..

.. SLIDES BY
..
.. John Loucks
..
.. St. Edward’s University
Modified by
Danny Cho

© 2023 Cengage Learning. All Rights Reserved.

Slide
1
 Special Chapter A

Sampling Distributions
and
Interval Estimation (Confidence Interval)

© 2023 Cengage Learning. All Rights Reserved.

Slide
2
 Sampling Distributions

 Selecting a Sample
 Point Estimation
 Introduction to Sampling Distributions
 Sampling Distribution ofx
 Sampling Distribution ofp

© 2023 Cengage Learning. All Rights Reserved.

Slide
3
 Introduction

An element is the entity on which data are collected.

A variable/attribute is a characteristic of interest for the elem

Observation is the set of measurements for a particular elem

A population is a collection of all the elements of interest.

A sample is a subset of the population.

The sampled population is the population from which the

sample is drawn.

© 2023 Cengage Learning. All Rights Reserved.

Slide
4
 Data Elements

Variable/ Observatio Elemen

Attribute n t

© 2023 Cengage Learning. All Rights Reserved.

Slide
5
 Introduction

The reason we select a sample is to collect data to

answer a research question about a population.

The sample results provide only estimates of the

values of the population characteristics.

The reason is simply that the sample contains only

a portion of the population.

With proper sampling methods, the sample results

can provide “good” estimates of the population
characteristics.
© 2023 Cengage Learning. All Rights Reserved.
Slide
6
 Point Estimation

Point estimation is a form of statistical inference.

In point estimation we use the data from the sample

to compute a value of a sample statistic that serves
as an estimate of a population parameter.

We refer tox as the point estimator of the population

mean .

s is the point estimator of the population standard

deviation .

p is the point estimator of the population proportion p

© 2023 Cengage Learning. All Rights Reserved.

Slide
7
 Point Estimation
 Example: St. Andrew’s College
St. Andrew’s College received 900
applications
from prospective students. The application form
contains a variety of information including the
individual’s Scholastic Aptitude Test (SAT) score
and
whether or not the individual desires on-campus
At a meeting in a few hours, the Director of
housing.
Admissions would like to announce the average
SAT
score and the proportion of applicants that
want to
live on campus, for the population of 900
© 2023 Cengage Learning. All Rights Reserved.
applicants. Slide
8
 Point Estimation
 Example: St. Andrew’s College
However, the necessary data on the
applicants have
not yet been entered in the college’s
computerized
database. So, the Director decides to estimate
the
values of the population parameters of interest
based
on sample statistics. The sample of 30
applicants is
© 2023selected usingAllcomputer-generated
Cengage Learning. Rights Reserved. random
Slide
9
numbers.
 Point Estimation Using Excel
 Excel Value Worksheet
(Sorted) A B C D
Applicant Random SAT On-Campus
1 Number Number Score Housing
2 12 0.00027 1207 No
3 773 0.00192 1143 Yes
4 408 0.00303 1091 Yes
5 58 0.00481 1108 No
6 116 0.00538 1227 Yes
7 185 0.00583 982 Yes
8 510 0.00649 1363 Yes
9 394 0.00667 1108 No
Note: Rows 10-31 are not shown.
© 2023 Cengage Learning. All Rights Reserved.
Slide
10
 Point Estimation
 x as Point Estimator of 
x
 x 32,910

i
1097
30 30
 s as Point Estimator of 

s
 i
(x  x )2


163,996
75.2
29 29
 p as Point Estimator of
p p 20 30 .67
Note: Different random numbers would have
identified a different sample which would have
resulted in different point estimates.
© 2023 Cengage Learning. All Rights Reserved.
Slide
11
 Point Estimation

Once all the data for the 900 applicants were

entered
in the college’s database, the values of the
population
 Population Mean SAT Score
parameters of interestxwere calculated.
   i 1090
900
 Population Standard Deviation for SAT Score


 i
(x   )2

80
900
 Population Proportion Wanting On-Campus
Housing 648
p .72
900
© 2023 Cengage Learning. All Rights Reserved.
Slide
12
 Summary of Point Estimates
Obtained from a Simple Random Sample

Population Parameter Point Point

Parameter Value Estimator Estimate
m = Population mean 1090 x= Sample mean 1097
SAT score SAT score

s = Population std. 80 s = Sample stan- 75.2

deviation for dard deviation
SAT score for SAT score

p = Population pro- .72 p= Sample pro- .67

portion wanting portion wanting
campus housing campus housing

© 2023 Cengage Learning. All Rights Reserved.

Slide
13 23
 Population Parameters vs. Point
Incomplete
Estimators

Population Point
Parameter Estimator
∑ 𝑥𝑖 ∑ 𝑥𝑖
m = Population mean =
𝑁
x= Sample mean = 𝑛

¿ √ ∑ ¿¿¿¿ ¿ √ ∑ ¿¿¿¿
s = Population s = Sample
SD SD

p = Population =∑ h𝑖 p= Sample =∑ h𝑖
proportion 𝑁 proportion 𝑛

© 2023 Cengage Learning. All Rights Reserved.

Slide
14 23
 Sampling Distribution ofx
 Process of Statistical Inference

Population A simple random sample

with mean of n elements is selected
m=? from the population.

The value of x is used to The sample data

make inferences about provide a value for
the value of m. the sample meanx .

© 2023 Cengage Learning. All Rights Reserved.

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15
 Sampling Distribution ofx

The sampling distribution of x is the probability

distribution of all possible values of the sample
mean x.
• Expected Value ofx

E( x) = 

where:  = the population mean

When the expected value of the point estimator
equals the population parameter, we say the point
estimator is unbiased.

© 2023 Cengage Learning. All Rights Reserved.

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16
 Sampling Distribution ofx
• Standard Deviation ofx

We will use the following notation to define the

x of
standard deviation of the sampling distribution
.
s = x
x the standard deviation of
s = the standard deviation of the population
n = the sample size
N = the population size

© 2023 Cengage Learning. All Rights Reserved.

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17
 Sampling Distribution ofx
• Standard Deviation ofx
Finite Population Infinite Population

N n  
x  ( ) x 
N1 n n
• A finite population is treated as being
infinite if n/N < .05.
• ( N  n ) / ( N  1) is the finite population
correction factor.
•  x is referred to as the standard
error of the
sample mean.
© 2023 Cengage Learning. All Rights Reserved.
Slide
18
 Sampling Distribution ofx

When the population has a normal distribution, the

sampling distribution of x is normally distributed
for any sample size.

In most applications, the sampling distribution xof

can be approximated by a normal distribution
whenever the sample is size 30 or more.

In cases where the population is highly skewed or

outliers are present, samples of size 50 may be
needed.

© 2023 Cengage Learning. All Rights Reserved.

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19
 Sampling Distribution ofx

The sampling distribution of x can be used to

provide probability information about how close
the sample mean x is to the population mean m .

© 2023 Cengage Learning. All Rights Reserved.

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20
 Central Limit Theorem

When the population from which we are selecting

a random sample does not have a normal distribution,
the central limit theorem is helpful in identifying the
shape of the sampling distribution ofx .

CENTRAL LIMIT THEOREM

In selecting random samples of size n from a
population, the sampling distribution of the sample
mean x can be approximated by a normal
distribution as the sample size becomes large.

© 2023 Cengage Learning. All Rights Reserved.

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21
 Sampling Distribution ofx
 Example: St. Andrew’s College

Sampling
Distribution  80
of x x   14.6
n 30
for SAT
Scores Note: We use this because
n/N = 30/900 = 0.033 < .05

x
E(x) 1090

© 2023 Cengage Learning. All Rights Reserved.

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22
 Sampling Distribution ofx
 Example: St. Andrew’s College
What is the probability that a simple
random
sample of 30 applicants will provide an
estimate of
the population mean SAT score that is within
In other words, what is the probabilityxthat
+/-10
will
of the actual population mean  ? x
be between 1080 and 1100? (i.e.,  = E( ) =
1090)

© 2023 Cengage Learning. All Rights Reserved.

Slide
23
 Sampling Distribution of x for SAT Scores
 Example: St. Andrew’s College

Sampling  x 14.6
Distribution
of x
for SAT Probability
Scores of this Area = ?

x
1080 1090 1100

© 2023 Cengage Learning. All Rights Reserved.

Slide
24
 Sampling Distribution ofx
 Example: St. Andrew’s College
Step 1: Calculate the z-value at the upper endpoint of
the interval.
z = (-µ)/ =x (1100 - 1090)/14.6 = .68
Step 2: Find the area under the curve to the left of the
upper endpoint.
P(z < .68) = .7517

© 2023 Cengage Learning. All Rights Reserved.

Slide
25
 Sampling Distribution ofx
 Example: St. Andrew’s College
Cumulative Probabilities for
the Standard Normal
Distribution
z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09
. . . . . . . . . . .
.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224
.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549
.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852
.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133
.9 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389
. . . . . . . . . . .

© 2023 Cengage Learning. All Rights Reserved.

Slide
26
 Sampling Distribution ofx
 Example: St. Andrew’s College

Sampling
Distribution  x 14.6
of x
for SAT
Scores

Area = .7517

x
1090 1100

© 2023 Cengage Learning. All Rights Reserved.

Slide
27
 Sampling Distribution ofx
 Example: St. Andrew’s College
Step 3: Calculate the z-value at the lower endpoint of
the interval.
z = (-µ)/ =x (1080 - 1090)/14.6= - .68
Step 4: Find the area under the curve to the left of the
lower endpoint.
P(z < -.68) = 1 - P(z < .68)
= 1 - .7517
= .2483

© 2023 Cengage Learning. All Rights Reserved.

Slide
28
 Sampling Distribution of x for SAT Scores
 Example: St. Andrew’s College

Sampling
Distribution  x 14.6
of x
for SAT
Scores

Area = .2483

x
1080 1090

© 2023 Cengage Learning. All Rights Reserved.

Slide
29
 Sampling Distribution of x for SAT Scores
 Example: St. Andrew’s College
Step 5: Calculate the area under the curve between
the lower and upper endpoints of the interval.
P(-.68 < z < .68) = P(z < .68) - P(z < -.68)
= .7517 - .2483
= .5034
The probability that the sample mean SAT
score will
be between 1080 and 1100 is:
P(1080 < x< 1100) = .5034

© 2023 Cengage Learning. All Rights Reserved.

Slide
30
 Sampling Distribution of x for SAT Scores
 Example: St. Andrew’s College

Sampling  x 14.6
Distribution
of x
for SAT Probability of
Scores this Area
= .5034

x
1080 1090 1100

© 2023 Cengage Learning. All Rights Reserved.

Slide
31
 Relationship Between the Sample Size
x of
and the Sampling Distribution
 Example: St. Andrew’s College
• Suppose we select a simple random sample of 100
applicants instead of the 30 originally considered.
• E(x ) = m regardless of the sample size. In
our x
• example,
Whenever the E( ) remains
sample size isatincreased,
1090. the standard
error of the mean  x is decreased. With the increase
in the sample size to n = 100, the standard error of
the mean is decreased from 14.6 to:
N  n   900  100  80 
x       .94333(8.0) 7.55
N  1 n  900  1  100 

© 2023 Cengage Learning. All Rights Reserved.

Slide
32
 Relationship Between the Sample Size
x of
and the Sampling Distribution
 Example: St. Andrew’s College

With n = 100,
 x  7.55

With n = 30,
 x 14.6

x
E(x) 1090
© 2023 Cengage Learning. All Rights Reserved.
Slide
33
 Relationship Between the Sample Size
x of
and the Sampling Distribution
 Example: St. Andrew’s College

Sampling  x 7.55
Distribution
of x
for SAT
Scores
Area = .8147

x
108010901100
© 2023 Cengage Learning. All Rights Reserved.
Slide
34
 Relationship Between the Sample Size
x of
and the Sampling Distribution
 Example: St. Andrew’s College
• Recall that when n = 30, P(1080 < x < 1100) = .5034
• We follow the same steps to solve for P(1080 x<
< 1100) when n = 100 as we showed earlier when
n = 30.
• Now, with n = 100, P(1080 < x < 1100) = .8147.
• Because the sampling distribution with n = 100 has a
smaller standard error, the values ofx have less
variability and tend to be closer to the population
mean than the values of x with n = 30.

© 2023 Cengage Learning. All Rights Reserved.

Slide
35
 Sampling Distribution ofp
 Making Inferences about a Population
Proportion
Population A simple random sample
with proportion of n elements is selected
p=? from the population.

The value of p is used The sample data

to make inferences provide a value for
about the value of p. the p
sample
proportion .
© 2023 Cengage Learning. All Rights Reserved.
Slide
36
 Sampling Distribution ofp

The sampling distribution of p is the probability

distribution of all possible values of the sample
proportion .p
• Expected Value ofp

E ( p)  p

where:
p = the population proportion

© 2023 Cengage Learning. All Rights Reserved.

Slide
37
 Sampling Distribution ofp
• Standard Deviation ofp
Finite Population Infinite Population

N  n p(1  p) p (1  p )
p  p 
N1 n n

•  p is referred to as the standard

error of
• the sample proportion.
( N  n ) / ( N  1) is the finite population
correction factor.
• A finite population is treated as being
infinite if n/N < .05.
© 2023 Cengage Learning. All Rights Reserved.
Slide
38
 Form of the Sampling Distribution ofp

The sampling distribution of p can be approximated

by a normal distribution whenever the sample size
is large enough to satisfy the two conditions:

np > 5 and n(1 – p) > 5

. . . because when these conditions are satisfied, the

probability distribution of x in the sample proportion,
p
= x/n, can be approximated by normal distribution
(and because n is a constant).

© 2023 Cengage Learning. All Rights Reserved.

Slide
39
 Sampling Distribution ofp
 Example: St. Andrew’s College
Recall that 72% of the prospective students
applying
to St. is
What Andrew’s Collegethat
the probability desire on-campus
a simple random sample
housing.
of 30 applicants will provide an estimate of the
population proportion of applicant desiring on-campus
housing that is within plus or minus .05 of the actual
population proportion?

© 2023 Cengage Learning. All Rights Reserved.

Slide
40
 Sampling Distribution ofp
 Example: St. Andrew’s College

Sampling
Distribution
of p
Probability
of this Area = ?

p
.67 .72 .77

© 2023 Cengage Learning. All Rights Reserved.

Slide
41
 Sampling Distribution ofp
 Example: St. Andrew’s College
For our example, with n = 30 and p = .72,
the
normal distribution is an acceptable
approximation
because: np = 30(.72) = 21.6 > 5
and
n(1 - p) = 30(.28) = 8.4 > 5

© 2023 Cengage Learning. All Rights Reserved.

Slide
42
 Sampling Distribution ofp
 Example: St. Andrew’s College

Sampling .72(1  .72)

Distribution p  .082
30
of p

p
E(p) .72

© 2023 Cengage Learning. All Rights Reserved.

Slide
43
 Sampling Distribution ofp
 Example: St. Andrew’s College
Step 1: Calculate the z-value at the upper endpoint
(i.e., .72+.05) of the interval.
z = (- p)/  =
p (.77 - .72)/.082 = .61
Step 2: Find the area under the curve to the left of
the upper endpoint.
P(z < .61) = .7291

© 2023 Cengage Learning. All Rights Reserved.

Slide
44
 Sampling Distribution ofp
 Example: St. Andrew’s College
Cumulative Probabilities for
the Standard Normal
Distribution
z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09
. . . . . . . . . . .
.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224
.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549
.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852
.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133
.9 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389
. . . . . . . . . . .

© 2023 Cengage Learning. All Rights Reserved.

Slide
45
 Sampling Distribution ofp
 Example: St. Andrew’s College

Sampling  p .082
Distribution
of p

Area = .7291

p
.72 .77

© 2023 Cengage Learning. All Rights Reserved.

Slide
46
 Sampling Distribution ofp
 Example: St. Andrew’s College
Step 3: Calculate the z-value at the lower endpoint
(i.e., .72 - .05) of the interval.
z = (- p)/ = p (.67 - .72)/.082 = - .61
Step 4: Find the area under the curve to the left of the
lower endpoint.
P(z < -.61) = 1 - P(z < .61)
= 1 - .7291
= .2709

© 2023 Cengage Learning. All Rights Reserved.

Slide
47
 Sampling Distribution ofp
 Example: St. Andrew’s College

Sampling  p .082
Distribution
of p

Area = .2709

p
.67 .72

© 2023 Cengage Learning. All Rights Reserved.

Slide
48
 Sampling Distribution ofp
 Example: St. Andrew’s College
Step 5: Calculate the area under the curve between
the lower and upper endpoints of the interval.
P(-.61 < z < .61) = P(z < .61) - P(z < -.61)
= .7291 - .2709
= .4582
The probability that the sample proportion of applicants
wanting on-campus housing will be within +/-.05 of the
actual population proportion :

P(.67 < p< .77) = .4582

© 2023 Cengage Learning. All Rights Reserved.

Slide
49
 Sampling Distribution ofp
 Example: St. Andrew’s College

Sampling  p .082
Distribution
of p
Probability of
this Area = .4582

p
.67 .72 .77

© 2023 Cengage Learning. All Rights Reserved.

Slide
50
 Interval Estimation (Confidence Interval)
 Population Mean: s Known
 Population Mean: s Unknown
 Determining the Sample Size

© 2023 Cengage Learning. All Rights Reserved.

Slide
51
 Margin of Error and the Interval Estimate

A point estimator cannot be expected to provide the

exact value of the population parameter.

An interval estimate can be computed by adding and

subtracting a margin of error to the point estimate.

Point Estimate +/- Margin of Error

The purpose of an interval estimate is to provide

information about how close the point estimate is to
the value of the parameter.

© 2023 Cengage Learning. All Rights Reserved.

Slide
52
 Margin of Error and the Interval Estimate

The general form of an interval estimate of a

population mean is

x  Margin of Error

© 2023 Cengage Learning. All Rights Reserved.

Slide
53
 Interval Estimate of a Population Mean: s
Known
 In order to develop an interval estimate of a
population mean, the margin of error must
be computed using either:
• the population standard deviation s , or
• the sample standard deviation s
 s is rarely known exactly, but often a good
estimate can be obtained based on historical
data or other information.
 We will consider the following two cases:
 s is known
 s is unknown

© 2023 Cengage Learning. All Rights Reserved.

Slide
54
 Interval Estimate of a Population Mean: s
Known
 Interval Estimate of m (also called Confidence
Interval)

x z /2
n

where: x is the sample mean

1 - is the confidence coefficient (level)
z/2 is the z value providing an area of
/2 in the upper tail of the standard
normal probability distribution
s is the population standard deviation
n is the sample size

© 2023 Cengage Learning. All Rights Reserved.

Slide
55
 Interval Estimate of a Population Mean: s
Known
There is a 1 -  probability (or confidence
level) that the value of a sample mean will
x
provide za /2margin of error of or less.

Sampling
distribution
of x

/2 1 -  of all /2

x values

x

z /2  x z /2  x
© 2023 Cengage Learning. All Rights Reserved.
Slide
56
 Interval Estimate of a Population Mean: s
Known

Sampling
distribution
of x
1 -  of all
/2 /2
x values
interval
does x

not z /2  x interval
z /2  x
include includes
m [------------------------- x -------------------------] m
[------------------------- x -------------------------]
[------------------------- x -------------------------]
interval
© 2023 Cengage Learning. All Rights Reserved.
includes Slide
57
 Interval Estimate of a Population Mean: s
Known
 Values of za/2 for the Most Commonly
Used Confidence Levels

Confidence Table
Level a a/2 Look-up Area
za/2
90% .10 .05 .9500
1.645
95% .05 .025 .9750
1.960
99% .01 .005 .9950
2.576

© 2023 Cengage Learning. All Rights Reserved.

Slide
58
 Finding a z/2 value
Pep
Zone
5w-20
Motor Oil

 Solving for the Reorder Point

Find
Find the
the z-value
z-value that
that cuts
cuts off
off an
an area
area of
of .05
.05 or
or .025
.025
in
in the
the right
right tail
tail of
of the
the standard
standard normal
normal
distribution.
distribution.
zz .00
.00 .01
.01 .02
.02 .03
.03 .04
.04 .05
.05 .06
.06 .07
.07 .08
.08 .09
.09
.. .. .. .. .. .. .. .. .. .. ..
1.5
1.5 .9332
.9332 .9345
.9345 .9357
.9357 .9370
.9370 .9382
.9382 .9394
.9394 .9406
.9406 .9418
.9418 .9429
.9429 .9441
.9441
1.6
1.6 .9452
.9452 .9463
.9463 .9474
.9474 .9484
.9484 .9495
.9495 .9505
.9505 .9515
.9515 .9525
.9525 .9535
.9535 .9545
.9545
1.7
1.7 .9554
.9554 .9564
.9564 .9573
.9573 .9582
.9582 .9591
.9591 .9599
.9599 .9608
.9608 .9616
.9616 .9625
.9625 .9633
.9633
1.8
1.8 .9641
.9641 .9649
.9649 .9656
.9656 .9664
.9664 .9671
.9671 .9678
.9678 .9686
.9686 .9693
.9693 .9699
.9699 .9706
.9706
1.9
1.9 .9713
.9713 .9719
.9719 .9726
.9726 .9732
.9732 .9738
.9738 .9744
.9744 .9750
.9750 .9756
.9756 .9761
.9761 .9767
.9767
.. .. .. .. .. .. .. .. .. .. ..

© 2023 Cengage Learning. All Rights Reserved.

Slide
59
 Meaning of Confidence

Because 90% of all the intervals constructed using

x 1.645 x will contain the population mean,
we say we are 90% confident that the interval
x 1.645 x includes the population mean m.

We say that this interval has been established at the

90% confidence level.

The value .90 is referred to as the confidence

coefficient or confidence level.

© 2023 Cengage Learning. All Rights Reserved.

Slide
60
 Interval Estimate of a Population Mean: 
Known
 Example: Discount Sounds
Discount Sounds has 260 retail outlets
throughout
the United States. The firm is evaluating a
potential
location for a new outlet, based in part, on the
mean
A sample
annual income of of
sizethe n individuals
= 36 was taken;
in the the
sample
marketing
mean
area ofincome
the newis $41,100. The population is
location.
not
believed to be highly skewed. The
population
standard deviation is estimated to be
$4,500, and the
© 2023confidence coefficient
Cengage Learning. to be used in the
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61
interval
 Interval Estimate of a Population Mean:
 Known
 Example: Discount Sounds
95% of the sample means that can be observed
x
are within + 1.96 of the population mean .

The margin of error is:

  4,500 
z / 2 1.96   1,470
n  36 

Thus, at 95% confidence,

the margin of error is $1,470.

© 2023 Cengage Learning. All Rights Reserved.

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62
 Interval Estimate of a Population Mean: 
Known
 Example: Discount Sounds
Interval estimate of  is:

x  Margin of Error

$41,100 + $1,470
or
$39,630 to $42,570

We are 95% confident that the interval contains the

population mean.

© 2023 Cengage Learning. All Rights Reserved.

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63
 Interval Estimate of a Population Mean: 
Known Exam
 Example: Discount Sounds Question
Confidence Margin confirmed
Level of Error Interval Estimate
90% 1,234 39,866 to
42,334
95% 1,470 39,630
to 42,570
99% 1,932 39,168 to
43,032
In order to have a higher degree of confidence,
the margin of error and thus the width of the
confidence interval must be larger.
If you have to increase confidence level, you have to increase the
width of the interval estimate.
© 2023 Cengage Learning. All Rights Reserved.
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64
 Interval Estimate of a Population Mean: s
Known
 Adequate Sample Size

In most applications, a sample size of n = 30 is

adequate.

If the population distribution is highly skewed or

contains outliers, a sample size of 50 or more is
recommended.

© 2023 Cengage Learning. All Rights Reserved.

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65
 Interval Estimate of a Population Mean: s
Unknown
 If an estimate of the population standard
deviation s cannot be developed prior to
sampling, we use the sample standard
 deviation
This is the ss to s.
estimatecase.
unknown
 In this case, the interval estimate for m is based
on the t distribution.
 (We’ll assume for now that the population is
normally distributed.)

© 2023 Cengage Learning. All Rights Reserved.

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66
 t Distribution

A specific t distribution depends on a parameter

known as the degrees of freedom (= n - 1).

As the degrees of freedom increases, the difference

between the t distribution and the standard
normal probability distribution becomes smaller
and smaller.

© 2023 Cengage Learning. All Rights Reserved.

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67
 t Distribution

t
Standard distribution
normal (20 degrees
distribution of freedom)

t
distribution
(10 degrees
of
freedom)
z, t
0

© 2023 Cengage Learning. All Rights Reserved.

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68
 t Distribution

For more than 100 degrees of freedom, the standard

normal z value provides a good approximation to
the t value.

The standard normal z values can be found in the

infinite degrees () row of the t distribution table.

In other words, t distribution becomes a standard

normal distribution when the sample size ⇒ .

© 2023 Cengage Learning. All Rights Reserved.

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69
 t Distribution

Degrees Area in Upper Tail

of Freedom .20 .10 .05 .025 .01 .005
. . . . . . .
50 .849 1.299 1.676 2.009 2.403 2.678
60 .848 1.296 1.671 2.000 2.390 2.660
80 .846 1.292 1.664 1.990 2.374 2.639
100 .845 1.290 1.660 1.984 2.364 2.626
 .842 1.282 1.645 1.960 2.326 2.576

Standard
normal
z values
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70
 Interval Estimate of a Population Mean:
s Unknown
 Interval Estimate

s
x t/2/2
n

where: 1- = the confidence coefficient

t/2 = the t value providing an area of /2
in the upper tail of a t distribution
with n - 1 degrees of freedom
s = the sample standard deviation

© 2023 Cengage Learning. All Rights Reserved.

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71
 Interval Estimate of a Population Mean: s
Unknown
 Example: Apartment Rents
A reporter for a student newspaper is
writing an
article on the cost of off-campus housing. A
sample
of 16 efficiency apartments within a half-mile
of
Let us provide a 95% confidence interval
campus resulted in a sample mean of $750
estimate
per month
of the mean rent per month for the population
and a sample standard deviation of $55.
of
efficiency apartments within a half-mile of
campus.
© 2023 Cengage Learning. All Rights Reserved.
We will assume this population to be normallySlide
72
 Interval Estimate of a Population Mean: s
Unknown
At 95% confidence,  = .05, and /2 = .025.
t.025 is based on n - 1 = 16 - 1 = 15 degrees of freedom.
In the t distribution table we see that t.025 = 2.131.
Degrees Area in Upper Tail
of Freedom .20 .100 .050 .025 .010 .005
15 .866 1.341 1.753 2.131 2.602 2.947
16 .865 1.337 1.746 2.120 2.583 2.921
17 .863 1.333 1.740 2.110 2.567 2.898
18 .862 1.330 1.734 2.101 2.520 2.878
19 .861 1.328 1.729 2.093 2.539 2.861
. . . . . . .
© 2023 Cengage Learning. All Rights Reserved.
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73
 Interval Estimate of a Population Mean: s
Unknown
 Interval Estimate
s
x t.025 Margin
n of Error

55
750 2.131 750 29.30
16

We are 95% confident that the mean rent per mon

for the population of efficiency apartments within a
half-mile of campus is between $720.70 and $779.30

© 2023 Cengage Learning. All Rights Reserved.

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74
 Interval Estimate of a Population Mean: s
Unknown
 Adequate Sample Size

In most applications, a sample size of n = 30 is

adequate when using the expression
develop an interval estimate of a population mean.

If the population distribution is highly skewed or

contains outliers, a sample size of 50 or more is
recommended.

© 2023 Cengage Learning. All Rights Reserved.

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 Summary of Interval Estimation
Procedures
for a Population Mean
Can the
Yes No
population standard
deviation s be assumed s Unknown
known ? Case
s Known Use the sample
Case standard deviation
s to estimate s

Use Use
 s
x z / 2 x t / 2
n n
© 2023 Cengage Learning. All Rights Reserved.
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76
 Sample Size for an Interval Estimate
of a Population Mean

Let E = the desired margin of error.

E is the amount added to and subtracted from the

point estimate to obtain an interval estimate.

Given a desired margin of error, which is selecte

prior to sampling, we could determine the sample
size necessary to satisfy the margin of error.

The Necessary Sample Size equation requires a

value for the population standard deviation s .

© 2023 Cengage Learning. All Rights Reserved.

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 Sample Size for an Interval Estimate
of a Population Mean
 Margin of Error

E z /2
n

 Necessary Sample Size

( z / 2 ) 2  2
n
E2

From this equation, what else can you talk

about?

© 2023 Cengage Learning. All Rights Reserved.

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 Sample Size for an Interval Estimate
of a Population Mean

The Necessary Sample Size equation requires a

value for the population standard deviation s .

If s is unknown, a preliminary or planning value

for s can be used in the equation.

© 2023 Cengage Learning. All Rights Reserved.

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79
 Sample Size for an Interval Estimate
of a Population Mean
 Example: Discount Sounds
Recall that Discount Sounds is evaluating a
potential location for a new retail outlet, based
in
part, on the mean annual income of the
individuals in
Suppose that Discount Sounds’
the marketing area
management teamof the new location.
wants an estimate of the population mean such
that
there is a .95 probability that the sampling
How large a sample size is needed to meet
error is
the
$500 or less.
required precision?
© 2023 Cengage Learning. All Rights Reserved.
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80
 Sample Size for an Interval Estimate
of a Population Mean

z /2 500
n
At 95% confidence, z.025 = 1.96. Recall that =
4,500. (1.96)2 (4,500)2
n 2
311.17  312
(500)

A sample of size 312 is needed to reach a desired

precision of + $500 at 95% confidence.

© 2023 Cengage Learning. All Rights Reserved.

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81
 Confidence Level/Margin of Error/Sample
Size
Example: US National 2020 Presidential
Election Polls
 Election Day: November 3, 2020

Election Polls:

https://www.270towin.com/2020-polls-biden-tru
mp/national/

© 2023 Cengage Learning. All Rights Reserved.

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