need some help on a statistics challenge

[JSD - NYC]

tickersu wrote:

JSD - NYC wrote:
I certainly do not discount what tickersu is saying and he definitely seems to know what he is talking about. However, I just can’t replicate the results he seems to be getting at. It appears he is using a test against a single mean set forth in Section 3.1 of Reading 11. When I try to calculate the test out, I get very large t-value, which appears to be a consequence of the very large sample size.
Question to mlwl8521 - Can you let us know where the question came from and how the authors took it on?

I’m conducting a test for a single proportion, not for the mean number of outcomes. I don’t believe the CFA curriculum explicitly covers hypothesis testing (or CIs) for proportions.
The way I read this question was that a sample of 10,000 bernoulli trials was taken. This would be a binomial sampling of n=1 (since a binomial random variable is the sum of a bunch of bernoulli trials). In other words, I viewed it as a single observation of a binomial random variable where the number of bernoulli trials in each binomial observation is 10,000, and we observed 120 successes (this is one of many values in the binomial distribution for the variable, but it’s all we sampled).
You can replicate my results by using the information I provided. Try reading this website a bit to see what I am describing. http://stattrek.com/hypothesis-test/proportion.aspx?Tutorial=AP
Be sure to check out the sample problems they walk you through as well, so you see the calculation for a hypothesis test of proportion.

Great stats site!
Ok, I have come up with the 2.01 z-score (2.010076 really) using the method outlined. I then got a little mad that the concept wasn’t coming to me, so I brute-force coded the cumulative Bernoulli distribution. Turns out that the cumulative probability for a z-score of 2.010076 is 0.977788. My cumulative probability for the Bernoulli distribution of p=.01 at <=120 positives =0.977866, a difference of .0000097 most likely due to Window’s inability to deal with really small and large numbers (i.e., the small probabilities like .01^10000 and huge combinatorics like 10000C120, which becomes more problematic when you are trying to mutilitply them and then sum up a bunch of them together).
I then re-jiggered the distribution to p=.012 and checked the cumulative distribution for >=100 and got .033853, which compares to a predicted cumulative probability based on a z-score of -1.836796 (by the process I used above) of .033120 or a difference of .000733, again probably small enough for that this looks like a Window’s issue.
So I am getting the idea that both hypothesis tests work.
Thoughts?
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EDIT: Of course the reason for the small differences could also be the fact that Bernoulli distributions are discrete, not continuous. So we would not expect a perfect match in any case.
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[tickersu]

JSD - NYC wrote:

tickersu wrote:

JSD - NYC wrote:
I certainly do not discount what tickersu is saying and he definitely seems to know what he is talking about. However, I just can’t replicate the results he seems to be getting at. It appears he is using a test against a single mean set forth in Section 3.1 of Reading 11. When I try to calculate the test out, I get very large t-value, which appears to be a consequence of the very large sample size.
Question to mlwl8521 - Can you let us know where the question came from and how the authors took it on?

I’m conducting a test for a single proportion, not for the mean number of outcomes. I don’t believe the CFA curriculum explicitly covers hypothesis testing (or CIs) for proportions.
The way I read this question was that a sample of 10,000 bernoulli trials was taken. This would be a binomial sampling of n=1 (since a binomial random variable is the sum of a bunch of bernoulli trials). In other words, I viewed it as a single observation of a binomial random variable where the number of bernoulli trials in each binomial observation is 10,000, and we observed 120 successes (this is one of many values in the binomial distribution for the variable, but it’s all we sampled).
You can replicate my results by using the information I provided. Try reading this website a bit to see what I am describing. http://stattrek.com/hypothesis-test/proportion.aspx?Tutorial=AP
Be sure to check out the sample problems they walk you through as well, so you see the calculation for a hypothesis test of proportion.

Great stats site!
Ok, I have come up with the 2.01 z-score (2.010076 really) using the method outlined. I then got a little mad that the concept wasn’t coming to me, so I brute-force coded the cumulative Bernoulli distribution. Turns out that the cumulative probability for a z-score of 2.010076 is 0.977788. My cumulative probability for the Bernoulli distribution of p=.01 at <=120 positives =0.977866, a difference of .0000097 most likely due to Window’s inability to deal with really small and large numbers (i.e., the small probabilities like .01^10000 and huge combinatorics like 10000C120, which becomes more problematic when you are trying to mutilitply them and then sum up a bunch of them together).
I then re-jiggered the distribution to p=.012 and checked the cumulative distribution for >=100 and got .033853, which compares to a predicted cumulative probability based on a z-score of -1.836796 (by the process I used above) of .033120 or a difference of .000733, again probably small enough for that this looks like a Window’s issue.
So I am getting the idea that both hypothesis tests work.
Thoughts?

I think what matters is using the appropriate methodology for the given question. You could do a hypothesis test for the population mean number of successes, or you could do hypothesis test for the population proportion of successes. You’ll have different calculations in each case. So, each will work, provided you’re answering the question you asked.
What we should be looking at now is the p-value for the hypothesis test. What does everyone think?

 

[JSD - NYC]

tickersu wrote:
I think what matters is using the appropriate methodology for the given question. You could do a hypothesis test for the population mean number of successes, or you could do hypothesis test for the population proportion of successes. You’ll have different calculations in each case. So, each will work, provided you’re answering the question you asked.

Right, I think we were appropriately testing alternatively the sample proportion and the population mean. If I have some more time, I might try to see if I can figure out an algebraic equivilance between the two tests, but for now, I am just going to let this one rest.
In any case, thanks for putting me on to StatTrek. I’m definitely going to be using that one.