Tuesday, September 20, 2016

Hypothesis tests pertaining to intra-day and inter-day ETF price changes.




Question   Using data below on intra-day and inter-day price changes of a large-cap ETF test the following.

Test the hypothesis that the variance of the intra-day price changes is equal to the variance of the inter-day price changes.

Test the hypothesis that the average intra-day price change is equal to the average inter-day 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 Intra-day Change for
 a Large Cap ETF
Daily Change
Intra-day Change
Inter-day 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 intra-day price change is the percentage difference between the closing price and opening price on the current day.  The inter-day 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 p-value 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 -0.0015-0.00044 or -0.00059.


STD_ERR=((0.00973)2/3179 + (0.00696)2/3179)) 0.5 Or 0.00021.


The t-statistic is -2.76.  


The p-value 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 t-statistic so plug in the absolute value.)

We reject null hypothesis at a 0.01 level of significance.

Would a one-tailed 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.




No comments:

Post a Comment