Ken Black QA ch08

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


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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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XZ

0 2

Z 2

Z

2

2.5

2

.52

2

z

2

z


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XZ

0 2

Z 2

Z

2

2 12

1

2

2

z

2

z


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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


25/43

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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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


31/43

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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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


38/43


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


39/43

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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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

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Ken Black QA ch08

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