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8/3/2019 Ken Black QA ch08
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Business Statistics, 5th ed.
by Ken Black
Chapter 8
Statistical Inference:Estimation for
Single Populations
Discrete Distributions
PowerPoint presentations prepared by Lloyd Jaisingh,Morehead State University
8/3/2019 Ken Black QA ch08
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Learning Objectives
Know the difference between point and intervalestimation.
Estimate a population mean from a sample meanwhen s is known.
Estimate a population mean from a sample meanwhen s is unknown. Estimate a population proportion from a sample
proportion. Estimate the population variance from a sample
variance. Estimate the minimum sample size necessary to
achieve given statistical goals.
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Statistical Estimation
Point estimate -- the single value of a statisticcalculated from a sample which is used toestimate a population parameter
Interval Estimate -- a range of values calculatedfrom a sample statistic(s) and standardizedstatistics, such as thez Selection of the standardized statistic is
determined by the sampling distribution. Selection of critical values of the standardizedstatistic is determined by the desired level ofconfidence.
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100(1 - )%Confidence Interval toEstimate when s is Known
Point estimate
Interval
Estimate
n
xx
nzx
nzx
or
nzx
s
s
s
2/2/
2/
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Distribution of Sample Means
for (1-)% Confidence
X
Z0
2Z
2Z
2
2
2
z
2
z
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Distribution of Sample Means
for (1-)% Confidence
XZ
0 2
Z 2
Z
2
2.5
2
.52
2
z
2
z
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Distribution of Sample Means
for (1-)% Confidence
XZ
0 2
Z 2
Z
2
2 12
1
2
2
z
2
z
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Distribution of Sample Means
for 95% Confidence
.4750 .4750
X
95%
.025.025
Z1.96-1.96 0
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95% Confidence Interval for 96.1,85,46,510 2/ s znx
78.51922.500
78.951078.9510
85
4696.1510
85
4696.1510
2/2/
s
s
n
zx
n
zx
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95% Confidence Intervals for
X
95%
XX
X
X
X
X
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95% Confidence Intervals for
X
95%
XX
X
XX
X
Is our interval,
500.22519.78, in the
red?
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Demonstration Problem 8.1
365.12545.8910.1455.10910.1455.10
44
7.7645.1455.10
44
7.7645.1455.10
2/2/
s
s
nzxnzx
645.1confidence%90
.44,7.7,455.10
2/
sz
nx
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Demonstration Problem 8.2
85.3675.31
554.230.34554.230.341800
50800
50
833.230.34
1800
50800
50
833.230.34
112/2/
s
s
N
nN
nzx
N
nN
nzx
33.2confidence%98
.50and,800=,8,30.34
z
nNx s
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Confidence Interval to Estimate whenn is Large and s is Unknown
nszx
nszx
orn
szx
2/2/
2/
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Car Rental Firm Example
2.908.807.45.857.45.85
110
3.19575.25.85
110
3.19575.25.85
n
szx
n
szx
575.2confidence%99.110and,3.19S,5.85
z
nx
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Z Values for Some of the More
Common Levels of Confidence
90%
95%
98%
99%
ConfidenceLevel
z/2 Value
1.645
1.96
2.33
2.575
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Estimating the Mean of a Normal
Population: Unknown s The population has a normal distribution.
The value of the population standarddeviation is unknown.
z distribution is not appropriate for theseconditions
tdistribution is appropriate
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Thet Distribution
Developed by British statistician, WilliamGosset
A family of distributions -- a unique
distribution for each value of its parameter,degrees of freedom (d.f.)
Symmetric, Unimodal, Mean = 0, Flatterthan az
tformula
n
s
xt
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Comparison of Selectedt Distributions
to the Standard Normal
-3 -2 -1 0 1 2 3
Standard Normal
t (d.f. = 25)
t (d.f. = 1)
t (d.f. = 5)
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Table of Critical Values oft
df t0.100 t0.050 t0.025 t0.010 t0.0051 3.078 6.314 12.706 31.821 63.656
2 1.886 2.920 4.303 6.965 9.925
3 1.638 2.353 3.182 4.541 5.841
4 1.533 2.132 2.776 3.747 4.6045 1.476 2.015 2.571 3.365 4.032
23 1.319 1.714 2.069 2.500 2.807
24 1.318 1.711 2.064 2.492 2.79725 1.316 1.708 2.060 2.485 2.787
29 1.311 1.699 2.045 2.462 2.756
30 1.310 1.697 2.042 2.457 2.750
40 1.303 1.684 2.021 2.423 2.704
60 1.296 1.671 2.000 2.390 2.660
120 1.289 1.658 1.980 2.358 2.617
1.282 1.645 1.960 2.327 2.576
t
With df = 24 and= 0.05,
t = 1.711.
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Confidence Intervals for of a NormalPopulation: Unknown s
1
1,2/1,2/
1,2/
ndf
n
stx
n
stx
or
n
stx
nn
n
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Solution for Demonstration Problem 8.3
18.310.1
04.114.204.114.2
14
29.1012.314.2
14
29.1012.314.2
1,2/1,2/
n
stx
n
stx nn
012.3
005.02
99.1
2
131,14,29.1,14.2
13,005.
t
ndfnsx
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MINITAB Solution for Demonstration
Problem 8.3
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Comp Time: Excel Normal View
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Comp Time: Excel Formula View
A B C D E F
1 Comp Time Data
2 6 21 17 20 7 0
3 8 16 29 3 8 12
4 11 9 21 25 15 16
56 n = =COUNT(A2:F4)
7 Mean = =AVERAGE(A2:F4)
8 S = =STDEV(A2:F4)
9 Std Error = =B8/SQRT(B6)
10
11 = 0.112 df = =B6-1
13 t = =TINV(B11,B12)
14
15 =B7-B13*B9 =B7+B13*B9
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Confidence Interval to Estimate
the Population Proportion
sizesample=
proportionpopulation=
-1=
proportionsample=:
2/2/
n
p
pq
pwhere
n
qpzpp
n
qpzp
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Solution for Demonstration Problem 8.5
20.012.0
04.016.004.016.0
212
)84.0)(16.0(645.116.0
212
)84.0)(16.0(645.116.0
2/2/
p
p
p
n
qpzpp
n
qpzp
645.1%90
84.016.01-1=
16.0212
34
,34,212
zConfidence
pq
n
x
pxn
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Population Variance
Variance is an inverse measure of the groupshomogeneity.
Variance is an important indicator of total qualityin standardized products and services. Managers
improve processes to reduce variance. Variance is a measure of financial risk. Variance of
rates of return help managers assess financial andcapital investment alternatives.
Variability is a reality in global markets.Productivity, wages, and costs of living varybetween regions and nations.
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Estimating the Population Variance
Population Parameter s
Estimator ofs
formula for Single Variance
1
)(2
2
n
xxs
1-freedomofdegrees
)1(2
2
2
n
sn
s
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Confidence Interval for s2
confidenceoflevel1
1
11
2
2/1
22
2
2/
2
s
ndf
snsn
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Selected 2 Distributionsdf = 3
df = 5
df = 10
0
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2 Table
0 5 10 15 20
0.10
df = 5
9.23635
df 0.975 0.950 0.100 0.050 0.0251 9.82068E-04 3.93219E-03 2.70554 3.84146 5.023902 0.0506357 0.102586 4.60518 5.99148 7.377783 0.2157949 0.351846 6.25139 7.81472 9.348404 0.484419 0.710724 7.77943 9.48773 11.143265 0.831209 1.145477 9.23635 11.07048 12.832496 1.237342 1.63538 10.6446 12.5916 14.4494
7 1.689864 2.16735 12.0170 14.0671 16.01288 2.179725 2.73263 13.3616 15.5073 17.53459 2.700389 3.32512 14.6837 16.9190 19.0228
10 3.24696 3.94030 15.9872 18.3070 20.4832
20 9.59077 10.8508 28.4120 31.4104 34.169621 10.28291 11.5913 29.6151 32.6706 35.478922 10.9823 12.3380 30.8133 33.9245 36.780723 11.6885 13.0905 32.0069 35.1725 38.075624 12.4011 13.8484 33.1962 36.4150 39.364125 13.1197 14.6114 34.3816 37.6525 40.6465
70 48.7575 51.7393 85.5270 90.5313 95.023180 57.1532 60.3915 96.5782 101.8795 106.628590 65.6466 69.1260 107.5650 113.1452 118.1359
100 74.2219 77.9294 118.4980 124.3421 129.5613
With df = 5 and =0.10, 2 = 9.23635
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Two Table Values of2
0 2 4 6 8 10 12 14 16 18 20
df = 7
.05
.05
.95
2.16735 14.0671
df 0.950 0.0501 3.93219E-03 3.841462 0.102586 5.991483 0.351846 7.814724 0.710724 9.487735 1.145477 11.070486 1.63538 12.59167 2.16735 14.0671
8 2.73263 15.50739 3.32512 16.919010 3.94030 18.3070
20 10.8508 31.410421 11.5913 32.670622 12.3380 33.924523 13.0905 35.172524 13.8484 36.4150
25 14.6114 37.6525
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90% Confidence Interval for s2
007146.001101.16735.2
0022125).18(
0671.14
0022125).18(
)1()1(
______________________________________
16735.2
0671.14
10.,71,8,0022125.
2
2
2
2/1
22
2
2/
2
2
95.
2
2/1.01
2
2/1
2
05.
2
2/1.0
2
2/
2
s
s
s
snsn
ndfns
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Solution for Demonstration Problem 8.6
4277.27648.0
4011.12
)2544.1(125
3641.39
)2544.1(125
11
2
2
2
2/1
22
2
2/
2
s
s
s
snsn
4011.12
3641.39
05.,241,25,2544.1
2
975.
2
2/05.01
2
2/1
2
025.
2
2/05.0
2
2/
2
ndfns
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Determining Sample Size
when Estimating z formula
Error of Estimation(tolerable error)
Estimated Sample Size
Estimated s
n
xz
s
xE
E
z
E
zn
ss 2
2
2
22
2
s1
4range
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Sample Size When Estimating : Example
44or30.43
12
)4( 2)645.1( 2
2
22
2
E
zn
s
645.1confidence%90
4,1
z
E s
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Solution for Demonstration Problem 8.7
38or52.37
22
)25.6( 2)96.1( 2
2
22
E
zn s
25.6254
1
4
1:
96.1confidence%95
25,2
rangeestimated
z
rangeE
s
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Determining Sample Size
when Estimatingp
zformula
Error of Estimation (tolerableerror)
Estimated Sample
Size
n
qp
ppZ
ppE
E
pqzn
2
2
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Solution for Demonstration Problem 8.8
448,1or7.447,1
)003(.)60.0)(40.0()33.2(
2
2
2
2
E
pqzn
E
Confidence Z
estimated P
Q P
0 03
98% 2 33
0 40
1 0 60
.
.
.
.
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Determining Sample Size when
Estimatingp with No Prior Information
P
n
0
50
100
150
200250
300
350
400
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
z = 1.96
E = 0.05
E
zn
2
2
4
1
p
0.5
0.4
0.3
0.2
0.1
pq
0.25
0.24
0.21
0.16
0.09
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Example: Determining n when
Estimatingp with No Prior Information
2716.270
)05(.
)50.0)(50.0()645.1(2
2
2
2
or
pqn
Ez
50.01
50.0,estimate
645.1%90
05.0
pq
pusepofpriornowith
zConfidence
E
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Copyright 2008 John Wiley & Sons, Inc.All rights reserved. Reproduction or translation
of this work beyond that permitted in section 117of the 1976 United States Copyright Act without
express permission of the copyright owner isunlawful. Request for further information shouldbe addressed to the Permissions Department, JohnWiley & Sons, Inc. The purchaser may makeback-up copies for his/her own use only and notfor distribution or resale. The Publisher assumes
no responsibility for errors, omissions, or damagescaused by the use of these programs or from theuse of the information herein.