Tests of Hypotheses Involving Two Populations

12.1 Tests for the Differences of MeansComparison of two means: and The method of comparison depends on the

design of the experiment

The samples will either be Independent or Dependent

1 2

Independent Samples – implies data values obtained from one sample are unrelated to values from the other sample

 Dependent Samples – implies subjects are paired

so that they are as much alike as possible The purpose of pairing is to explain subject to

subject variability. (In some studies we apply both treatments to the same subject)

 If subject to subject variability is large relative to the

expected treatment differences then a dependent sample design should be considered

12.3 Tests for Differences of Means of Paired SamplesObserve n pairs of observations

observation for treatment i in pair j

difference between treatments 1 and 2 in

pair j

ijx

jjj xxd 21

Pair Treatment 1 Treatment 2 Difference

1

2

3

. . . .

. . . .

. . . .

n

Sample Mean

Sample Variance

Sample Standard Deviation

11x

2ds

22s

21s

1x 2x 21 xxd

21x 21111 xxd

12x 22x 22122 xxd

13x 23x 23133 xxd

nx1 nx2 nnn xxd 21

1s 2s ds

For dependent sample design all analysis is based on differences:

The differences are a sample from a distribution with mean and an unknown variance

We can calculate Confidence Intervals and perform Hypothesis Tests in the same way as with one sample

ndddd ,...,,, 321

21 d2d

Point Estimate for is

The Test Statistic is

which has a t-distribution with n-1 degrees of freedom

Confidence Interval for

21 d 21 xxd

nsd

Td

d

21 d

nstE d

Ed

ExampleA marketing expert wishes to prove that a new product display will increase sales over a traditional display

 

Treatment 1: New Method

Treatment 2: Traditional Method

 12 stores are available for the studyThere is considerable variability from store to storeWe will divide the 12 stores into six pairs such that within each pair the stores are as alike as possibleMeasurement will be cases sold in a one month period

PairNew MethodTreatment 1

Traditional Method

Treatment 2Difference

1 13 11 2

2 31 29 2

3 20 21 -1

4 19 17 2

5 42 39 3

6 26 22 4

Sample Mean

Sample Variance

Sample Standard Deviation

2d

8.22 ds

6733.1ds

Perform a Hypothesis test usingConclude with 95% confidence that the new method produces larger sales.

Perform and interpret a 95% confidence intervalWe are 95% confident that the mean increase in sales is between 0.244 cases and 3.756 cases using the new product display.

6n2d

05.

6733.1ds

12.2 Tests of Differences of Means Using Small Samples from Normal Populations When the Population Variances Are Equal but UnknownNotation for Two Independent Samples

Sample 1 (Treatment 1)

Sample 2 (Treatment 2)

Sample Size

Data

Sample Mean

Sample Variance

Sample Standard Deviation

1n

11131211 ,...,,, nxxxx

1x21s

1s

2n

22232221 ,...,,, nxxxx

2x22s

2s

The point Estimate for (Independent Samples)

is

This is the same as what we did with dependent samples

The sampling distribution for

21

21 xx

2121 xx

21 xx

2

22

1

212

21 nnxx

If we sample from populations with means of and ,

and standard deviations of and respectively, then

is approximately standard normal for large n.

The form of a confidence interval and a hypothesis test for two independent samples depends on what we assume

12

1 2

2

22

1

21

2121

nn

xxZ

If we sample from populations 1 and 2, the samples are

independent, and then

has a t-distribution with

Where is pooled estimate of

This means

22

21

2.. 21 nnfd

2

2

1

2

2121

nS

nS

xxT

pp

2

11

21

222

2112

nn

snsns p

22

21

2pp ss

Thus for a Hypothesis Test the Test Statistic is

and a Confidence Interval is calculated using

2

2

1

2

2121

nS

nS

xxT

pp

2

2

1

2

nS

nS

tE pp

Exx 21

ExampleA study is designed to compare gas mileage with a fuel additive to gas mileage without the additive.  A group of 10 Ford Mustangs are randomly divided into two groups and the gas mileage is recorded for one tank of gas.

Treatment 1(with additive)

Treatment 2(without additive)

Sample Size

Data 26.3, 27.4, 25.1, 26.8, 27.1

24.5, 25.4, 23.7, 25.9, 25.7

Sample Mean

Sample Variance

51 n

54.261 x

813.021 s

52 n

04.252 x

848.022 s

Do we have enough evidence at to prove that the

Additive increases gas mileage. Assume

Conclude with 95% confidence that the additiveimproves gas mileage

Compute and interpret a 95% Confidence Interval for

We are 95% confident that the additive will increase gas

mileage by an amount between 0.177 and 2.823 miles.

22

21

05.

21

What if we do not assume ?It then makes no sense to compute

A reasonable variable is

Unfortunately the distribution of T is unknown.

22

21 2ps

2

22

1

21

2121 )(

ns

ns

xxT

For large and , T is approximately standard normal.

For small and , the distribution of T can be

approximated by a t-distribution using a complicated formula for the degrees of freedom.

1n 2n

1n 2n