7
Sampling Chapter and Sampling Distributions
LEARNING OBJECTIVES
The two main objectives for Chapter 7 are to give you an appreciation for
the proper application of sampling techniques and an understanding of the
sampling distributions of two statistics, thereby enabling you to:
1.
2.
3.
Contrast sampling to census and differentiate among different methods
of sampling, which include simple, stratified, systematic, and cluster
random sampling; and convenience, judgment, quota, and snowball
nonrandom sampling, by assessing the advantages associated with each
Describe the distribution of a samples mean using the central limit
theorem, correcting for a finite population if necessary
Describe the distribution of a samples proportion using the z formula
for sample proportions
CHAPTER TEACHING STRATEGY
Virtually every analysis discussed in this text deals with sample data. It is
important, therefore, that students are exposed to the ways and means that
samples are gathered. The first portion of Chapter 7 deals with sampling.
Reasons for sampling versus taking a census are given. Most of these reasons are
tied to the fact that taking a census costs more than sampling if the same
measurements are being gathered. Students are then exposed to the idea of
random versus nonrandom sampling. Random sampling appeals to their concepts
of fairness and equal opportunity. This text emphasizes that nonrandom samples
are non-probability samples and cannot be used in inferential analysis because
levels of confidence and/or probability cannot be assigned. It should be
emphasized throughout the discussion of sampling techniques that as future
business managers (most students will end up as some sort of
supervisor/manager) students should be aware of where and how data are
gathered for studies. This will help to assure that they will not make poor
decisions based on inaccurate and poorly gathered data.
The central limit theorem opens up opportunities to analyze data with a
host of techniques using the normal curve. Section 7.2 begins by showing one
population (randomly generated and presented in histogram form) that is
uniformly distributed and one that is exponentially distributed. Histograms of the
means for random samples of varying sizes are presented. Note that the
distributions of means pile up in the middle and begin to approximate the
normal curve shape as sample size increases. Note also by observing the values
on the bottom axis that the dispersion of means gets smaller and smaller as sample
size increases thus underscoring the formula for the standard error of the mean ().
As the student sees the central limit theorem unfold, he/she begins to see that if
the sample size is large enough sample means can be analyzed using the normal
curve regardless of the shape of the population.
Chapter 7 presents formulas derived from the central limit theorem for
both sample means and sample proportions. Taking the time to introduce these
techniques in this chapter can expedite the presentation of material in chapters 8
and 9.
CHAPTER OUTLINE
7.1 Sampling
Reasons for Sampling
Reasons for Taking a Census
Frame
Random Versus Nonrandom Sampling
Random Sampling Techniques
Simple Random Sampling
Stratified Random Sampling
Systematic Sampling
Cluster (or Area) Sampling
Nonrandom Sampling
Convenience Sampling
Judgment Sampling
Quota Sampling
Snowball Sampling
Sampling Error
Nonsampling Errors
7.2
Sampling Distribution of
Sampling from a Finite Population
7.3
Sampling Distribution of
KEY TERMS
Central Limit Theorem
Cluster (or Area) Sampling
Convenience Sampling
Disproportionate Stratified Random Sampling
Quota Sampling
Random Sampling
Sample Proportion
Sampling Error
Finite Correction Factor
Frame
Judgment Sampling
Nonrandom Sampling
Nonrandom Sampling Techniques
Nonsampling Errors
Proportionate Stratified Random Sampling
Simple Random Sampling
Snowball Sampling
Standard Error of the Mean
Standard Error of the Proportion
Stratified Random Sampling
Systematic Sampling
Two-Stage Sampling
SOLUTIONS TO PROBLEMS IN CHAPTER 7
7.1
a)
i.
ii.
A union membership list for the company.
A list of all employees of the company.
b)
i.
ii.
White pages of the telephone directory for Utica, New York.
Utility company list of all customers.
c)
i.
ii.
Airline company list of phone and mail purchasers of tickets from the
airline during the past six months.
A list of frequent flyer club members for the airline.
d)
i.
ii.
List of boat manufacturer's employees.
List of members of a boat owners association.
e)
i.
ii.
Cable company telephone directory.
Membership list of cable management association.
7.4
Size of motel (rooms), age of motel, geographic location.
b)
Gender, age, education, social class, ethnicity.
c)
Size of operation (number of bottled drinks per month), number of
employees, number of different types of drinks bottled at that location,
geographic location.
d)
Size of operation (sq.ft.), geographic location, age of facility, type of
process used.
a)
Under 21 years of age, 21 to 39 years of age, 40 to 55 years of age, over 55
years of age.
b)
Under $1,000,000 sales per year, $1,000,000 to $4,999,999 sales per year,
$5,000,000 to $19,999,999 sales per year, $20,000,000 to $49,000,000 per
year, $50,000,000 to $99,999,999 per year, over $100,000,000 per year.
c)
Less than 2,000 sq. ft., 2,000 to 4,999 sq. ft.,
5,000 to 9,999 sq. ft., over 10,000 sq. ft.
d)
East, southeast, midwest, south, southwest, west, northwest.
e)
7.5
a)
Government worker, teacher, lawyer, physician, engineer, business person,
police officer, fire fighter, computer worker.
f) Manufacturing, finance, communications, health care, retailing, chemical,
transportation.
7.6
n = N/k = 100,000/200 = 500
7.7
N = n k = 75(11) = 825
7.8
k = N/n = 3,500/175 = 20
Start between 0 and 20. The human resource department probably has a list of
company employees which can be used for the frame. Also, there might be a
company phone directory available.
7.9
a)
i.
ii.
Counties
Metropolitan areas
b)
i.
ii.
States (beside which the oil wells lie)
Companies that own the wells
c)
i.
ii.
States
Counties
7.10
Go to the district attorney's office and observe the apparent activity of various
attorneys at work. Select some who are very busy and some who seem to be
less active. Select some men and some women. Select some who appear to
be older and some who are younger. Select attorneys with different ethnic
backgrounds.
7.11
Go to a conference where some of the Fortune 500 executives attend.
Approach those executives who appear to be friendly and approachable.
7.12
Suppose 40% of the sample is to be people who presently own a personal
computer and 60% with people who do not. Go to a computer show at the city's
conference center and start interviewing people. Suppose you get enough people
who own personal computers but not enough interviews with those who do not.
Go to a mall and start interviewing people. Screen out personal computer owners.
Interview non personal computer owners until you meet the 60% quota.
7.13
= 50,
= 10,
n = 64
a) P(> 52):
z = = 1.6
from Table A.5, Prob. = .4452
P( > 52) = .5000 - .4452 = .0548
b) P(< 51):
z = = 0.80
from Table A.5 prob. = .2881
P(< 51) = .5000 + .2881 = .7881
c) P(< 47):
z = = -2.40
from Table A.5 prob. = .4918
P(< 47) = .5000 - .4918 = .0082
d) P(48.5 < < 52.4):
z = = -1.20
from Table A.5 prob. = .3849
z = = 1.92
from Table A.5 prob. = .4726
P(48.5 < < 52.4) = .3849 + .4726 = .8575
e) P(50.6 < < 51.3):
z = = 0.48
from Table A.5, prob. = .1844
z=
from Table A.5, prob. = .3508
P(50.6 < < 51.3) = .3508 - .1844 = .1644
7.14
= 23.45
= 3.8
a) n = 10, P( > 22):
z=
= -1.21
from Table A.5, prob. = .3869
P( > 22) = .3869 + .5000 =
.8869
b) n = 4, P( > 26):
z = = 1.34
from Table A.5, prob. = .4099
P( > 26) = .5000 - .4099 =
7.15
n = 36
= 278
.0901
P(< 280) = .86
.3600 of the area lies between = 280 and = 278. This probability is
associated with z = 1.08 from Table A.5. Solving for :
z=
1.08 =
1.08 = 2
= = 11.11
7.16
n = 81
= 12
P( > 300) = .18
.5000 - .1800 = .3200 and from Table A.5, z.3200 = 0.92
Solving for :
z=
0.92 =
0.92 = 300 -
1.2267 = 300 -
7.17
a)
= 300 - 1.2267 = 298.77
N = 1,000 n = 60
= 75
=6
P( < 76.5):
z = = 2.00
from Table A.5, prob. = .4772
P(< 76.5) = .4772 + .5000 = .9772
b) N = 90
n = 36
= 108
= 3.46
P(107 < < 107.7):
z=
= -2.23
from Table A.5, prob. = .4871
z=
= -0.67
from Table A.5, prob. = .2486
P(107 < < 107.7) = .4871 - .2486 =
c) N = 250
P(
>
z=
n = 100
= 35.6
.2385
= 4.89
36):
= 1.05
from Table A.5, prob. = .3531
P(
36) = .5000 - .3531 = .1469
>
d) N = 5000
n = 60
= 125
P( < 123):
z=
= -1.16
from Table A.5, prob. = .3770
P( < 123) = .5000 - .3770 = .1230
= 13.4
7.18
= 30
= 99.9
n = 38
a) P( < 90):
z=
= -2. 03
from table A.5, area = .4788
P( < 90) = .5000 - .4788 = .0212
b) P(98 < < 105):
z=
= 1.05
from table A.5, area = .3531
z=
= -0.39
from table A.5, area = .1517
P(98 < < 105) = .3531 + .1517 = .5048
c) P( < 112):
z=
= 2.49
from table A.5, area = .4936
P(< 112) = .5000 + .4936 = .9936
d) P(93 < < 96):
z=
= -1.42
from table A.5, area = .4222
z=
= -0.80
from table A.5, area = .2881
P(93 < < 96) = .4222 - .2881 = .1341
7.19
N = 1500
n = 100
= 177,000
= 8,500
P( > $185,000):
z=
= 9.74
from Table A.5, prob. = .5000
7.20
P( > $185,000) = .5000 - .5000 = .0000
= $65.12
= $21.45
n = 45
P( > ) = .2300
Prob. lies between and = .5000 - .2300 = .2700
from Table A.5, z.2700 = 0.74
Solving for :
z=
0.74 =
2.366 = - 65.12
7.21
and
= 11.8
= 50.4
= 65.12 + 2.366 = 67.486
n = 42
a) P( > 52):
z = = 0.88
from Table A.5, the area for z = 0.88 is .3106
P(> 52) = .5000 - .3106 = .1894
b) P( < 47.5):
z=
= -1.59
from Table A.5, the area for z = -1.59 is .4441
P( < 47.5) = .5000 - .4441 = .0559
c) P( < 40):
z=
= -5.71
from Table A.5, the area for z = -5.71 is .5000
P( < 40) = .5000 - .5000 = .0000
d) 71% of the values are greater than 49. Therefore, 21% are between the
sample mean of 49 and the population mean, = 50.4.
The z value associated with the 21% of the area is -0.55
z.21 = -0.55
z=
-0.55 = =
16.4964
7.22
p = .25
a) n = 110
z=
P(< .21):
= -0.97
from Table A.5, prob. = .3340
P(< .21) = .5000 - .3340 = .1660
b) n = 33
z=
P( > .24):
= -0.13
from Table A.5, prob. = .0517
P( > .24) = .5000 + .0517 = .5517
c)
n = 59
z=
P(.24 < < .27):
= -0.18
from Table A.5, prob. = .0714
z = = 0.35
from Table A.5, prob. = .1368
P(.24 < < .27) = .0714 + .1368 = .2082
d) n = 80
P(> .30):
z = = 1.03
from Table A.5, prob. = .3485
P( > .30) = .5000 - .3485 =
e) n = 800
.1515
P( > .30):
z = = 3.27
from Table A.5, prob. = .4995
P( > .30) = .5000 - .4995 = .0005
7.23
p = .58
n = 660
a) P(> .60):
z = = 1.04
from table A.5, area = .3508
P(> .60) = .5000 - .3508 = .1492
b) P(.55 < < .65):
z=
= 3.64
from table A.5, area = .4998
z=
= 1.56
from table A.5, area = .4406
P(.55 < < .65) = .4998 + .4406 = .9404
c)
P( > .57):
z = = -0.52
from table A.5, area = .1985
P( > .57) = .1985 + .5000 = .6985
d)
P(.53 < < .56):
z = = -1.04
z = = -2.60
from table A.5, area for z = -1.04 is .3508
from table A.5, area for z = -2.60 is .4953
P(.53 < < .56) = .4953 - .3508 = .1445
e)
P( < .48):
z = = -5.21
from table A.5, area = .5000
P( < .48) = .5000 - .5000 = .0000
7.24
p = .40
P( > .35) = .8000
P(.35 < < .40) = .8000 - .5000 = .3000
from Table A.5, z.3000 = -0.84
Solving for n:
z=
-0.84 =
=
8.23 =
n = 67.73 68
7.25
p = .28
n = 140
P( < ) = .3000
P( < < .28) = .5000 - .3000 = .2000
from Table A.5, z.2000 = -0.52
Solving for :
z=
-0.52 =
-.02 = - .28
7.26
P(x > 150): n = 600
= .28 - .02 = .26
p = .21 x = 150
= = .25
z=
= 2.41
from table A.5, area = .4920
P(x > 150) = .5000 - .4920 = .0080
7.27
p = .48
n = 200
a) P(x < 90):
= = .45
z = = -0.85
from Table A.5, the area for z = -0.85 is .3023
P(x < 90) = .5000 - .3023 = .1977
b) P(x > 100):
= = .50
z = = 0.57
from Table A.5, the area for z = 0.57 is .2157
P(x > 100) = .5000 - .2157 = .2843
c) P(x > 80):
= = .40
z = = -2.26
from Table A.5, the area for z = -2.26 is .4881
P(x > 80) = .5000 + .4881 = .9881
7.28
p = .19
n = 950
a) P( > .25):
z = = 4.71
from Table A.5, area = .5000
P( > .25) = .5000 - .5000 = .0000
b) P(.15 < < .20):
z = = -3.14
z=
= 0.79
from Table A.5, area for z = -3.14 is .4992
from Table A.5, area for z = 0.79 is .2852
P(.15 < < .20) = .4992 + .2852 = .7844
c) P(133 < x < 171):
=
= .14
=
= .18
P(.14 < < .18):
z = = -3.93
z=
= -0.79
from Table A.5, the area for z = -3.93 is .49997
the area for z = -0.79 is .2852
P(133 < x < 171) = .49997 - .2852 = .21477
7.29
= 76,
a) n = 35,
= 14
P( > 79):
z = = 1.27
from table A.5, area = .3980
P( > 79) = .5000 - .3980 = .1020
b) n = 140, P(74 < < 77):
z = = -1.69
z = = 0.85
from table A.5, area for z = -1.69 is .4545
from table A.5, area for 0.85 is .3023
P(74 < < 77) = .4545 + .3023 = .7568
c) n = 219,
z=
P( < 76.5):
= 0.53
from table A.5, area = .2019
P( < 76.5) = .5000 + .2019 = .7019
7.30 p = .46
a) n = 60
P(.41 < < .53):
z = = 1.09
from table A.5, area = .3621
z = = -0.78
from table A.5, area = .2823
P(.41 < < .53) = .3621 + .2823 = .6444
b) n = 458
P(< .40):
z = = -2.58
from table A.5, area = .4951
P( < .40) = .5000 - .4951 = .0049
c) n = 1350
P(> .49):
z = = 2.21
from table A.5, area = .4864
P(> .49) = .5000 - .4864 = .0136
7.31
7.32
Under 18
18 25
26 50
51 65
over 65
p = .55
n = 600
250(.22) =
250(.18) =
250(.36) =
250(.10) =
250(.14) =
n=
55
45
90
25
35
250
x = 298
= = .497
P(< .497):
z = = -2.61
from Table A.5, Prob. = .4955
P(< .497) = .5000 - .4955 =
.0045
No, the probability of obtaining these sample results by chance from a population
that supports the candidate with 55% of the vote is extremely low (.0045). This is
such an unlikely chance sample result that it would cause the researcher to
probably reject her claim of 55% of the vote.
7.33 a) Roster of production employees secured from the human
resources department of the company.
b) Alpha/Beta store records kept at the headquarters of
their California division or merged files of store
records from regional offices across the state.
c) Membership list of Maine lobster catchers association.
7.34
= $ 17,755
P(
= $ 650
n = 30
N = 120
< 17,500):
z = = -2.47
from Table A.5, the area for z = -2.47 is .4932
P(
< 17,500) = .5000 - .4932 = .0068
7.35
Number the employees from 0001 to 1250. Randomly sample from the random
number table until 60 different usable numbers are obtained. You cannot use
numbers from 1251 to 9999.
7.36
= $125
n = 32
= $110
P( > $110):
z = = -3.70
from Table A.5, Prob.= .5000
P( > $110) = .5000 + .5000 =
1.0000
2 = $525
P( > $135):
z = = 2.47
from Table A.5, Prob.= .4932
P( > $135) = .5000 - .4932 = .0068
P($120 < < $130):
z = = -1.23
z = = 1.23
from Table A.5, Prob.= .3907
P($120 < < $130) = .3907 + .3907 = .7814
7.37 n = 1100
a) x > 810,
p = .73
=
z=
= 0.48
from table A.5, area = .1844
P(x > 810) = .5000 - .1844 = .3156
b) x < 1030,
p = .96,
= = .9364
z = = -3.99
from table A.5, area = .49997
P(x < 1030) = .5000 - .49997 = .00003
c) p = .85
P(.82 < < .84):
z = = -2.79
from table A.5, area = .4974
z = = -0.93
from table A.5, area = .3238
P(.82 < < .84) = .4974 - .3238 = .1736
7.38
1)
2)
3)
4)
5)
6)
7)
8)
9)
The managers from some of the companies you are interested in
studying do not belong to the American Managers Association.
The membership list of the American Managers Association is not up-todate.
You are not interested in studying managers from some of the companies
belonging to the American Management Association.
The wrong questions are asked.
The manager incorrectly interprets a question.
The assistant accidentally marks the wrong answer.
The wrong statistical test is used to analyze the data.
An error is made in statistical calculations.
The statistical results are misinterpreted.
7.39
Divide the factories into geographic regions and select a few factories to represent
those regional areas of the country. Take a random sample of employees from
each selected factory. Do the same for distribution centers and retail outlets.
Divide the United States into regions of areas. Select a few areas. Take a random
sample from each of the selected area distribution centers and retail outlets.
7.40
N = 12,080
n = 300
k = N/n = 12,080/300 = 40.27
Select every 40th outlet to assure n > 300 outlets.
Use a table of random numbers to select a value between 0 and 40 as a starting
point.
7.41
p = .54
n = 565
a) P(x > 339):
= = .60
z = = 2.86
from Table A.5, the area for z = 2.86 is .4979
P(x > 339) = .5000 - .4979 = .0021
b) P(x > 288):
=
= .5097
z = = -1.45
from Table A.5, the area for z = -1.45 is .4265
P(x > 288) = .5000 + .4265 = .9265
c) P( < .50):
z = = -1.91
from Table A.5, the area for z = -1.91 is .4719
P( < .50) = .5000 - .4719 = .0281
7.42
= $550
n = 50
= $100
P( < $530):
z = = -1.41
from Table A.5, Prob.=.4207
P(x < $530) = .5000 - .4207 = .0793
7.43
= 56.8
a) P( > 60):
z = = 1.86
n = 51
= 12.3
from Table A.5, Prob. = .4686
P( > 60) = .5000 - .4686 = .0314
b) P( > 58):
z = = 0.70
from Table A.5, Prob.= .2580
P( > 58) = .5000 - .2580 = .2420
c) P(56 < < 57):
z=
= -0.46
z = = 0.12
from Table A.5, Prob. for z = -0.46 is .1772
from Table A.5, Prob. for z = 0.12 is .0478
P(56 < < 57) = .1772 + .0478 = .2250
d) P(< 55):
z = = -1.05
from Table A.5, Prob.= .3531
P( < 55) = .5000 - .3531 = .1469
e) P( < 50):
z = = -3.95
from Table A.5, Prob.= .5000
P( < 50) = .5000 - .5000 = .0000
7.45 p = .73 n = 300
a) P(210 < x < 234):
=
= .70
z = = -1.17
=
= .78
z = = 1.95
from Table A.5, the area for z = -1.17 is .3790
the area for z = 1.95 is .4744
P(210 < x < 234) = .3790 + .4744 = .8534
b) P(> .78):
z = = 1.95
from Table A.5, the area for z = 1.95 is .4744
P(> .78) = .5000 - .4744 = .0256
c) p = .73
n = 800
P(> .78):
z = = 3.19
from Table A.5, the area for z = 3.19 is .4993
P(> .78) = .5000 - .4993 = .0007
7.46
n = 140
P(x > 35):
= = .25
p = .22
z = = 0.86
from Table A.5, the area for z = 0.86 is .3051
P(x > 35) = .5000 - .3051 = .1949
P(x < 21):
= = .15
z = = -2.00
from Table A.5, the area for z = 2.00 is .4772
P(x < 21) = .5000 - .4772 = .0228
n = 300
p = .20
P(.18 < < .25):
z = = -0.87
from Table A.5, the area for z = -0.87 is .3078
z = = 2.17
from Table A.5, the area for z = 2.17 is .4850
P(.18 < < .25) = .3078 + .4850 = .7928
7.47
By taking a sample, there is potential for obtaining more detailed information.
More time can be spent with each employee. Probing questions can
be asked. There is more time for trust to be built between employee and
interviewer resulting in the potential for more honest, open answers.
With a census, data is usually more general and easier to analyze because it is in a
more standard format. Decision-makers are sometimes more comfortable with a
census because everyone is included and there is no sampling error. A census
appears to be a better political device because the CEO can claim that everyone in
the company has had input.
7.48
p = .75
n = 150
x = 120
P(> .80):
z = = 1.41
from Table A.5, the area for z = 1.41 is .4207
P( > .80) = .5000 - .4207 = .0793
7.49 a) Switzerland: n = 40
= $ 30.67
=$3
P(30 < < 31):
from Table A.5, the area for z = 0.70 is .2580
the area for z = -1.41 is .4207
P(30 < < 31) = .2580 + .4207 = .6787
b) Japan: n = 35
= $3
= $ 20.20
P( > 21):
from Table A.5, the area for z = 1.58 is .4429
P( > 21) = .5000 - .4429 = .0571
c) U.S.: n = 50
= $ 23.82
=$3
P( < 22.75):
from Table A.5, the area for z = -2.52 is .4941
P(< 22.75) = .5000 - .4941 = .0059
7.50
a)
b)
c)
d)
7.51
Age, Ethnicity, Religion, Geographic Region, Occupation, UrbanSuburban-Rural, Party Affiliation, Gender
Age, Ethnicity, Gender, Geographic Region, Economic Class
Age, Ethnicity, Gender, Economic Class, Education
Age, Ethnicity, Gender, Economic Class, Geographic Location
= $281
n = 65
= $47
P( > $273):
z = = -1.37
from Table A.5 the area for z = -1.37 is .4147
P( > $273) = .5000 + .4147 = .9147