Hypothesis testing procedures are applied to interday and intraday stock price changes.
Question Using data below on intraday and interday
price changes of a largecap ETF test the following.
Test the hypothesis that the variance of the intraday price
changes is equal to the variance of the interday price changes.
Test the hypothesis that the average intraday price change
is equal to the average interday price change?
Data: The descriptive statistics used in this post
were generated in a post that I wrote in my finance blog. The financial issues can be found at the
link below.
Daily Change and
Intraday Change for
a Large Cap ETF


Daily Change

Intraday Change

Interday Change


Average

0.00036

0.00015

0.00044

Median

0.00078

0.00034

0.00066

Stand deviation

0.01195

0.00973

0.00696

skew

0.08159

0.20989

0.29673

kurtosis

11.12289

10.00726

9.94566

Min

0.08943

0.07945

0.05864

Max

0.11458

0.07478

0.05259

10th

0.03533

0.02781

0.02102

90th

0.01184

0.00902

0.00697

Count

3180

3180

3180

The intraday price change is the percentage difference
between the closing price and opening price on the current day. The interday price change is the percentage
difference between the opening price on the current day and the closing price
on the previous day.
Analysis:
The test statistic for equality of variances is the ratio of
the variances. The test statistic in
this case is (0.00973)2/(0.00696)2 or 1.95
The test statistic follows the F distribution with 3,179
degrees of freedom for both the numerator and the denominator. Using Fdist(1.95,3179,3179) we get a pvalue
that is essentially 0.
We reject the null hypothesis that the variances are equal.
Our test of the
hypothesis the averages are equal assumes unequal variances. The test statistic used in the difference in
the averages divided by the standard error where the standard error.
The difference in the averages is 0.00150.00044 or 0.00059.
STD_ERR=((0.00973)^{2}/3179 + (0.00696)^{2}/3179))
^{0.5} Or 0.00021.
The tstatistic is 2.76.
The pvalue was obtained using the TDIST function in Excel t.dist(2.76,3180,2) which gives us
0.0058. (Note the TDIST function does
not allow for negative values to the tstatistic so plug in the absolute
value.)
We reject null hypothesis at a 0.01 level of significance.
Would a onetailed test result in the rejection of the null hypothesis at a
0.01 level of significance?
I am planning to create more problems of this type and list
them as a quiz in my statistical resources blog.
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