Note 11 of 5E

Statistics with Economics and Statistics with Economics and Business ApplicationsBusiness Applications

Chapter 7 Estimation of Means and Proportions

Small-Sample Estimation of a Population Mean

Note 11 of 5E

ReviewReview

I. I. What’s in last lecture?What’s in last lecture?

Large-Sample Estimation Chapter 7

II. II. What's in this lecture?What's in this lecture?

Small-Sample Estimation of a Population Mean Read Chapter 7

Note 11 of 5E

IntroductionIntroduction• When the sample size is small, the

estimation procedures in Lecture note 10 are not appropriate.

• Point estimators remain the same• There are small sample interval

estimators/confidence intervals for , the mean of a normal populationthe difference between two

normal population means

Note 11 of 5E

The Sampling DistributionThe Sampling Distribution of the Sample Mean of the Sample Mean

• When we take a sample from a normal population, the sample mean has a normal distribution for any sample size n, and

has a standard normal distribution. • But if is unknown, and we must use s to estimate it,

the resulting statistic is not normalis not normal.

n

xz

/

n

xz

/

normal!not is / ns

x normal!not is / ns

x

x

Note 11 of 5E

Student’s t DistributionStudent’s t Distribution• Fortunately, this statistic does have a

sampling distribution that is well known to statisticians, called the Student’s t Student’s t distributiondistribution, , with nn-1 degrees of freedom-1 degrees of freedom..

ns

xt

/

ns

xt

/

• We can use this distribution to create estimation procedures for the population mean .

Note 11 of 5E

Properties of Student’s Properties of Student’s tt

• Shape depends on the sample size n or the degrees of freedom, degrees of freedom, nn-1-1..

• As n increases the shapes of the t and z distributions become almost identical.

•Mound-shapedMound-shaped and symmetric about 0.

•More variable than More variable than zz, with “heavier tails”

Note 11 of 5E

t distributiont distribution

-4 -3 -2 -1 0 1 2 3 4

df = 2

df = 5

df = 10

df = 25

Normal

Comparison of normal and t distibutions

Note 11 of 5E

Using the Using the tt-Table-Table• Table 4 gives the values of t

that cut off certain critical values in the right tail of the t distribution.

• Use index dfdf and the appropriate tail area aa to find ttaa, , the value of t with area a to its right.

For a random sample of size n = 10, find a value of t that cuts off .025 in the right tail.

Row = df = n –1 = 9

t.025 = 2.262

Column subscript = a = .025

aa

ttaa

Note 11 of 5E

.1on with distributi- ta of right tail in the

/2 area off cuts that of value theis wheren

stx

is mean population theof

interval confidence )100%-(1 Sample-Small

2/

α/2

ndf

tt

.1on with distributi- ta of right tail in the

/2 area off cuts that of value theis wheren

stx

is mean population theof

interval confidence )100%-(1 Sample-Small

2/

α/2

ndf

tt

Small Sample Confidence Interval for Small Sample Confidence Interval for Population Mean Population Mean

Assumption: population must be normal

Note 11 of 5E

ExampleExample

Ten randomly selected students were each asked to list how many hours of television they watched per month. The results are

82 66 90 84 7588 80 94 110 91

Find a 90% confidence interval for the true mean number of hours of television watched per month by students.

Note 11 of 5E

Example ContinuedExample Continued

We find the critical t value of 1.833 by looking on the t table in the row corresponding to df = 9, in the column with label t.050. The 90% confidence interval for is

Calculating the sample mean and standard deviation we have 842.11sand,86x,10n Calculating the sample mean and standard deviation we have 842.11sand,86x,10n

n

s tx

10

842.11)833.1(86 86.686

)86.92,14.79(

2/

Note 11 of 5E

Estimating the Difference Estimating the Difference between Two Meansbetween Two Means

•You can also create a 100(1-)% confidence interval for 1-2.

2 freedom of degrees

t with of valuecritical theis tand

2

)1()1( with

21

/2

21

222

2112

nn

nn

snsns

2 freedom of degrees

t with of valuecritical theis tand

2

)1()1( with

21

/2

21

222

2112

nn

nn

snsns

21

22/21

11)(

nnstxx

21

22/21

11)(

nnstxx

Remember the three assumptions:

1. Original populations normal

2. Samples random and independent

3. Equal population variances

Remember the three assumptions:

1. Original populations normal

2. Samples random and independent

3. Equal population variances 2

22

1

Note 11 of 5E

ExampleExample

A student recorded the mileage he obtained while commuting to school in his car. He kept track of the mileage for twelve different tanks of fuel, involving gasoline of two different octane ratings. Compute the 95% confidence interval for the difference of mean mileages. His data follow:

87 Octane 90 Octane26.4, 27.6, 29.7 30.5, 30.9, 29.228.9, 29.3, 28.8 31.7, 32.8, 29.3

Note 11 of 5E

ExampleExampleLet 87 octane fuel be the first group and 90 octane fuel the second group, so we have n1 = n2 = 6 and

1 1 2 2x =28.45, s 1.228, x =30.73, s 1.392

d.f.=n1+n2-2=10. The critical value of t is 2.228.

431).(-4.129,-. is interval confidence 95% The

849.128.2

)5/15/1(723.1228.273.3045.28

)/1/1(

723.110

938.15508.15

2

)1()1(

212

2/21

21

222

2112

nnstxx

nn

snsns

431).(-4.129,-. is interval confidence 95% The

849.128.2

)5/15/1(723.1228.273.3045.28

)/1/1(

723.110

938.15508.15

2

)1()1(

212

2/21

21

222

2112

nnstxx

nn

snsns

Note 11 of 5E

Key ConceptsKey ConceptsI. One Population MeanI. One Population Mean

.1on with distributi- ta of right tail in the

/2 area off cuts that of value theis wheren

stx

is mean population theof interval

confidence )100%-(1 ,population normalFor

2/

α/2

ndf

tt

.1on with distributi- ta of right tail in the

/2 area off cuts that of value theis wheren

stx

is mean population theof interval

confidence )100%-(1 ,population normalFor

2/

α/2

ndf

tt

Note 11 of 5E

Key ConceptsKey ConceptsII.II. Two Population MeansTwo Population Means

1. Both populations are normal. 2. Two samples are independent.

3. Two populations have a common variance.

.2

on with distributi- ta of right tail in the

/2 area off cuts that of value theis where

/1/1tx-x

is of interval confidence )100%-(1

21

2/

21α/221

21

nndf

tt

nns

.2

on with distributi- ta of right tail in the

/2 area off cuts that of value theis where

/1/1tx-x

is of interval confidence )100%-(1

21

2/

21α/221

21

nndf

tt

nns