Ch 8-Statistical Inference Estimation for Single Population

8/3/2019 Ch 8-Statistical Inference Estimation for Single Population

1/42

Chapter 8

Statistical Inference:

Estimation for Single Populations


2/42

This chapter broadly Focus on below points

The difference between point and intervalestimation.

Estimation of a population mean from a samplemean when s is known.

Estimation of a population mean from a samplemean when s is unknown.


3/42

Estimating the Population Mean

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

Interval estimate - a range of values within whichthe analyst can declare, with some confidence, thepopulation lies


4/42

Confidence Interval to Estimate when s is Known

Point estimate

Interval Estimate

n

xx

nzx

nzx

or

nzx

s

s

s


5/42

Distribution of Sample Means for 95% Confidence

.4750 .4750

X

95%

.025.025

Z

1.96-1.96 0


6/42


For a 95% confidence interval = .05

/2 = .025

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

.5000 - .0250 = .4750

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


7/42


is used to locate the Z value in constructing theconfidence interval

The confidence interval yields a range withinwhich 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 ofnumbers when the data are normally distributed


8/42

95% Confidence Intervals for

X

95%

X

X

X

X

X

X


9/42

96.1,85,46,510 2/ s znx

78.51922.500

78.951078.9510

85

4696.1510

85

4696.1510

2/2/

s

s

nzxnzx

95% Confidence Interval for


10/42

Demonstration Problem 8.1

A survey was taken of U.S. companies that dobusiness with firms in India. One of the questionson the survey was: Approximately how many yearshas your company been trading with firms in India?

A random sample of 44 responses to this question yieldeda mean of 10.455 years. Suppose the population standarddeviation for this question is 7.7 years. Using thisinformation, construct a 90% confidence interval for themean number of years that a company has been tradingin India for the population of U.S. companies trading withfirms in India.


11/42

365.12545.8

91.1455.1091.1455.1044

7.7645.1455.10

44

7.7645.1455.10

ss

nzx

nzx

645.1confidence%90

.44,7.7,455.10

z

nx s



12/42


A study is conducted in a company that employs800 engineers. A random sample of 50 engineersreveals that the average sample age is 34.3 years.Historically, the population standard deviation of

the age of the companys engineers isapproximately 8 years. Construct a 98%confidence interval to estimate the average age ofall the engineers in this company.


13/42

85.3675.31

554.23.34554.23.341800

50800

50

833.23.34

1800

50800

50

833.23.34

11

s

s

N

nN

nzx

N

nN

nzx

33.2confidence%98

.50and,800=,8,3.34

z

nNx s



14/42

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 inthe estimation process.

z distribution is not appropriate for theseconditions when the Population Std Dev isunknown, tdistribution is appropriate, and youuse the Sample Std Dev in the t formula


15/42

tDistribution

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

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

tdistribution is used instead of the z distribution for

doing inferential statistics on the population mean whenthe population Std Dev is unknown and the populationis normally distributed

With the t distribution,use the Sample Std Dev


16/42

ns

xt

tDistribution

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 tformula


17/42

tDistribution Characteristics

tdistribution flatter in middle and have morearea in their tails than the normal distribution

tdistribution approach the normal curve as n becomeslarger

t distribution is to be used when the population varianceor population Std Dev is unknown, regardless of the sizeof the sample


18/42

Reading the tDistribution

ttable uses the area in the tail of the distribution Emphasis in the ttable is on , and each tail of the

distribution contains /2 of the area under the curvewhen confidence intervals are constructed

tvalues are located at the intersection of the dfvalue and the selected /2 value


19/42

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


20/42

Table of Critical Values oft

t

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

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 4.604

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


21/42

1

ndf

n

s

txn

s

tx

or

n

stx

Confidence Intervals for of a Normal Population:

Unknown s


22/42


The owner of a large equipment rental company wants to make arather 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. Fourteendifferent 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


23/42

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

ndfnsx

Solution for Demonstration Problem 8.3


24/42

MINITAB Solution for Demonstration Problem 8.3


25/42

Comp Time: Excel Normal View


26/42

sizesample=

proportionpopulation=

-1=

proportionsample=

:

22

n

ppq

p

where

n

qpzpp

n

qpzp

Confidence Interval to Estimate the Population

Proportion

Estimating the population proportion oftenmust be made


27/42


A clothing company produces mens jeans. The jeans aremade and sold with either a regular cut or a boot cut. Inan 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 fromthe companys two Oklahoma City retail outlets. Only 34of the sales were for boot-cut jeans. Construct a 90%confidence interval to estimate the proportion of thepopulation in Oklahoma City who prefer boot-cut jeans.


28/42

20.012.0

04.016.004.016.0

212

)84.0)(16.0(645.116.0212

)84.0)(16.0(645.116.0

p

p

p

n

qpzpp

n

qpzp

645.1%90

84.016.01-1=

16.021234,34,212

zConfidence

pq

nxpxn



29/42

1

)(2

2

n

xxs

1-freedomofdegrees

)1(

2

22

n

sn

s

Estimating the Population Variance

Population Parameter s

Estimator ofs

formula for Single Variance


30/42

confidencoflevel1

1

112

21

22

2

2

2

s

ndf

snsn

Confidence Interval for s2


31/42

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 0.050

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 15.50739 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


32/42

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.

1

2

21

2

05.

2

2

1.

2

2

2

s

s

s

snsn

ndfns


33/42


The U.S. Bureau of Labor Statistics publishes data on the hourlycompensation 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 workersin 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.


34/42

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

2

1

2

025.

2

2

05.

2

2

2

ndfns



35/42

Determining Sample Size when Estimating

It may be necessary to estimate the sample size whenworking on a project

In studies where is being estimated, the size of thesample can be determined by using the z formula for

sample means to solve for n Difference between and is the error of

estimation Error of Estimation = ( - )

x

x


36/42

Determining Sample Size when Estimating

z formula

Error of Estimation (tolerable error)

Estimated Sample Size

Estimated s

n

x

z s

xE

E

z

E

zn

ss 2

2

2

22

2

s1

4range


37/42

44or30.43

12

)4( 2)645.1( 2

2

22

2

E

zn

s

645.1confidence%904,1

zE s

Sample Size When Estimating : Example


38/42


Suppose you want to estimate the average age of allBoeing 737-300 airplanes now in active domestic U.S.service. You want to be 95% confident, and you wantyour estimate to be within one year of the actual figure.The 737-300 was first placed in service about 24 yearsago, but you believe that no active 737-300s in the U.S.domestic fleet are more than 20 years old. How large ofa sample should you take?


39/42

38or52.37

22

)25.6( 2)96.1( 2

2

22

Ezn s

25.6254

1

4

1:

96.1confidence%9525,2

rangeestimated

zrangeE

s



40/42

Determining Sample Size when Estimatingp

z formula

Error of Estimation (tolerable error)

Estimated Sample Size

n

qp

ppZ

ppE

E

pqzn

2

2


41/42


Hewitt Associates conducted a national survey todetermine the extent to which employers are promotinghealth and fitness among their employees. One of thequestions asked was, Does your company offer on-siteexercise classes? Suppose it was estimated before thestudy that no more than 40% of the companies wouldanswer Yes. How large a sample would Hewitt Associateshave to take in estimating the population proportion toensure a 98% confidence in the results and to be within

.03 of the true population proportion?


42/42

448,1or7.447,1

)003(.

)60.0)(40.0()33.2(2

2

2

2

E

pqzn

60.01

40.0

33.2%9803.0

PQ

Pestimated

ZConfidenceE


Ch 8-Statistical Inference Estimation for Single Population

Documents