Homework 7
due Mar. 21
861 (962)
868 (969)
94 (104)
920 (1022)
861.
Mean = np = 6(0.25) = 1.5
Value
0
1
2
3
4
Observed
4
21
10
13
2
Expected
8.8989
17.7979
14.8315
6.5918
1.6479
The expected frequency for value 4 is less than 3.
Combine this cell with value 3:
Value
0
1
2
34
Observed
4
21
10
15
Expected
8.8989
17.7979
14.8315
8.2397
The degrees of freedom are k

p

1 = 4

0

1 = 3
a) 1) The variable of interest is the form of the distribution for the random variable X.
2) H
0
: The form of the distribution is binomial with n = 6 and p = 0.25
3) H
1
: The form of the distribution is not binomial with n = 6 and p = 0.25
4)
α
= 0.05
5) The test statistic is
(
29
χ
0
2
2
1
=

=
∑
O
E
E
i
i
i
i
k
6) Reject H
0
if
χ
χ
o
2
0 05 3
2
7 81
=
.
,
.
7)
(
29
(
29
χ
0
2
2
2
4
88989
88989
15
8 2397
8 2397
=

+
+

=
.
.
.
.
10.39
8) Since 10.39 > 7.81 reject H
0
. We can conclude that the distribution is not binomial with n = 6 and p =
0.25 at
α
= 0.05.
b) Pvalue = 0.0155 (found using Minitab)
868.
1. The variable of interest is failures of an electronic component.
2. H
0
: Type of failure is independent of mounting position.
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 Spring '08
 Perevalov
 Statistics, Null hypothesis, Statistical hypothesis testing

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