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Unformatted text preview: Conﬁdence Intervals for µ1 − µ2
• Compare responses in two groups
• Can arise in two ways
1. 2. 1 2 Population
1
2 Sample
Size
n1
n2 Variable Mean St. dev. X
Y µ1
µ2 σ1
σ2 Parameter of interest
θ=
Natural estimator
Θ=
Then
¯¯
E[X −Y ] = 3 Assume independent samples. Then
¯¯
Var [X −Y ] =
Finally assume both populations are normal. Then
¯¯
X −Y ∼ Then standardized
Z= Note. Make sure you understand why
means, variances, and Z statements hold.
These are properties of linear combinations or random variables and normals.
4 Therefore C.I. construction reduces to
that for a single population in which
Z=
replaces the previous
¯
X −µ
Z=
√
σ/ n
As before, start with
P (−zα/2 ≤ Z ≤ zα/2) = 1 − α
and use simple algebra to rewrite this as
¯¯
P (X − Y − zα/2 2
σ1
σ2
¯¯
+ 2 ≤ µ1 − µ2 ≤ X − Y + zα/2
n1
n2 2
σ1
σ2
+ 2 = 1−α
n1
n2 to end with the conﬁdence interval
¯¯
X − Y ± zα/2 2
2
σ1
σ2
+
n1
n2
5 What if population cannot be assumed normal? For large n1, n2 Central Limit Theorem. Replace σ1 with
S1 and σ2 with S2 where S1, S2 are
the sample standard deviations for X, Y
samples respectively. to get approxi mate 100(1 − α)% conﬁdence interval
2
2
S1
S2
¯¯
+
X − Y ± zα/2
n1
n2 6 Example 4.25. Comparison of breaking strength
of two fabrics. Find 95% C.I. for diﬀerence in
population means. Fabric A
Fabric B Sample
Size
35
30 Sample
Mean
25.2
28.5 Sample
St. dev.
5.2
5.9 x − y = 25.2 − 28.5 = −3.3
¯¯
2
s2
1 + s2
M E = z0.025
n1
n2 1/2 5.22
5.92
= 1.96
+
35
30
= 1.96(1.3903) = 2.725
C.I. = −3.3 ± 2.725 = (−6.025, −0.575)
Because C.I. is on the left side of the origin, we
conclude that Fabric A has a smaller breaking
strength than Fabric B.
7 Suppose want to be more certain of conclusion, say 99% conﬁdent. What is the C.I. now? x − y = 25.2 − 28.5 = −3.3
¯¯
s2
s2
1+ 2
M E = z0.005
n1
n2 1/2 5.22
5.92
= 2.57
+
35
30
= 2.57(1.3903) = 3.5731
C.I. = −3.3 ± 3.573 = (−6.873, +0.273)
Now, data don’t show signiﬁcant diﬀerence.
8 ...
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This note was uploaded on 01/03/2012 for the course EE 1244 taught by Professor Drera during the Fall '10 term at Conestoga.
 Fall '10
 drera

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