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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