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Statistics
The difference of two means
April 27, 2009
1. Assumptions: X1 , . . . , Xm and Y1 , . . . , Yn are independent random samples from populations that have a normal
distribution with unknown means µX , µY and unknown variances.
2. The key fact is this: under these assumptions,
X − Y − (µX − µY )
q 2
∼ Norm(0, 1) .
2
σX
σY
+
m
n
This follows from the fact that variances and means add and that the sum of independent normal random
variables is normal.
3. We replace σX and σY by the corresponding sample standard deviations to get that this random variable
X − Y − (µX − µY )
q 2
2
SX
SY
m + n
has approximately a t-distribution with ν degrees of freedom where ν is
ν=
2
SX
m
2 /m)2
(SX
m−1
2
SY
n
+
+
2
2 /n)2
(SY
n−1
4. Insert long story here about “old-fashioned” practice and the Behrens-Fisher problem.
5. Confidence intervals for µ1 − µ2 :
r
∗
x−y±t
s2
s2X
+ Y
m
n
!
t∗ = tα/2,ν
6. Robustness.
Homework - due Thursday, April 30, 2009
1. Read Section 7.2.
2. Do problems 7.4,5,6.
> s=getdata(’subliminal’)
> s[c(1,2),]
group before after
1
T
18
24
2
T
18
25
> t.test(after-before~group,data=s)
Welch Two Sample t-test
data:
after - before by group
t = -1.9136, df = 13.919, p-value = 0.07647
alternative hypothesis: true
difference in means is not equal to 0
95 percent confidence interval:
-6.6825702 0.3825702
sample estimates:
mean in group C mean in group T
8.25
11.40