1. ## p-value measurement

My gratitude for anyone who can help me, a statistics novice educated in the last hour by Wikipedia and other sources, with the following problem.

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I have been asked to find some suitable p-value which applies to the following data:

I have 2 groups.
The first group of 10 people are, e.g., male, and have an average value of 52% for a given property X, with a standard deviation of 29.8%.
The second group of 45 people are, e.g, female and have an average of 35% for the same property X, with a standard deviation of 21.9%.
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Now, I have read between the lines and presumed that the "null hypothesis" which I have learnt about in the last hour would be that there is no correlation between gender and property X.

I figure that, for the null hypothesis to be true, both groups should have a mean of 38.1% (the sample mean of both groups). Already, I feel that that last point is a bit weak, since the presence of correlation and unequal sample size means that the sample mean in the null hypothesis will lean towards the larger group.... but anyways, it is the best I have to work with.

I guess my question is, does anyone know of any measure that will provide a p-value given the above data, to represent the probabililty of the given observed data if there was no correlation between the group variable (gender) and the measured property X.

Having had a look around Wikipedia, and being a complete statistics novice, I am overwhelmed by the amount of jargon I cannot digest and different avenues of approach for p-value measurement. I would normally be a bit more patient, however I do need some answer for a p-value within the next couple of hours. Hence I post here and hope that someone has their brain wired in a way that I don't, and that benefits solving my problem.

I think perhaps the most promising approach I found was the Student's t-test, and more specifically Welch's t-test.

Any help appreciated!!!

2. Apologies for my deadline induced hastiness, which I hope will not be interpreted as laziness. I found the following post further down.

I'm hoping the Excel file posted there will solve my problems? Inputting the following:

Avg 52.000 35.000
SD 29.800 21.900
n 10 45

I get the following results:

Pooled Std Dev 9.973
Computed t Statistic 1.705
Critcal Value of t 2.006
Probability of Computed t 0.094

and am hoping I can interpret the 0.094 value as my p-value. If so, a warm thanks will be heading in JohnM's direction.

3. I am surprised nobody answered this one. That is not quite right.

We assume the Null possibility is true and ask ourselves whether what we are viewing is consistant with that possibility.

In this case we are assuming that the characteristic of interest has some constant probability of being possessed by individuals in a population and that that probability is independent of some other catagorical covariate such as s.ex or group.

Because we are doing two draws from a binomial random variable under the null, we know about our supposed variance then you might expect from statistical theory. The t-test usually corresponds to an assumption of normality with unknown variance so it is not an appropriate test.

Your test is described as testing for two independent proportions. Or a two sample proportion test or some variant on those words.

http://faculty.vassar.edu/lowry/propdiff_ind.html

That worksheet will do the math for you. Sample A is the females, and sample B is the males.

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