confidence intervals final

8/10/2019 Confidence Intervals Final

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Confidence intervals

Resi Emina

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Lesson Content

Confidence interval

For population mean

For variance

For population proportion

For regresion coefiicients

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Confidence interval for the

mean

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Confidence interval for the mean

We want to estimate the population mean(which does not change) using the samplemean (which will change from sample tosample)

The population mean will be in the range:

error)inargm(samplingx

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Samples means will vary fromsample to sample

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Confidence interval for thepopulation mean

Standard deviation from population isknown

z distribution

Standard deviation from population isntknown

t distribution


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population mean, known

where: is the sample mean

z is the upper ( ) critical value for thestandard normal distribution and depends onrequired confidence

is the mean standard error

X

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X n

2 ( ) 1 1X XP X z X z F z

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Adequate Sample Size,known

In most applications, a sample size ofn =30 isadequate.

If the population distribution is highly skewed orcontains outliers, a sample size of 50 or more isrecommended.

If the population is not normally distributed but isroughly symmetric, a sample size as small as 15will suffice.

If the population is believed to be at leastapproximately normal, a sample size of less than15 can be used.

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Example Discount Sounds, IntervalEstimate of Population Mean,

known

Discount Sounds has 260 retail outlets throughoutthe United States. The firm is evaluating apotential location for a new outlet, partially basedon the mean of an annual income of the individualswho live in the marketing area of the new location.

A sample of sizen = 36 was taken. The sample mean income is $31,100. The population is not believed to be highly skewed. The population standard deviation is estimated to

be $4,500, and the confidence coefficient to beused in the interval estimate is 0.95.

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95% of the sample means that can be observed

are within + 1.96 standard errors of the population mean.

The margin of error is:

Thus, at 95% confidence, the margin of error is $1,470.

Example Discount Sounds, IntervalEstimate of Population Mean, known, cont.

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Interval estimate of is:

Example Discount Sounds, Interval

Estimate of Population Mean, known, cont.

We are 95% confident that the given interval contains the

population mean.

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population mean,unknown

If standard deviation from population isntknown, unbiased estimator is:

whereS is standard deviation from sample.

2

1

i ix X f

Sn


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population mean,unknown, cont.

where: is the mean from the sample

t is the upper ( ) critical value for thetdistribution with (n-1) degrees of freedom,

is approximation or estimation ofmean standard error

Ifn>30, than we can replacet distribution withznormal distribution.

1 1 12 ( ) 1 1n n nX XP X t S X t S S t

X1

2

1nt X

SSn

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Question 1

One hundred students passedthe Statistics exam. In arandom sample of 15 studentswe get the following grades:

6, 7, 8, 6, 7, 9, 9, 10, 7, 8, 9,7, 8, 7, 6.

We wish to rate the averageStatistics grade for allstudents with 95% probability.To determine estimation forthe mean standard errorin thiscase we will use the followingformula:

a.

b.

c.

d.

S

n

S

n

1n

n

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Question 2

In the sample of 22 elements, wecalculated the mean of 54 and the varianceof 24,8. We wish to determine the intervalfor population mean, with 99% certainty.What frequency distribution do we need to

apply? Fisher's

Chi-square

Student's

Normal16

Example 1

It is assumed that the basic set has normaldistribution. We took a sample of 56 elementsand calculated the arithmetic mean of 12.5 with

standard deviation of 2.

What interval will contain population mean withthe type I error of 4%?

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Solution

n>30, unknown standard deviation forpopulation, we know only standarddeviation for samplez distribution

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12.5

2

0.04

n

X

S

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Solution, cont.

Interval that will contain population mean with the type Ierror of 4% is 11.95 13.05.

X XX z S X z S

( ) 1 0.98 2.062 from tables

F z z

2 212.5 2.06 12.5 2.06

56 56

11.95 13.05


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Example 2

Suppose a random sample of 14 students waschosen, and each student was asked thenumber of hours he or she studies each week.The resulting statistics were:

Determine confidence interval for averagehours he or she studies each week, ifconfidence level is 99%.

9.2 and 0.3X S

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Solution

n


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5

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Question 3

Thet distribution is used when:

a. The standard deviation from population isunknown and the sample is large

b. The standard deviation from population is knownand the sample is large

c. The standard deviation from population is knownand the sample is small

d. The standard deviation from population isunknown and the sample is small

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Summary of Interval EstimationSummary of Interval Estimation

ProceduresProcedures

for a Population Meanfor a Population Mean

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variance

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Confidence interval for thepopulation variance

Depending on whether the sample issmall or large for the determination ofconfidence interval for the populationvariance we use:

chi-square distribution for small samplesor

normal distribution for large sample

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Confidence interval for thepopulation variance, small sample

2 2

2

2 2

1,1 1,2 2

1

n n

n S n S P

2 2

11,1

2

2 2

11,

2

( ) 12

( )2

nn

nn

P

P

This is not usual formfor confidence interval.

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Confidence interval for thepopulation variance, large sample

2 2

2

2 2

2 22 ( ) 1

2 3 2 3

n S n S P F z

n z n z

( ) 12

F z z


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Example 4

Suppose a student is measuring the boilingtemperature of a certain liquid observes thereadings (in degrees Celsius):

102.5, 101.7, 103.1, 100.9, 100.5, and 102.2

on 6 different samples of the liquid.

What is the confidence interval for thepopulation variance at a 97% confidence level?

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Solution

It is necessary first to determine variancefrom the sample:

It is a small sample (n=6) and we use a chi-square distribution for=0.03.

Now we can complete the term for theconfidence interval:

0.9697S

2 22

2 2

1,1 1,2 2

1

n n

n S n S P

2 2 2

1 6 1 5, 0,9851,1

2

2 2 2

1 6 1 5, 0,0151,

2

( ) 1 14, 0982

( ) 0, 0622

nn

from tables

nn

from tables

P

P

2

2

6 0.9697 6 0.9697

14.098 0.662

0.413 8.789

Confidence interval for variance ofvariable temperature boiling withreliability 97% read (0.413-8.789).

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Example 5

According to report for 2009. year, we have dataabout predicted Recovery rate in cent per dollarafter closing business (fromhttp://www.doingbusiness.org/CustomQuery/,predictions for 2009) for sample with 33 countries.We have data in Excel sheet and we get thesample mean 52.91 and sample variance 541.16.

We have to construct confidence interval forvariance of variable Recovery rate for populationof all countries with the type I error equal to 4%.

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Solution

It is a large sample (n=33) and we use anormal distribution for=0.04.

Now we can complete the term for confidenceinterval:

2 22

2 2

2 2

2 3 2 3

n S n S

n z n z

2

2 2

2

2 33 541.16 2 33 541.16

2 33 3 2.05 2 33 3 2.05

3 53 .6 2 1, 0 30 .4 9

0.04 ( ) 1 0.982

2.05from tables

F z

z

Confidence interval for variance ofvariable Recovery rate for populationof all countries with the type I errorequal to 4% is (353.62-1,030.49).

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Confidence interval for theproportion

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Confidence interval for thepopulation proportion

Applying the general formula for a confidence interval, theconfidence interval for a population proportion,p, is

where:

is the proportion in the sample,

z depends on the desired level of confidence, and

, the standard error of a proportion, is equal to:

AA A pp p z

1

A

A A

p

p p

n

Ap

p
http://www.doingbusiness.org/CustomQuery/http://www.doingbusiness.org/CustomQuery/


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Confidence interval for thepopulation proportion, cont.

Sincep for population is not known,p fromsample is used to estimate it. Therefore theestimated value of standard error of aproportion is:

and than will be:

1

A

A A

p

p pS

n

AA A pp p z S 38

Example 6

The manager of a bank that has 1,000 depositorsin a small city wants to determine the proportionof its depositors with more than one account atthe bank.

Set up a 90% confidence interval estimate of thepopulation proportion of the banks depositorswho have more than one account at the bank if arandom sample of 100 depositors is selectedwithout replacement and 35 of them state thatthey have more than one account at the bank.

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Solution

p from sample =35/100=0.35 andn=100

1 0.35 0.650.048

100AA A

p

p pS

n

0.10 ( ) 1 0.95 1.65

2 from tablesF z z

0.35 1.65 0.048 0.35 1.65 0.048

0.2708 0.4292 27.08 42.92%

AA A pp p z S

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Question 4

If we decide to use 99% confidenceinterval rather than 95% confidenceinterval, we would expect the confidenceinterval to become:

a. Wider

b. Stay the same

c. Smaller

d. Increase by 4%.

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Determining sample size

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Determining sample size forestimating population mean

The margin of errorE is the maximum differencebetween the observed sample mean and the truevalue of the population mean:

where:

is known as the critical value, the positivez valuethat is at the vertical boundary for the area of in theright tail of the standard normal distribution.

is the population standard deviation.

n is the sample size.

2 2

XE z z

n

12

z


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Determining sample size forestimating population mean, cont.

Sample size necessary to produce resultsaccurate to a specified confidence andmargin of error:

Or if we dont know population standarddeviation and we have standard deviationfor sample:

2

12

n zE

2

12

Sn z

E

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Example 7

A survey is planned to determine the mean annualfamily medical expenses of employees in a largecompany. The management of the companywishes to be 99% confident that the sampleaverage is correct within $50 of the true averageannual family medical expenses. A pilot studyindicates that the standard deviation is estimatedto be $400.

How large a sample size is necessary if sampling isdonewithout replacement?

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Solution

A sample size sholud ben=426.

400

50

0.01

?

E

n

0.01 ( ) 1 0.995 2.582 from tablesF z z

2 2

12

4002.58 426

50n z

E

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Determining sample size forestimating population proportion

The margin of errorE is the maximum differencebetween the observed sample proportion and thetrue value of the population proportion:

where:

is known as the critical value, the positivevalue that is at the vertical boundary for the areaof in the right tail of the standard normaldistribution.

pA is the proportion from population.

n is the sample size.

1

A

A A

p

p pE z z

n

12

z

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Determining sample size forestimating population proportion, cont.

Sample size necessary to produce resultsaccurate to a specified confidence andmargin of error.

If we dont know our population proportionand we have a proportion from sample:

2

2

1A Az p p

nE

2

2

1A A

z p pn

E

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Example 8

An automobile dealer wants to estimate theproportion of customers who still own the cars theypurchased 5 years earlier. Sales records indicatethat the population of owners is 4,000. A randomsample of 200 customers selectedwithoutreplacement from the automobile dealers recordsindicate that 82 still own cars that were purchased 5years earlier.

What sample size is necessary to estimate the trueproportion to be within 0.025 with 95%confidence?


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Solution

A sample size sholud ben=1,487

0.05

82/ 200 0.41

0.025

?

Ap

E

n

0.05 ( ) 1 0.975 1.962 from tables

F z z

2 22 2

1 1.96 0.41 0.591, 487

0.025

A Az p pn

E

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Confidence interval forregession coefficients

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Estimation Process

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Confidence Interval for 1

We can use a 95% confidence intervalfor1 to test the hypotheses just usedin thet test.

H0 is rejected if the hypothesized valueof 1 is not included in the confidenceinterval for 1.

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Confidence Interval for 1 , cont.

The form of a confidence interval for1 is:

Where:

b1 is the point estimate

is the margin of error

is thet value providing an area of (/2) in the uppertail of at distribution with (n 2) degrees of freedom

11 / 2 bb t s 11 / 2 bb t s

1/2 bt s

1/2 bt s

2/t 2/t

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Example 9

The evil Swindler hasbeen collecting data onthe effect radiationexposure has on CaptainAmazings super powers.Here is the number ofminutes of exposure toradiation, paired with thenumber of tons CaptainAmazing is able to lift: 67

7.56.5

86

9.55.5

85

104.50

124

Weight(tons)

Radiationexposure(minutes)


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Scatter plot

4

5

6

7

8

9

10

11

4 4 , 5 5 5 , 5 6 6 , 5 7 7 , 5

r a d i a t i o n e x p o s u r e

we

igh

Linearreression

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Excel solution for Regression

7Observations

0,899Standard Error

0,789Adjusted R

Square

0,8234R Square

0,908Multiple R

Regression Statistics

22,9286Total

0,8074,0365Residual

0,00472323,40718,89218,8921Regression

SignificanceFFMSSSdf

ANOVA

-0,770-2,5160,005-4,8380,340-1,643

Radiationexposure(minutes)

22,63012,8700,0009,3511,89817,750Intercept

Upper95%

Lower95%P-valuet Stat

StandardErrorCoefficients

t value for (n-2)=5 andtipe I error=0.05 isequal 2.6

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References

Curwin J. and Slater R., Quantitative Methodsfor Business Decisions, Thomson Learning fifth edition 2002.

Levine D.M. and others, Statistics forManagers Using Microsoft Excel, Prentice Hall,

NY, 2005.

Somun-Kapetanovi R., Statistika u ekonomijii menadmentu, Ekonomski fakultet u

Sarajevu, Sarajevo 2006.

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Thank you for your attention!

confidence intervals final

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