO 6
© 2012 McGraw-Hill Ryerson Limited
57
Example – Sample Size for a Population
Proportion
The study in the previous example also estimates the
proportion of tress planted. An NGO wants the estimate to
be within 0.15 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?
LO 6
© 2012 McGraw-Hill Ryerson Limited
58
Solution – Sample Size for a Population
Proportion
 1.645 
n  (0.5)(1  0.5) 

0.15


 30.06
n  30 trees
LO 6
2
© 2012 McGraw-Hill Ryerson Limited
59
Sample Size for a Population Proportion In
Excel
LO 6
© 2012 McGraw-Hill Ryerson Limited
60
You Try It Out!
The arithmetic mean of grade point average (GPA) of all
graduating seniors during the past 10 years is to be
estimated. GPAs range between 3.0 and 5.0. The mean
GPA is to be estimated within plus or minus 0.05 of the
population mean. The standard deviation is estimated to be
0.389. Use the 99 percent level of confidence.
LO 5
© 2012 McGraw-Hill Ryerson Limited
61
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
© 2012 McGraw-Hill Ryerson Limited
62
Chapter Summary
a. To determine the confidence limits when the
population standard deviation is known, we
use the z distribution. The formula is:
X z

n
8  1
b. To determine the confidence limits when the
population standard deviation is unknown, we
use the t distribution. The formula is:
X t
s
n
8  2 
© 2012 McGraw-Hill Ryerson Limited
63
Chapter Summary
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.
© 2012 McGraw-Hill Ryerson Limited
64
Chapter Summary
IV. A proportion is a ratio, fraction, or percent that indicates
the part of the sample or population that has the
particular characteristic.
A. A sample proportion is found by X, the number of
successes, divided by n, the number of
observations.
B. The standard error of the sample proportion reports
the variability in the distribution of sample
proportions. It is found by
sp 

p 1 p
n

8  4 
© 2012 McGraw-Hill Ryerson Limited
65
Chapter Summary
C. We construct a confidence interval for a population
proportion from the following formula.
V. For the finite population, the standard error is adjusted
by the factor NN  1n . .
© 2012 McGraw-Hill Ryerson Limited
66
Chapter Summary
VI. 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 population mean.
1. the desired level of confidence, usually expressed by z
2. the maximum allowable error, E
3. the variation in the population, expressed by σ
The formula to determine the sample size for the mean is:
 z 
n

 E 
2
© 2012 McGraw-Hill Ryerson Limited
8  9 
67
Chapter Summary
B. There are three factors that determine the sample
size when we wish to estimate a population
proportion.
1. the desired level of confidence, which is usually
expressed by z.
2. the maximum allowable error, E
3. an estimate of the population proportion. If no
estimate is available, use .50.
The formula to determine the sample size for a
proportion is:
z
n  p(1  p)  
E
2
8  10
© 2012 McGraw-Hill Ryerson Limited
68
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