Lec 08_Estimation and Confidence Interval(2)

15/06/2015

1

RECOMMENDED READINGCustomised Text, Adapted from ‘Statistical Techniques in Business & Economicsby Lind, Marchal 16th Edition’ McGraw Hill

Chapter 8, Page 241 – 259, 262 – 264, 266 - 276 excluding CI for PopulationProportion, sample size to estimate population proportion

Explain the difference between a point estimate and aninterval estimate

Use normal distribution to construct a confidenceinterval for population mean

Use t distribution to construct a confidence interval forpopulation mean

Decide whether normal or t distribution should used inconstructing confidence interval for population mean

Determine a sample size at specific level of confidence

Using information included in a sample todraw inferences about the population

Point Estimate &Interval Estimate

CI: Known σ orlarge sample

CI: Unknown σor small sample

Determinesample size, n

Student’s tdistribution

Estimator• A sample statistic serves the purpose of statistical

estimation regarding a population parameter

Estimate• The corresponding observed value of a sample statistic

15/06/2015

2

A single numerical value of a statistic that isused to estimate the true value ofcorresponding population parameter

E.g. Sample mean is a point estimate of thepopulation mean






A range of values within which the true valueof the population parameter is likely to beincluded

E.g. Confidence Interval






Parameter Estimator

Population Mean () Sample mean ( )Population variance (2) Sample variance (s2)

Population standarddeviation()

Sample standard deviation (s)






15/06/2015

3

Point estimate – either right or wrong Interval estimate – serves purpose better Lies over a range of values where true population

parameter is likely to be in Provide information about possible error margin Gives degree of certainty (confidence level) to the

estimation











Confidence Interval

Sample Statistic(Point Estimate)

Confidence Limit(Lower)

Confidence Limit(Upper)

Level of confidence (1 – α) X 100% Expressed in percentage terms Degree of certainty associated with an interval

estimate

Confidence Interval An interval estimate with designated level of

confidence to include the true value of populationparameter






15/06/2015

4






Image source: http://garfield.dale.ro/garfield-1999-march.html






Confidence Interval

Confidence Limit(Lower)

Confidence Limit(Upper)

Confidence Level ( 1- )%. Eg. 95% Interpretation (1-α)% of the

time, theintervalestimate willinclude thetruepopulationparameter






X

1

X

1

Z

Z

A positive z – value

A negative z – value

15/06/2015

5

100(1 -)% of intervals contain ; 100% do not contain






X

1

X

2

2

Suppose = 0.025, find the value of Z0.025






X

X

2 0.0250.4750.475

2 0.025

0.95 = 95%

Suppose = 0.025, find the value of Z0.025






X

X

2 0.0250.4750.475

2 0.025

0.95 = 95%

.-

15/06/2015

6

Confidence Level Z-Value

90% 1.645

95% 1.96

99% 2.58






*Note: ALWAYS remember that C.I is TWO-sided

1. Underlying population is normal with a knownσ (regardless of size of n)

2. Underlying population is unknown BUT n≥30(Use s (sample s.d.) when σ is unknown)

According to sampling theory: is normally distributed

Mean µ = µ, S.D. σ or * Point Estimate &Interval Estimate





According to sampling theory: is normally distributed

Mean µ = µ, S.D. σ or * Z =

~ N(0,1)






15/06/2015

7






X

X

22 1 - α

Deriving the C.I.

P(- < Z < ) = 1 – α

P(- < < ) = 1 - α

P(- < < ) = 1 – α

- < <

- < < - < < + The (1-α)*100% C.I. is: ±






To compute a (1 – )100% confidenceinterval for 1. Determine n, and . Also determine the

distribution of 2. Find such that P(- < < ) = 1 – α

3. Calculate the (1-α)*100% C.I. by: ± Z






15/06/2015

8

A study was conducted to estimate the mean amount spent onChristmas gifts for a typical family with 2 children. A sample size of75 was taken and the mean amount spend was $195. Assume thepopulation standard deviation is $35. Find a 95% confidenceinterval for the true mean for all such families.

1. Determine n, and . Also determine the distribution of n = 75, = 195, = = = 4.041

Underlying population unknown but n is large▪ is normally distributed






2. Find such that P(- < < ) = 1 – α

▪ 95% CI 95% ÷ 2 = 47.5% = 0.475

▪ . 1.96 (Read 0.475 from Z table)

3. Calculate the (1-α)*100% C.I. by: ±▪ 195 ± 1.96 * 4.041 = (195 – 1.96*4.041, 195+1.96*4.041)

= (187.08, 202.92)






95% C.I. (187.08, 202.92)

Interpretation

I am 95% confident that the mean amount spent onChristmas gift for such families is between $187.08 and$202.92.






(1 – α) *100%

Confidence

Interval

15/06/2015

9

The mean monthly income of a sample of 60 families living inPandan Valley is $4,800 with a standard deviation of $900.Compute a 90% confidence interval of the mean monthly incomeof all families living in Pandan Valley.

1. Determine n, and . Also determine the distribution of n = 60, = $4,800, = = = 116.190

Underlying population unknown but n is largeσ unknown so use s

▪ is normally distributed






2. Find such that P(- < < ) = 1 – α

▪ 90% CI 90% ÷ 2 = 45% = 0.45

▪ . 1.645 (Read 0.45 from Z table)

3. Calculate the (1-α)*100% C.I. by: ±▪ 4800 ± 1.645 * 116.190

= (4800 – 1.645*116.190, 4800+1.645*116.190)= (4608.87, 4991.13)






90% C.I. (4608.87, 4991.13)

Interpretation

I am 90% confident that the mean monthly income offamilies living in Pandan Valley is between $4608.87 and$4991.13.






(1 – α) *100%

Confidence

Interval

15/06/2015

10

Factors affecting the confidence interval± = ±

Level of confidence (α)▪ The higher the confidence level, the wider the C.I.

Population standard deviation (σ)▪ The bigger the σ, the wider the C.I.

Sample size (n)▪ The smaller the n, the wider the C.I.






Unbiasedness: When expected value ofestimator is equal to population parameterbeing estimated

Consistency: As sample size increases, valueof estimator comes close to value ofpopulation parameter being estimated

Efficient: Estimator has a low variancerelative to other estimators. Estimator has aminimized variance.

Sufficiency/Robust: If no other estimatorwill be able to get any additional informationabout the population parameter beingestimated

15/06/2015

11

1. Underlying population is normal with aunknown σ AND n <30

According to sampling theory: follows student’s t distribution with n-1 degrees

of freedom

Mean µ = µ, S.D. s t = ~






Continuous distribution symmetric about themean = 0, and ranges from – to

A family of distributions specified by ‘degreesof freedom’

Bell-shaped – but ‘flatter’ and more ‘spreadout’ than normal distribution

As n increases, t distribution converges tonormal











0

t (df = 5 n =6)

StandardNormal(Z)

t (df = 13 n=14)t - distribution

t-curve tend to z-curvewhen n

15/06/2015

12

Using the t distribution table to determinethe probability of t exceeds a certain value P(t > ) = α

1. Find out degrees of freedom, n-1

2. Specify the right tail probability, α

3. Look up the t distribution table for , withcorresponding degree of freedom and right tailprobability






degreesoffreedom

Confidence Level ( 1 - )0.80 0.90 0.95 0.98 0.99 0.999

Level of Significance for One-Tailed Test ( )0.10 0.05 0.025 0.01 0.005 0.0005

Level of Significance for Two-Tailed Test ( )0.20 0.10 0.05 0.02 0.01 0.001






Right-tailed test Left-tailed test 1 -

Two-tailed test

Right-tailed test

Two-tailed test

degreesoffreedom

Confidence Level ( 1 - )0.80 0.90 0.95 0.98 0.99 0.999

Level of Significance for One-Tailed Test ( )0.10 0.05 0.025 0.01 0.005 0.0005

Level of Significance for Two-Tailed Test ( )0.20 0.10 0.05 0.02 0.01 0.001

1 3.078 6.314 12.706 31.821 63.657 636.6192 1.886 2.920 4.303 6.965 9.925 31.5993 1.638 2.353 3.182 4.541 5.841 12.9244 1.533 2.132 2.776 3.747 4.604 8.610






80% CI with df = 3= 1.638

15/06/2015

13

Find the value of t for the confidence level of90% with a sample size of 10

Confidence level of 90% Degree of freedom:n – 1 = 10 – 1 = 9 = 1.833






t n-10 t

/2= 0.05

t = 1.833

To compute a (1 – )100% confidence interval for 1. Determine n, and . Also determine the distribution

of 2. Find ,3. Calculate the (1-α)*100% C.I. by:

± t = ± Point Estimate &Interval Estimate





A new flavor of soft drink will be test-marketed for onemonth at stores of a large supermarket chain. Sales amountsfrom a random sample of 16 stores showed an average salesof $1,200 with a standard deviation of $180. Set up a 99%confidence interval estimate for the mean sales of this newflavor of soft drink.

1. Determine n, and . Determine the distribution of n = 16, = $1,200, = = 45

Unknown σ and n < 30 ~ =






15/06/2015

14

2. Find ,▪ 99% CI▪ , . , 2.947

3. Calculate the (1-α)*100% C.I. by: ± ▪ 1200 ± 2.947 * 45

= (1200 – 2.947*45, 1200+2.947*45)= (1067.39, 1332.62)






99% C.I. (1067.39, 1332.62)

Interpretation

I am 99% confident that the mean sales of this new flavoursoft drink is between $1067.39 and $1332.62.






(1 – α) *100%

ConfidenceInterval






Is OriginalPopulationNormally

distributed withknown σ

is normally distributedregardless of sample size

n ≥ 30 ?

approximatelynormal (by CLT)

: t

YES

YESNO

NO

15/06/2015

15

The (1 – )100% confidence interval for is±

Mid point of interval = Error Margin, ε = ±






The Error Margin, ε is the maximum tolerantsampling error under the (1-α)X100% level of C.I. E.g. 50kg±3kg Error Margin = Max tolerance = 3kg






Confidence Level ( 1- )%.

ε = =

ε =

σ ε =

n =






Given Error Margin , ε Level of C.I , α S.D , σ

We can determinesample size, n

15/06/2015

16

An economist wants to estimate the mean annual income ofhouseholds in Clementi. It is assumed that the populationstandard deviation is $4,000. The economist wants to have a95% confidence that the sample mean will be within $500 ofthe true mean. How large a sample is required?

Given σ = $4,000, ε = $500, α = 0.05 = 1.96

n = ( ∗ε = ( . ∗ = 245.86 ≈ 246






Factors affecting the sample sizen = ( ∗ε

Level of confidence (α)▪ The higher the confidence level, the larger the sample

size

Population standard deviation (σ)▪ The bigger the σ, the larger the sample size

Error Margin (ε)▪ The smaller the ε, the larger the sample size






Lec 08_Estimation and Confidence Interval(2)

Documents

Transcript of Lec 08_Estimation and Confidence Interval(2)