# Interpreting z-score of an Output

#### Freddy19911

##### New Member
Hi,
I am struggling to interpret the z-values for the attached output. t values are given in the brackets with index a, z-score is given in the brackets with index b. Unfortunately, I have no clue what the z score tells me in this case. According to the t score, there are abnormal returns of -0,35%, which are however not significant on any level (t value of -1,37). What exactly does the z score of -1,96 tells me with the respective value of 40.4? #### noetsi

##### Fortran must die
A z score tells you how likely a value is relative to the mean and distribution of the data (through the standard deviation). As the absolute value of the z score gets higher the value is less likely. Beyond that I am not sure enough of this diagram to interpret it.

#### Freddy19911

##### New Member
A z score tells you how likely a value is relative to the mean and distribution of the data (through the standard deviation). As the absolute value of the z score gets higher the value is less likely. Beyond that I am not sure enough of this diagram to interpret it.
Thanks for your reply. Unfortunately, I am still struggling to get what the purpose of this output is in this specific case.
The author states the following:

"The estimated excess returns for bonds are slightly skewed so we also report both median excess bond returns and the proportion of positive window-spanning excess bond returns. A binomial test (a Z-statistic) indicates whether the observed proportion of positive window-spanning excess bond returns is reliably different from 50%.
The Z-Statistic on the proportion positive is based on a binomial test of whether that proportion is >50%"

So the t-test with the corresponding t-value proofs the significance of mean excess re
turns and in the sample case "All", the Mean excess returns are -0.35 but the t-value suggest, that they are not significance different from 0 and so I reject the hypothesis and conclude that, in the mean, there are no significant excess returns. Since this is what the author wants to test, I am not sure why he provides the output with the z-score. Any suggestions?
Again, thanks for you help!

#### Junes

##### Member
From what I understand he's worried that the assumptions of the t-test are not met. He says the distribution is "slightly skewed". However, it's not the distribution of the variable that's assumed to be normal but the sample mean. If the distribution of bond bond excess is slight skewed than probably the sample mean is normally distributed if n is 215.

He's using the binomial test as an alternative, basically as a direction test to determine if the percentage is different from 50% (expected from chance). This ignores the magnitude of each excess, which the t-test does take into account. In theory, the number of positive excesses could be >50% while the mean is negative.

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#### Freddy19911

##### New Member
Thanks June. After doing some research I hopefully got it right now: Basically I have a number of 215 corporate bonds and for some of those bonds the author estimated excess returns after the event of a Credit Rating Downgrade. The distribution of those estimated returns are slightly skewed and may not follow a normal distribution. However, the t-test does only assumes that the sample mean is normally distributed and because of the Central limit theorem I can assume that this is the case, given a big enough sample size of 215 observations. Therefore, the Binomial test is basically neglectable when interpreting the results.
If I want to interpret the Z-score I would say: Given the sample size "All", in 40.4% of the cases there are excess notable returns. The binomial test is testing, whether the observed 40.4% differ from 50% (Is there an explanation why the authors choose 50% in this case?). If they do differ significantly, one would conclude that the 40.4% excess returns are not there by chance but are significant. Therefore this would be "additional significance" to the t test results. However, the Z score does not tell me whether the excess returns are positiv or negative.

Did I get it right or are there some mistakes in my understanding?

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#### Junes

##### Member
Completely correct. I assume the 50 percent is chosen because if there's no effect then you would expect half to be higher than zero and half lower, just by chance. But I don't know the context well enough to know if that makes sense.

#### Freddy19911

##### New Member
All right, one last question: I want to get the corresponding probability to the z score of -1.96. Do I have to look it up in this table:

http://www.z-table.com

or in the table for binomial test? I am a bit confused because of the term z score and binomial test in the same sentence. I thought these two are completely different things.

Again, thanks a lot for your help!

#### Junes

##### Member
A z table should be fine.

A binomial test can be done using an exact test, which uses combinatrics to calculate the p value. However the binomial distribution is approximately normal for large n, with parameters m = np and s^2 = np(1 - p), where n is the number of trials (the number of bonds) and p is the probability of "success" (here 50%) So you can also use a z test using the number of positive/negative excess returns.

#### Junes

##### Member
If you wonder why one would use a z test over an exact test, part of it is historical: those are simpler to calculate. However there can be other advantages, such as easily being able to calculate a confidence interval or power calculations.

#### Junes

##### Member
Oh and I'm not sure if this is correct: "However, the Z score does not tell me whether the excess returns are positiv or negative." It does tell you something like that: if it's positive then there are more positive excess returns than negative ones. However, since it doesn't take into account the magnitude, it could still be the case that the mean excess return swings the other way. If you meant that, then it's correct.

#### Freddy19911

##### New Member
Makes Sense so far, thanks. I just Wonder why the Headline in the Output is „percent of Excess returns > 0“
Since the authors measure the Effect of Rating downgrades (in this case one would expect excess returns <0, so negative excess Returns) i would have Expected that they test the percentage of excess returns < 0.but Mayen „percent of Excess returns > 0“ refers to any case of excess returns (positive as well as negative). I am just Confused by the Term „Proportion of positive excess returns“ which they use to describe the Test in the paper.
Anyway, this isn‘t Really a statistical question and you really helped me out here. Thanks a lot.