Go to index Two Sample Inference for Means Farrokh Alemi Ph.D Kashif Haqqi M.D.
6.3 Two-Sample Inference for Means
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Transcript of 6.3 Two-Sample Inference for Means
6.3 Two-Sample Inference for Means
November 17, 2003
Paired Differences
Matched Pairs Designexperimental plan where the
experimental units are divided into halves and two treatments are randomly assigned to the halves
Attempting to determine if there is a significant difference between the mean responses of the treatments
Procedure
Obtain the difference between responses for each experimental unit
Analyze the differences using a one-sample approachIf a large sample is obtained, use
critical values from the Standard Normal distribution (z)
Otherwise, use critical values from the corresponding t distribution
Example 9 pg 370
Students worked with a company on the monitoring of the operation of an end-cut router in the manufacture of a wood product. They measured the critical dimensions of a number of pieces of a type as they came off the router. Both a leading-edge and a trailing-edge measurement were made on each piece. Both were to have a target value of .172 in.
Piece Leading Edge
Trailing Edge
1 .168 .169
2 .170 .168
3 .165 .168
4 .165 .168
5 .170 .168
Piece Leading Edge
Trailing Edge
Difference
1 .168 .169 -.001
2 .170 .168 .002
3 .165 .168 -.003
4 .165 .168 -.003
5 .170 .168 .002
Confidence Interval
n
stdor
n
szd dd
0023.0008. dsd
Hypothesis Test
Is there a significant difference between the measurements at α =.01?
Ho: µ = 0 Ha: µ ≠ 0
n
sd
dT
#
Independent Samples
The goal of this type of inference is to compare the mean response of two variables (or treatments) when the data are not paired or matched
A key assumption that will be made is that the separate samples used to collect information concerning the two variables are independent One sample does not influence the other
sample in any way Furthermore, one must assume that both
sampled populations are normally distributed
Large Sample Comparison
The quantity of interest is a linear combination of population means, namely µ1 - µ2
The above quantity will be estimated by As a result, various quantities of the sampling
distribution of the difference, under the assumption of equality between the means, need to be developed
Ho: µ1 - µ2 = #
Ha: µ1 - µ2 ≠ #
21 xx
Test Statistic
2
22
1
21
#)( 21
ns
ns
xxZ
Confidence Interval
2
22
1
21)( 21 n
snsZxx
Example A company research effort involved finding
a workable geometry for molded pieces of a solid. A comparison was made between the weight of molded pieces and the weight of irregularly shaped pieces that could be poured into the same container. A series of 30 attempts to pack both the molded and the irregular pieces of the solid were compared. Is there enough evidence to suggest that the irregular pieces produced higher weights?
The Data
1 = molded n1 = 30
s1= 9.31 = 164.65
2=irregular n2 = 30
s2= 8.51 = 179.651x 2x
Small Samples
If at least one sample size is small, then use critical values from a t distribution for constructing confidence intervals and performing hypothesis tests with degrees of freedom obtained by Satterthwaite’s Approximation
Satterthwaite’s Approximation
2
2
22
2
2
1
21
1
2
2
22
1
21
11
11
ˆ
ns
nns
n
ns
ns
Test Statistic
21
11
21 #)(
nnPs
xxT
2
)1()1(
21
222
2112
nn
snsnsP
Confidence Interval
21
1121 )( nnPstxx
Example
The data shown gives spring lifetimes under two different levels of stress (900 and 950 N/mm2). Do the data give evidence of a significant difference at α = .05?
950 Level 900 Level
225 171 198 187 189 135 162 135 117 162
216 162 153 216 225 216 306 225 243 189
Minitab Output
Descriptive Statistics: 950, 900
Variable N Mean Median TrMean StDev SE Mean950 10 168.3 166.5 167.6 33.1 10.5900 10 215.1 216.0 211.5 42.9 13.6
Assignment
Page 385: #3, #4