Chapter 8 Ken Black

Copyright 2010 John Wiley & Sons, Inc. 1 Copyright 2010 John Wiley & Sons, Inc.

Business Statistics, 6th ed. by Ken Black

Chapter 8

Statistical Inference:

Estimation for Single Populations

Copyright 2010 John Wiley & Sons, Inc. 2

Learning Objectives

Know the difference between point and interval estimation.

Estimate a population mean from a sample mean when s is known.

Estimate a population mean from a sample mean when s is unknown.


Learning Objectives

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.


A point estimate is a static taken from a sample that is used to estimate a population parameter

Interval estimate - a range of values within which the analyst can declare, with some confidence, the population lies

Estimating the Population Mean


Confidence Interval to Estimate when s is Known

n

xx

nzx

nzx

or

nzx

s

s

s

Point estimate

Interval Estimate


Distribution of Sample Means for 95% Confidence

.4750 .4750

X

95%

.025 .025

Z 1.96 -1.96 0


is used to locate the Z value in constructing the confidence interval

The confidence interval yields a range within which the researcher feel with some confidence the population mean is located

Z score the number of standard deviations a value (x) is above or below the mean of a set of numbers when the data are normally distributed



For a 95% confidence interval

= .05

/2 = .025

Value of /2 or z.025 look at the standard normal distribution table under

.5000 - .0250 = .4750

From Table A5 look up .4750, and read 1.96 as the z value from the row and column



Confidence Intervals

Z values: 90% - 1.645

95% - 1.96

98% - 2.33

99% - 2.575


96.1 ,85,46,510 2/ s znx

78.51922.500

78.951078.9510

85

4696.1510

85

4696.1510

2/2/

s

s

nzx

nzx

95% Confidence Interval for


Example No. 1

A survey was taken of U.S. companies that do business with firms in India. One of the questions on the survey was: Approximately how many years has your company been trading with firms in India? A random sample of 44 responses to this question yielded a mean of 10.455 years. Suppose the population standard deviation for this question is 7.7 years. Using this information, construct a 90% confidence interval for the mean number of years that a company has been trading in India for the population of U.S. companies trading with firms in India.


Example No. 2

A study is conducted in a company that employs 800 engineers. A random sample of 50 engineers reveals that the average sample age is 34.3 years. Historically, the population standard deviation of the age of the companys engineers is approximately 8 years. Construct a 98% confidence interval to estimate the average age of all the engineers in this company.


Estimating the Mean of a Normal Population: Unknown s

The population has a normal distribution.

The value of the population Standard Deviation is unknown, then sample Std Dev must be used in the estimation process.

z distribution is not appropriate for these conditions when the Population Std Dev is unknown, t distribution is appropriate, and you use the Sample Std Dev in the t formula


t Distribution

A family of distributions -- a unique distribution for each value of its parameter, degrees of freedom (d.f.)

Symmetric, Unimodal, Mean = 0, Flatter than a z

t distribution is used instead of the z distribution for doing inferential statistics on the population mean when the population Std Dev is unknown and the population is normally distributed

With the t distribution, you use the Sample Std Dev


n

s

xt

A family of distributions - a unique distribution for each value of its parameter using degrees of freedom (d.f.)

Symmetric, Unimodal, Mean = 0, Flatter than a z

t formula

t Distribution


t distribution flatter in middle and have more area in their tails than the normal distribution

t distribution approach the normal curve as n becomes larger

t distribution is to be used when the population variance or population Std Dev is unknown, regardless of the size of the sample

t Distribution Characteristics


t table uses the area in the tail of the distribution Emphasis in the t table is on , and each tail of the distribution contains /2 of the area under the curve when confidence intervals are constructed

t values are located at the intersection of the df value and the selected /2 value

Reading the t Distribution


1

1,2/1,2/

1,2/

ndf

n

stx

n

stx

or

n

stx

nn

n

Confidence Intervals for of a Normal Population: Unknown s


Table of Critical Values of t

t

With df = 24 and = 0.05, t = 1.711.

df t0.100 t0.050 t0.025 t0.010 t0.005 1 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 4.604

5 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 2.797 25 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


1

ndf

n

stx

n

stx

or

n

stx

Confidence Intervals for of a Normal Population: Unknown s


Example # 3

The owner of a large equipment rental company wants to make a rather quick estimate of the average number of days a piece of ditch digging equipment is rented out per person per time. The company has records of all rentals, but the amount of time required to conduct an audit of all accounts would be prohibitive. The owner decides to take a random sample of rental invoices. Fourteen different rentals of ditch diggers are selected randomly from the files, yielding the following data. She uses these data to construct a 99% confidence interval to estimate the average number of days that a ditch digger is rented and assumes that the number of days per rental is normally distributed in the population.

3 1 3 2 5 1 2 1 4 2 1 3 1 1


18.310.1

04.114.204.114.2

14

29.1012.314.2

14

29.1012.314.2

n

stx

n

stx

012.3

005.02

99.1

2

131 ,14 ,29.1,14.2

13,005.

t

ndfn sx

Solution for Example 3


size sample =

proportion population =

-1=

proportion sample =

:

22

n

p

pq

p

where

n

qpzpp

n

qpzp

Confidence Interval to Estimate the Population Proportion

Estimating the population proportion often must be made


Example #4

A clothing company produces mens jeans. The jeans are made and sold with either a regular cut or a boot cut. In an effort to estimate the proportion of their mens jeans market in Oklahoma City that prefers boot-cut jeans, the analyst takes a random sample of 212 jeans sales from the companys two Oklahoma City retail outlets. Only 34 of the sales were for boot-cut jeans. Construct a 90% confidence interval to estimate the proportion of the population in Oklahoma City who prefer boot-cut jeans.


20.012.0

04.016.004.016.0

212

)84.0)(16.0(645.116.0

212

)84.0)(16.0(645.116.0

p

p

p

n

qpzpp

n

qpzp

645.1 %90

84.016.01-1=

16.0212

34 ,34 ,212

zConfidence

pq

n

xpxn

Example # 4


1

)( 22

n

xxs

1-freedom of degrees

)1(2

22

n

sn

s

Estimating the Population Variance

Population Parameter s

Estimator of s

formula for Single Variance


confidence of level 1

1

112

21

22

2

2

2

s

ndf

snsn

Confidence Interval for s2


Two Table Values of 2

0 2 4 6 8 10 12 14 16 18 20

df = 7

.05

.05

.95

2.16735 14.0671

df 0.950 0.050 1 3.93219E-03 3.84146 2 0.102586 5.99148 3 0.351846 7.81472 4 0.710724 9.48773 5 1.145477 11.07048 6 1.63538 12.5916 7 2.16735 14.0671 8 2.73263 15.5073 9 3.32512 16.9190

10 3.94030 18.3070

20 10.8508 31.4104 21 11.5913 32.6706 22 12.3380 33.9245 23 13.0905 35.1725 24 13.8484 36.4150 25 14.6114 37.6525


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

21

22

2

2

2

2

95.

2

2

1.1

2

21

2

05.

2

2

1.

2

2

2

s

s

s

snsn

ndfns


Example # 5

The U.S. Bureau of Labor Statistics publishes data on the hourly compensation costs for production workers in manufacturing for various countries. The latest figures published for Greece show that the average hourly wage for a production worker in manufacturing is $16.10. Suppose the business council of Greece wants to know how consistent this figure is. They randomly select 25 production workers in manufacturing from across the country and determine that the standard deviation of hourly wages for such workers is $1.12. Use this information to develop a 95% confidence interval to estimate the population variance for the hourly wages of production workers in manufacturing in Greece. Assume that the hourly wages for production workers across the country in manufacturing are normally distributed.


4277.27648.0

4011.12

)2544.1(125

3641.39

)2544.1(125

11

2

2

2

21

22

2

2

2

s

s

s

snsn

4011.12

3641.39

05. ,241 ,25 ,2544.1

2

975.

2

2

05.1

2

21

2

025.

2

2

05.

2

2

2

ndfns

Example # 5

Chapter 8 Ken Black

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