Homework 5 – solution
1.
Since
n
= 12, we should use no more than 4 bins.
Let the bins
correspond to the four seasons (with the first bin corresponding to winter).
Then we have
E
i
= 3 for
i
= 1
, . . . ,
4.
O
1
= 3,
O
2
= 6,
O
3
= 2, and
O
4
= 1.
The value of the test statistic is
χ
2
=
1
3
(0 + 9 + 1 + 4) = 4
.
67
The corresponding critical value is
χ
2
0
.
05
,
4

0

1
=
χ
2
0
.
05
,
3
= 7
.
81. Since
χ
2
<
χ
2
0
.
05
,
3
, we cannot reject the null hypothesis, and conclude that the data is
consistent with the uniform birthday hypothesis.
2. Let
x
be the number of female smokers in the sample. Then the table
of observed values looks like
smoking
nonsmoking
men
40

x
60 +
x
women
x
100

x
and the table of expected values is
smoking
nonsmoking
men
20
80
women
20
80
Computing the value of the test statistic, we obtain
χ
2
= (20

x
)
2
1
20
+
1
80
+
1
20
+
1
80
¶
=
(20

x
)
2
8
.
Equating it to the critical value, we obtain
(20

x
)
2
8
=
χ
2
0
.
05
,
1
= 3
.
84
.
Solving the latter for
x
, we get
x
= 20

√
8
·
3
.
84 = 14
.
46
.
1
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Therefore, the smallest number of female smokers for which the independence
hypothesis would not be rejected is 15.
3. Using Minitab we obtain that the test statistic value is
χ
2
= 14
.
19 and
the corresponding Pvalue is 0.116 meaning that there is only a moderate
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 Spring '08
 BEST
 Statistics, Normal Distribution, Probability theory, Statistical hypothesis testing, 3 degrees

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