Stat 401C
Lab#6 Solution Key
Fall 2009
Problem 1
a)
The normal probability plot (
see attached JMP output
) shows that the points fall approximately
on a straight line supporting the assumption that the paired differences
of PCB measurements
at the sites is a sample from a normal distribution. The box
plot also supports this conclusion.
See the
attached
Excel output
for the calculation of
d
s
and
d
s
.
b)
Let
1996
1982
µ
µ
µ
−
=
d
where
1982
µ
and
1996
µ
are assumed to be the means of PCB contents in
gull egg populations during those years.
Assuming that the paired differences in PCB content is
a sample from normal population with mean
d
µ
the following hypotheses can be tested:
0
:
0
≤
d
H
µ
vs.
0
:
>
d
a
H
µ
d
c
s
d
t
0
−
=
=
23
.
11
13
/
46
.
12
80
.
38
=
681
.
2
12
,
01
.
=
t
, that is R.R. is
681
.
2
>
t
Since
c
t
is in R.R. we rewject the null hypothesis at
01
.
=
α
and conclude that
there has been a significant decrease in mean PCB content in the herring gull
egg population.
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 Spring '10
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 Normal Distribution, Probability, Bartlett, 98%, Distributions Steel, 1 2 45 1 Miles

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