# Query - Two Tailed Hypothesis Testing

Discussion in 'P1.T2. Quantitative Methods (20%)' started by Avishek, Sep 30, 2007.

1. ### AvishekNew Member

Hi David,

For the below question patterns, would they be providing us the table or value?

In a two-tailed hypothesis test, Jack Olson computes a t-statistic of 2.7 based on a sample of 20 observations where the distribution is normal. If a 5 percent significance level is chosen, Olson should:

A) not make a conclusion pending additional observations.

B) reject the null hypothesis and conclude that the population mean is not significantly different from zero.

C) reject the null hypothesis and conclude that the population mean is significantly different from zero.

D) fail to reject the null hypothesis that the population mean is not significantly different from zero.

Regards, Avi

2. ### David Harper CFA FRMDavid Harper CFA FRM (test)Staff Member

Avi,

Based on a conversation I had about three weeks ago with Diane Beebe at GARP, on exactly this issue, she told me: Yes, you would be provided an extract from the t-Distribution table for such a question. Especially, in regard to this question, you need a table.

The question is instructive.

The t-stat of 2.7 says, basically, "the sample mean is 2.7 standard errors away from the hypothetical mean." Normally, without a lookup table, we often use 2.0 as a 'rule of thumb." That is, a t-stat > 2.0 is thought to be significant (i.e., unlikely due to random variation). However, the 2.0 is based roughly on a large sample (large deg freedom) and 95% confidence.

In this case, the lookup table for 19 degrees of freedom (20 - 1 = 19) gives you 2.09. Note, I did not know that from memory, of course not, I knew it was near 2.0, but I input into excel this function: =TINV(0.05,19) = 2.09. (Note, the =TINV() function is implicitly two-tailed, a lookup table might be one-tailed, in which case you would lookup 19 df @ 0.025 to get 2.09).

So, in this case 2.7 > 2.09 (two-tailed 95% confidence with 19 df)

Don't lose sight of what this simply means: 2.7 standard errors (deviations) away from the hypothetical mean is outside the area under the distribution that would include 95% of random occurrences.

So, given that, we'd reject the null and conclude the mean is sign. different than zero.

Let's now understand why we need a table:

What if the question were, "...If a 1% significance level is chosen...". Then, my critical t = TINV(0.01,19) = 2.86 and I must "fail to reject the null." If I want to be more confident (i.e., increase from 95% to 99% confidence), then must expand the area under the curve and my test statistic now falls under the 99% area that includes the possibility of random deviation from the hypothetical mean. See how the 2.7 implies reject @ 95% and accept @ 99%?

Okay, but while we are here, please DO MEMORIZE the critical z values because the test will assume you know them. See this post. You must memorize 1.645 and 2.33.

David

3. ### AvishekNew Member

Thanks a ton David for all the efforts.

Have memorized the below two: Is there anything more needed for the z-Table?

Two Tailed:
95%: 1.96
99%: 2.58

One- Tailed:
95%: 1.65
99%: 2.33

Thanks, Avi

4. ### David Harper CFA FRMDavid Harper CFA FRM (test)Staff Member

Avi,

No, these will be enough for the z-table.

Just be sure to know when to use each. VaR is always only one-tail (we care about losses, not gains) so 1.65/2.33 are more likely to be used on the test.

If the question is, "What is 95% VaR if volatility is 10%?" then we use one-tail : 95% VaR = (10%)(1.65) = 16.5%

If the question is, "We forecast a portfolio return of 10% with 5% volatility, what is the 95% confidence interval", the answer is two-tailed (confidence *interval*): 95% confidence interval = 10% +/- (1.96)(5%)

David

5. ### AvishekNew Member

David!

I got your point. You have explained this again and again in your videos. So I can't forget the application aspect going forward.

Warm wishes,
Avi