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# Comparing two means.pdf

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University of Alberta

Statistics

STAT368

Douglas Wiens

Winter

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24
3. Comparing two means
Testing whether or not certain treatments have
the same e ects often involves estimating the vari-
ance among the group averages, and comparing
this to an estimate of the underlying, residual
variation. The former is attributed to di erences
between the treatments, the latter to natural vari-
ation that one cannot control. Large values of
the ratio of these variances indicates true treat-
ment di erences. Mathematically, one computes
1
0 = 2
2
2
2 2 2 2
where for = 1 2 and , 1 ar2
independent r.v.s. Then 0 is the numerical value
of
2 1
1
2 2
2
where the two 2s are independent; this is the
denition of an r.v. on ( 1 2) degrees of
freedom: 1.
2 25
Inferences involving just one mean (or the di er-
ence between two means) can be based on the -
distribution. One can generally reduce the prob-
lem to the following. We observe 1
³ 2´
. Then
Ã !
2 2
¯ independently of 2 2 1
1
Here we use:
1. R.v.s 1 are jointly normally distributed if
and only if every linear combination of them is
also normal; this is the single most important
property of the Normal distribution.
2. In Normal samples the estimate of the mean and
the estimate of the variation around that mean
are independent. 26
Thus = ¯ is the numerical value of the
0
ratio of independent r.v.s
¯
= q , where (0 1);
2 ( 1)
1
this is the denition of Students on 1
d.f.
You should now be able to show that 2 follows
1
an 1distribution.
Return now to the mortar comparison problem.
n o
Let =1 be the samples ( = 1 for modied,
= 2 for unmodied; 1 = 2 = 10). Assume:
³ ´
11 1 1
1 ³ ´
2
21 2 2 2
2 2 2
First assume as well that 1 = 2 = , say.
Then
Ã Ã !!
¯ ¯ 2 1 1
1 2 +
1 2 27
2
The pooled estimate of is
2 2
2 ( 1 1) 1+ ( 2 1) 2
= + 2
1 2
2 1 + 2 1
2 1 2
1 + 2 2
2

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