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CHAPTER 9
Estimation from Sample Data
to accompany
Introduction to Business Statistics
fifth edition, by Ronald M. Weiers
Presentation by Priscilla Chaffe-Stengel
Donald N. Stengel
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Chapter 9 - Learning Objectives
• Explain the difference between a point and
an interval estimate.
• Construct and interpret confidence
intervals:
– with a z for the population mean or proportion.
– with a t for the population mean.
• Determine appropriate sample size to
achieve specified levels of accuracy and
confidence.
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Chapter 9 - Key Terms
• Unbiased estimator
• Point estimates
• Interval estimates
• Interval limits
• Confidence
coefficient
• Confidence level
• Accuracy
• Degrees of
freedom (df)
• Maximum likely
sampling error
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Unbiased Point Estimates
Population
Parameter
Sample
Statistic
• Mean, µ
x
• Variance,
s2
• Proportion, p
Formula
x

x = ni
s2
(x – x)2

i
s2 =
n –1
p
p = x successes
n trials
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Confidence Interval: µ, s Known
where x = sample mean
s = population standard
deviation
n = sample size
z = standard normal score
for area in tail = a/2
a 2
z:
x:
–z
s
x – z
n
ASSUMPTION:
infinite population
a
0
x
a 2
+z
s
x + z
n
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Confidence Interval: µ, s Unknown
where x = sample mean
s = sample standard
deviation
n = sample size
t = t-score for area
in tail = a/2
df = n – 1
a 2
t:
x:
–t
x –t s
n
ASSUMPTION:
Population
approximately
normal and
infinite
a
0
x
a 2
+t
x +t  s
n
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Confidence Interval on p
where p = sample proportion
n = sample size
ASSUMPTION:
n•p  5,
n•(1–p)  5,
and population
infinite
z = standard normal score
for area in tail = a/2
a 2
–z
p : p – z  p(1– p)
n
z:
a
0
p
a 2
+z
p + z  p(1– p)
n
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Summary: Computing Confidence
Intervals from a Large Population
• Mean:

s

x  za 
n 
2




• Proportion:






p
(
1
–
p
)
p  za 
n
2











s

x  ta 
n 
2
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Converting Confidence Intervals to
Accommodate a Finite Population
•Mean:
x  za  s  N – n
n
N –1
2
or












x  ta  s  N – n
n
N
–
1
2
•Proportion:













p  za 
2
p(1– p)  N – n
n
N –1













Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Interpretation of
Confidence Intervals
• Repeated samples of size n taken from the
same population will generate (1–a)% of
the time a sample statistic that falls within
the stated confidence interval.
OR
• We can be (1–a)% confident that the
population parameter falls within the
stated confidence interval.
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Sample Size Determination for µ
from an Infinite Population
• Mean: Note s is known and e, the bound
within which you want to estimate µ, is given.
– The interval half-width is e, also called the
maximum likely error: e = z  s
n
– Solving for n, we find:
2 s 2
z
n=
e2
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Sample Size Determination for µ
from a Finite Population
• Mean: Note s is known and e, the bound
within which you want to estimate µ, is given.
2
s
n =
e2 + s 2
z2
where
N
n = required sample size
N = population size
z = z-score for (1–a)% confidence
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Sample Size Determination for p
from an Infinite Population
• Proportion: Note e, the bound within which
you want to estimate p, is given.
– The interval half-width is e, also called the
maximum likely error: e = z  p(1– p)
n
– Solving for n, we find:
2
z
n = p(1– p)
e2
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Sample Size Determination for p
from a Finite Population
• Mean: Note e, the bound within which you
want to estimate µ, is given.
p(1– p)
n =
e2 + p(1– p)
N
z2
where n = required sample size
N = population size
z = z-score for (1–a)% confidence
p = sample estimator of p
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
An Example: Confidence Intervals
• Problem: An automobile rental agency has the
following mileages for a simple random sample of
20 cars that were rented last year. Given this
information, and assuming the data are from a
population that is approximately normally
distributed, construct and interpret the 90%
confidence interval for the population mean.
55
35
65
64
69
37
88
39
61
54
50
74
92
59
38
59
29
60
80
50
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
A Confidence Interval Example, cont.
• Since s is not known but the population is approximately
normally distributed, we will use the t-distribution to
construct the 90% confidence interval on the mean.
x = 57.9, s = 17.384
df = 20 –1 = 19, a / 2 = 0.05
So, t = 1.729
s
17.384
x  t   57.9  1.729 
n
20
a 2
t:
x:
–t
x –t s
n
 a
a 2
0
x
+t
x+t s
n
57.9  6.721  (51.179, 64.621)
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
A Confidence Interval Example, cont.
• Interpretation:
– 90% of the time that samples of
20 cars are randomly selected
from this agency’s rental cars,
the average mileage will fall
between 51.179 miles and
64.621 miles.
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
An Example: Sample Size
• Problem: A national political candidate
has commissioned a study to determine
the percentage of registered voters who
intend to vote for him in the upcoming
election. In order to have 95% confidence
that the sample percentage will be within
3 percentage points of the actual
population percentage, how large a simple
random sample is required?
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
A Sample Size Example, cont.
• From the problem we learn:
– (1 – a) = 0.95, so a = 0.05 and a /2 = 0.025
– e = 0.03
• Since no estimate for p is given, we will use 0.5
because that creates the largest standard error.
2( p)(1– p) 1.962 (0.5)(0.5)
z
=
= 1,067. 1
n=
e2
(0.03)2
To preserve the minimum confidence, the candidate
should sample n = 1,068 voters.
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.