Consider daily stock market closing prices for a company observed in
two sample periods. For Johnson & Johnson, a test of equal means in
the first and second half of 1999, was developed in the lecture notes
for Chapters 8.2 and 10.2. The test method assumed the same
variance in the two sample periods. If this variance assumption does
not describe the data then the test conclusions may be unreliable.
Is the assumption of equal variance in the two sample periods a
reasonable assumption ?
To answer this question, test the null
hypothesis of equal variances in the two samples against a twosided
alternative.
Summary statistics for the closing prices in the two sample periods
were reported as:
January to June
124
=
x
n
32.54
=
2
x
s
July to December
128
=
y
n
21.45
=
2
y
s
Econ 325 – Chapter 10.4
12
The test statistic is the variance ratio:
1.517
21.45
32.54
=
=
2
y
2
x
s
s
Note that the larger variance is placed in the numerator to give a
test statistic that exceeds one.
The test statistic can be compared with an Fdistribution with
numerator degrees of freedom (124 – 1) = 123 and
denominator degrees of freedom (128 – 1) = 127.
For a twotailed test, the pvalue is found as:
pvalue
)
F
(
P
)
127
,
123
(
1.517
2
>
⋅
=
The pvalue calculation is illustrated in the graph below.
PDF of
)
127
,
123
(
F
1.517
1
0.5
0
area = pvalue / 2
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Econ 325 – Chapter 10.4
13
To find an exact pvalue, with Microsoft Excel select Insert Function:
FDIST(1.517, 123, 127)
This gives the answer:
pvalue
/
2 =
0.0102
Therefore, the pvalue for the twotailed test is:
2 (0.0102) = 0.0204
At a 1% significance level the pvalue exceeds
α
= 0.01.
This gives evidence that, for the Johnson & Johnson daily closing
prices, the null hypothesis of equal variance in the two sample
periods is not rejected.
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 Fall '10
 WHISTLER
 Statistics, Variance, Null hypothesis, Statistical hypothesis testing, FC

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