Examples 3 - Methods of Applied Statistics - Spring 2009 | STAT 420, Study notes of Data Analysis & Statistical Methods

Material Type: Notes; Professor: Stepanov; Class: Methods of Applied Statistics; Subject: Statistics; University: University of Illinois - Urbana-Champaign; Term: Spring 2009;

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STAT 420
Examples #3
Spring 2009
1.
Assume that the distributions of Y
1
and Y
2
are
N
(
µ
1
,
σ
2
)
and
N
(
µ
2
,
σ
2
)
,
respectively. Given the
n
1
= 6 observations of Y
1
,
65, 68, 67, 66, 71, 68
and the
n
2
= 8 observations of Y
2
,
68, 70, 67, 71, 70, 72, 73, 69
test
H
0
:
µ
1
=
µ
2
vs.
H
1
:
µ
1
µ
2
using the ANOVA
F
test. Use a 5%
level of significance. What can you say about the p-value of this test?
Recall:
n
1
= 6,
n
2
= 8,
1
y
= 67.5,
2
y
= 70,
2
1
s
= 4.3,
2
2
s
= 4,
2
pooled
s
=
4.125, H
0
:
µ
1
=
µ
2
t
test statistic:
t
= –
2.279.
t
0.025
(
12
)
= 2.179.
J =
2
. N =
n
1 +
n
2 = 6 + 8 =
14
.
14
965
14
7085.676
...
2
2
1
1
===
+
+++
N
J
J
ynynyn
y
=
68.92857
.
SSB =
(
)
(
)
(
)
2 2
2
2
2
1
1
...
yynyynyyn
J
J+++
= 6
(
67.5 – 68.92857
)
2 + 8
(
70 – 68.92857
)
2 =
21.42857
.
MSB =
1
21.42857
1
SSB =
=
21.42857
.
pf3

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STAT 420 Examples #3 Spring 2009

1. Assume that the distributions of Y 1 and Y 2 are N ( μ 1 , σ 2 ) and N ( μ 2 , σ 2 ),

respectively. Given the n 1 = 6 observations of Y 1 ,

and the n 2 = 8 observations of Y 2 ,

test H 0 : μ 1 = μ 2 vs. H 1 : μ 1 ≠ μ 2 using the ANOVA F test. Use a 5%

level of significance. What can you say about the p-value of this test?

Recall:

n 1 = 6, n 2 = 8, y 1 = 67.5, y 2 = 70, s 12 = 4.3, s 22 = 4,

2

s pooled = 4.125, H 0 : μ 1 = μ 2 t test statistic: t = – 2.279.

t 0.025 ( 12 ) = 2.179.

J = 2. N = n 1 + n 2 = 6 + 8 = 14.

= 1 ⋅^1 +^2 ⋅^2 +...+ ⋅ =^6 ⋅^67.^5 +^8 ⋅^70 =

N

y n y n y nJ^ y^ J = 68.92857.

SSB = n (^) 1 ⋅( y 1 − y ) 2 + n 2 ⋅( y 2 − y ) 2 +... + nJ ⋅( yJy ) 2

MSB =

SSB =

J −

SSW = ( n (^) 1 − 1 ) ⋅ s 12 +( n 2 − 1 ) ⋅ s 22 +...+( nJ − 1 ) ⋅ sJ^2

MSW =

SSW = 49.

N − J

= 4.125. [ = s pooled^2 ]

SSTot = SSB + SSW = 70.92857.

F =

MSW

MSB (^) = = 5.1948. [ = t (^2) ]

ANOVA table:

Source SS DF MS F

Between 21.42857 1 21.42857 5.

Within 49.5 12 4.

Total 70.92857 13

F (^) 0.05 ( 1, 12 ) = 4.75. [ = ( t (^) 0.025 ( 12 ) ) 2 ]

Reject H 0 at αααα = 0.05.