P1.T2.209 T-statistic and confidence interval

[Dr. Jayanthi Sankaran]

Hi David,

209.2 Over the last two years, a fund produced an average monthly return of +3.0% but with monthly volatility of 10.0%. That is, assume the random sample size (n) is 24, with mean of 3.0% and sigma of 10.0%. Further, the population's returns are normal. Are the returns statistically significant, in other words, can we decide the true mean return is greater than Zero with 95% confidence?
a) No, the t-statistic is 0.85
b) No, the t-statistic is 1.47
c) Yes, the t-statistic is 2.55
d) Yes, the t-statistic is 3.83

n = 24, sample mean = 3.0% sigma = 10%, is mu>0 with 95%confidence?


t = Sample mean - population mean = 3% - 0% = 1.4697, t23, 95% = 1.71, two-tailed = 2.07

sigma/SQRT(n) 10%/SQRT(24)​

No, the t-statistic = 1.4697 (1.4697<1.71)
Even if we assumed normal one-sided, 95% critical Z is 1.645 (1.4697 <1.645)

In the above question, while we compute the t statistic why are we comparing it with 95% critical Z - I was under the impression that n>30 for such a comparison.

Thanks!
Jayanthi

 

[brian.field]

I am pretty sure that you can use the Z when the population is normal - or at least, that is why I would justify the use of Z here.

 

[Dr. Jayanthi Sankaran]

Hi Brian - granted that the population's returns are normal and that we can use the Z. However, why are we using the
t statistics? I culled this down from David's answer!

Thanks
Jayanthi

 

[brian.field]

I suppose the t statistics are being used since population variance is unknown.....
At the end of the day, I believe one can use either Z or t statistics when the population is normal and it shouldn't make a difference but I am unsure of this.

I also believe it is the case that critical t values will be greater (in abs value) than corresponding critical z values since the t distribution's tails are fatter.

So, David was probably using the Z critical value to emphasize the fact that you cannot reject the null in either case, i.e., in both cases, the statistic is less than the corresponding critical t value and critical z value.

 

[Dr. Jayanthi Sankaran]

You are right - while the sample variance is a given, there is no information about the population variance, here.
'I also believe it is the case that critical t values will be greater (in abs value) than corresponding critical z values since the t distribution's tails are fatter'. - a good point!

 

[lRRAngle]

My problem is at the end of the process. After we computed the Test Statistic at 1.469 and realized we need to compare it to 1.65 -> at this point, how do you determine if the answer to the question requires the Test Statistic to be smaller or greater than the critical value. It seems I am always stumped at this last part. Is there a rule of thumb?

Also, given I was confused on above, I went a head an constructed a confidence interval for the null hypothesis H: mean x <= 0;
and with 3 +/- 1.645* (10/24^0.5) I got CI [-0.37, 6.36] which I interpeted this to indicate that the null hypothesis could be rejected. But it looks like this gives me the wrong conclusion. So I totally messed this up and ehhh it feels like I am so close but yet so far

 

[David Harper CFA FRM]

@lRRAngle it's often forgotten that these values are standard deviations (the critical value of 1.65 is just the special case of the standard deviation for the "standard" unit variance distribution, is why I actually call them "standard standard deviations"). So when we observe a test statistic of 1.47 it literally means "the observed [aka, sample] mean is only 1.47 unit standard deviations away from [in distance] the hypothesized true (population) mean." So low values imply an observations that's near to the hypothesized null value: and if it's low then it's close and we shouldn't reject the null because, well, that could just happen with random (sampling) variation. High values indicate a greater distance and these are less likely to be random, and more likely to implicate our null as false. Hopefully that's helps!

 

[lRRAngle]

Thanks for the response in real time David. (You are hitting them down as soon as I put them up - How great is that?!

I still didn't lock in on the logic here though. The question is asking: "Can we decide the true mean return is greater than Zero with 95% confidence?"

Thinking out loud here: So the Null in this case, I would write is H(0): X =<0. (In this way, my thinking is, if we fail to reject, then we can assume that the mean return is in fact greater than Zero, and the answer is yes)
Now my test statistic was 1.469 which is less than the critical value of 1.645 ---> so you are saying that this could simply be due to randomness and having a 1.469 std from zero is not significant enough (or far enough from the zero) to indicate at a 95% confidence that the return is not in fact less than zero.

I guess, I shouldn't have constructed a CI because the CI approach is appropriate for when you are checking if a specific value is contained within the (random) CI, when this question is really asking if the mean can be greater than 0, hu?

 

[David Harper CFA FRM]

@lRRAngle

Yes, the implicit one-sided null here is H(0): µ ≤ 0 such that H(A): µ > 0. But this ("If we fail to reject, then we can assume that the mean return is in fact greater than Zero, and the answer is yes") is not quite correct; if we fail to reject, then we accept the null, which means that we decide per H(0) that "yes the true mean, µ, could be zero."

The given question asks "can we decide the true mean return is greater than Zero with 95% confidence?" and this translates into "can we reject the one-sided null in favor of the one-sided alternative which says H(A): µ>0"? The answer is, The 1.469 is not far enough away from zero, such that we cannot reject the null; i.e., we accept the null and we basically reject the alternative. Hence the correct answer is "no we cannot decide the true mean return is greater than zero ..." I hope that helps, i'll be back in the forum tomorrow (I need to get to other work now!). Thanks,

 

[lRRAngle]

This helps. TVM!