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STAT 350 – Fall 2008
Homework #12  Solution
covers through Lecture #19
1.
A biologist is studying a population of snarks.
She traps 25 snarks (consider this a random sample
from the population) and measures the length (in cm) and weight (in kg) of each snark.
Let
x
i
denote
the length of the
i
th snark and let
y
i
denote the weight of the
i
th snark.
She finds
25
1
i
i
x
=
∑
= 1635,
25
1
i
i
y
=
∑
=
3356,
25
2
1
i
i
x
=
∑
= 117325,
25
2
1
i
i
y
=
∑
= 468176, and
(
25
1
ii
i
)
x y
=
∑
= 230003.
The length of the snarks ranged
from 12.7 to 93.8 cm and their weight ranged from 74.0 to 181.4 kg.
Warning:
do not roundoff too much in intermediate calculations!
a. Calculate
SS
xy
.
()
1635 3356
230003
25
−
= 10520.60
b. Calculate
SS
xx
.
2
1635
117325
25
−
= 10396.00
c. Calculate
SS
yy
.
2
3356
468176
25
−
= 17666.56
e.
Calculate the sample standard deviation of snark weight.
17666.56
27.13128575
25 1
y
s
==
−
f.
Give a 95% CI for the true mean weight of snarks from this population.
3356/ 25 134.24
y
95% CI:
N
from Table IV
27.13128575
134.24
2.064
25
⎛
±
⎜
⎝⎠
⎞
⎟
→
(123.04, 145.44)
g.
The biologist set out a trap to catch another snark.
Give a 95% prediction interval for the weight
of the snark she will catch.
95% PI:
N
from Table IV
1
134.24
2.064
27.13128575
1
25
±+
→
(77.13, 191.35)
h.
Find the Pearson Correlation between the length and weight of these snarks.
10520.60
10396.00 17666.56
r
=
= 0.7763
i.
Find the least squares regression equation to predict the weight of a snark from the length of a
snark.
(give
a
and
b
to at least 5 decimal places)
b
= 10520.60/10396.00 = 1.01199
a
= (3356/25) – (1.01199)(1635/25) = 68.05616
= 68.05616 + 1.01199
x
ˆ
y
Homework #12  Solution
Page 1 of 6
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View Full Document j.
Create the ANOVA table for the regression analysis (use Excel to get the
p
value).
Total Sum of Squares (SSTo) = SS
yy
= 17666.56
Residual Sum of Squares (SSResid) = SSTo –
b
SS
xy
= 17666.56 – (1.01199)(10520.60)
= 7019.818006
Regression Sum of Squares (SSReg) = SSTo – SSResid = 17666.56 – 7019.818006
= 10646.74199
Source
df
Sum of Squares
Mean Square
F
p
Regression
1
10646.74199
10646.74199
34.8834
5.085
×
10
6
Error
23 7019.8180006
305.2094785
Total
24 17666.56
k.
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This note was uploaded on 02/23/2009 for the course STAT 350 taught by Professor Staff during the Fall '08 term at Purdue University.
 Fall '08
 Staff

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