T-Statistic

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FelixG

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Hey guys, I have got a statistical question about a paper from a financial journal article I just read.
In the article the author's investigaste the foreign exchange market and as part of that created a descriptive statistics table, the returns just represent simple daily buy-and-hold returns and Std Dev etc. are all for these, which looks roughly like this:

Exchange Rate A Exchange Rate B

Mean return (%) 0.268 0.336

Median return (%) 0.107 -0.007

Std Dev (%) 4.229 4.428

Skewness 0.781 0.885

Kurtosis 1.786 2.163

t-stat 0.993 1.192


This is just an example and there is around 10 different exchange rates on there but what I was wondering is how they calculated the t-stat? From what I have learned you calculate a t-stat by either comparing two samples, which would not make much sense since theres 10 different samples with varying degrees of correlations, or by having an assumed target mean against which to test, which as far as I know does not exist. Am I wrong about this or is there something I have missed?

Thanks for your time guys.
 
FelixG said:
Hey guys, I have got a statistical question about a paper from a financial journal article I just read.
In the article the author's investigaste the foreign exchange market and as part of that created a descriptive statistics table, the returns just represent simple daily buy-and-hold returns and Std Dev etc. are all for these, which looks roughly like this:

Exchange Rate A Exchange Rate B

Mean return (%) 0.268 0.336

Median return (%) 0.107 -0.007

Std Dev (%) 4.229 4.428

Skewness 0.781 0.885

Kurtosis 1.786 2.163

t-stat 0.993 1.192


This is just an example and there is around 10 different exchange rates on there but what I was wondering is how they calculated the t-stat? From what I have learned you calculate a t-stat by either comparing two samples, which would not make much sense since theres 10 different samples with varying degrees of correlations, or by having an assumed target mean against which to test, which as far as I know does not exist. Am I wrong about this or is there something I have missed?

Thanks for your time guys.
Click to expand...
OK - I already get suspicious of someone who uses kurtosis, so might as well be skeptical about quoting a value of the t-statistic as well. Looking at a single data set, such as Rate A, gathered over n days where n is a large number?? Skewness I presume is determined from the 3rd moment of the data around the mean, and kurtosis from the 4th moment. I have done a lot of statistical analysis of experimental (physics) data, but have never resorted to kurtosis or a t-statistic.

So how could t be calculated? Once you have determined the mean, the deviations of the data from the mean, \(\displaystyle X\), would have to be normalized by by an independent measure of the standard deviation compared to the observed standard deviation,\(\displaystyle Y/\sigma \). If \(\displaystyle X\) is normal and \(\displaystyle Y\) is \(\displaystyle \chi^2\), then

\(\displaystyle t = \dfrac{X\sqrt{n}}{Y}\)

is a sample from the t-distribution, which has mean 0 and Variance n/(n-2). Is there any clue what he chose for \(\displaystyle Y\)? Since it is already clear from the skewness of the distribution that \(\displaystyle X\) is not normal, I have no idea what the value of t near unity means.

I am not a statistician, but rather a physicist. Thus my task is to use statistics to assess the validity of data, without lying too much.
 
Thanks for the reply.

First, validating the data set is what he was trying to accomplish. I can only speculate that he has used the the standard deviation for Y, probably all calculating done through Excel.
As to the number of observations, the sample period consists of 264 (monthly) observations for each of the exchange rates.
But, I don't think I understand the formula used or the reasoning behind it. Would it not make more sense to calculate the p-value from what has been done up to the t-statistic?
 
FelixG said:
Thanks for the reply.

First, validating the data set is what he was trying to accomplish. I can only speculate that he has used the the standard deviation for Y, probably all calculating done through Excel.
As to the number of observations, the sample period consists of 264 (monthly) observations for each of the exchange rates.
But, I don't think I understand the formula used or the reasoning behind it. Would it not make more sense to calculate the p-value from what has been done up to the t-statistic?
Click to expand...
Looking in Excel, I find a function TTEST that will make a comparison of two distributions to find probability that they have a common mean, but I don't see an algorithm for the t-statistic of one distribution.

I also don't think I understand the formula used or the reasoning behind it. It doesn't convey any meaning to me in the context of describing a single non-normal distribution. Sorry about that!

Could it have anything to do with "standard deviation of the estimate" of the mean?
 
It has come to my attention that he has used the one-sample t-test formula and chose 0 as value for the population mean, therefore just using the mean, standard deviation and number of observations for this calculation.
I am not sure what this number is supposed to represent, if it represents anything at all.
 
FelixG said:
It has come to my attention that he has used the one-sample t-test formula and chose 0 as value for the population mean, therefore just using the mean, standard deviation and number of observations for this calculation.
I am not sure what this number is supposed to represent, if it represents anything at all.
Click to expand...
It measures whether the sample mean is a good representative of the population mean. If the sample mean does not even properly represent the known mean of the population, then we can be certain that the sample is not generally representative, no?

http://www.statisticssolutions.com/...ry-of-statistical-analyses/one-sample-t-test/
 
JeffM said:
It measures whether the sample mean is a good representative of the population mean. If the sample mean does not even properly represent the known mean of the population, then we can be certain that the sample is not generally representative, no?

http://www.statisticssolutions.com/...ry-of-statistical-analyses/one-sample-t-test/
Click to expand...
Thanks Jeff - that makes some level of sense, but since it is already clear the distribution is not normal, then this is not a valid test. How about just using the Sampling Theorem?

Consider case A. Given that the sample provides an estimator of the population standard deviation, 4.229%, then the distribution of sample means is normal and has a standard deviation 4.229%/sqrt(N) = 0.260%. The observed mean is 0.268%. That is only one standard deviation away from 0, so the hypothesis that the true mean is 0 can not be rejected.
 
DrPhil said:
Thanks Jeff - that makes some level of sense, but since it is already clear the distribution is not normal, then this is not a valid test. How about just using the Sampling Theorem?

Consider case A. Given that the sample provides an estimator of the population standard deviation, 4.229%, then the distribution of sample means is normal and has a standard deviation 4.229%/sqrt(N) = 0.260%. The observed mean is 0.268%. That is only one standard deviation away from 0, so the hypothesis that the true mean is 0 can not be rejected.
Click to expand...
Dr. Phil

It is always a surprise when I make even some semblance of sense. Digging way back into my hazy recollections, I think the t-test is preferred when both the mean and variance of the underlying population are unknown. As the size of the sample goes up, it is also my recollection that the distribution of the t-statistic is in the limit equal to the normal distribution. In other words, the t-statistic is to be used for testing the mean if the sample is relatively small.

I was really not trying to explain why statisticians consider the t-distribution to be superior to the normal distribution in certain circumstances, a task that is definitely beyond my powers; I was just trying to explain what I thought the test was being used to do.
 
JeffM said:
It is always a surprise when I make even some semblance of sense. Digging way back into my hazy recollections, I think the t-test is preferred when both the mean and variance of the underlying population are unknown. As the size of the sample goes up, it is also my recollection that the distribution of the t-statistic is in the limit equal to the normal distribution. In other words, the t-statistic is to be used for testing the mean if the sample is relatively small.

I was really not trying to explain why statisticians consider the t-distribution to be superior to the normal distribution in certain circumstances, a task that is definitely beyond my powers; I was just trying to explain what I thought the test was being used to do.
Click to expand...
Any clod of an experimental physicist can understand the normal distribution - but only the pros know about the t-distribution. My expertise extends through chi^2, and I once used an F-test in a published paper (to the amazement of my colleagues), but that's about it - except for what I learn trying to answer questions in this forum!

I appreciate the dialogue.
 
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