Why is p-value greater than 1 when it should not

After doing some searching online, I realized that p values should not be greater than 1. Somebody please tell me what I am doing wrong here:

Final Template2.xlsx

Kindly refer to problem 2 here:

http://stattrek.com/hypothesis-test/...px?Tutorial=AP

Change null and alternative from:

Null hypothesis: μ1 - μ2 >= 7
Alternative hypothesis: μ1 - μ2 < 7

To:

Null hypothesis: μ1 - μ2 = 0
Alternative hypothesis: μ1 - μ2 not equal

[.... deleted by me ....]

[.... deleted by me ....]

As near as I can tell, the calculation programmed into the spreadsheet is correct. I put those parameters (t=2.24 and df=145) into stattrek's t-distribution calculator (link in tutorial page) and into this calculator http://www.danielsoper.com/statcalc/...tor.aspx?id=41 and got the same answer as your function did. So this appears to be the correct result for the cumulative distribution function. This table of cumulative probabilities (http://www.sjsu.edu/faculty/gerstman...er/t-table.pdf ) shows many entries that are greater than 1. Are you certain that your calculation must be less than 1?

I have not thought through this fully, but is it possible that you intended to compute the probability density function (3rd argument of T.DIST() function is 0 or FALSE)? The probability density function will return values much closer to 0 (and probably cannot return a value greater than 1).

It has been a long time since I looked at t-tests, so I don't remember the details very well. I would suggest that you need to check your procedure and make sure you are clear on what the p value represents.

P values should not be greater than 1. They will mean probabilities greater than 100 percent.

Consider for example the P value in my attachment:1.973

If multiplied by 100 to convert to percentage you get 197.3%.

Is this realistic?

According to explanation here the p shouldn't be greater than 1:

https://socratic.org/questions/can-a...why-or-why-not

How can I perform the 3 t tests with the following as the alternative hypotheses:

1st - test for the equality of the means

2nd - if mean 1 is greater than mean 2

3rd - if mean one is less than mean two.

Originally Posted by Onditi

Consider for example the P value in my attachment:1.973 [....] Is this realistic?

No.

Aha! Part of the problem is: the old TDIST function, which I am used to, is not parameter-for-parameter compatible with the new T.DIST function.

Since you want to test u1-u2=0, you should use a two-tailed test. Enter one of the following formulas into P6:

=T.DIST.2T(ABS(O6),N6)

or

=TDIST(ABS(O6),N6,2)

or

=2*TDIST(ABS(O6),N6,1)

or

=2*T.DIST(-ABS(O6),N6,TRUE)

I can explain further later, if you wish. I have to leave now.

Originally Posted by joeu2004

Part of the problem is: the old TDIST function, which I am used to, is not parameter-for-parameter compatible with the new T.DIST function.

And the various descriptions of Student's t are inconsistent, not only among themselves, but also within themselves.

For example, consider the stattrek.com tutorial. On the one hand, it says that the Student's t-statistic is calculated by the formula t = [ (x1 - x2) - d ] / SE = -1.99. On the other hand, it says that the p-value is interpreted as P(t < -1.99) = 0.027; that is, the probability that t < -1.99. In the latter context, t refers to a random variable, not the Student's t-statistic (which is -1.99).

The stattrek.com Student's t calculator uses the correct nomenclature, namely: P(T < t). T is the random variable that has a Student's t-distribution with df degrees of freedom, and t is the Student's t-score for the sample data.

Originally Posted by joeu2004

Since you want to test u1-u2=0, you should use a two-tailed test. Enter one of the following formulas into P6:
=T.DIST.2T(ABS(O6),N6)
or
=TDIST(ABS(O6),N6,2)
or
=2*TDIST(ABS(O6),N6,1)
or
=2*T.DIST(-ABS(O6),N6,TRUE)

Your formula 2*T.DIST(O6,N6,2) works for the tutorial problem #1 only because the t-score in O6 is negative. (And because T.DIST interprets any non-zero numerical value as TRUE in the 3rd parameter.)

T.DIST(t,df,TRUE) returns P(T < t) for t <= 0 and t > 0. But for a two-tailed test and t > 0, we want P(T < -t) + P(t < T); that is, the sum of the cumulative tail probabilities. Since the Student's t-distribution is symmetrical, P(T < -t) = P(t < T), and P(T < -t) + P(t < T) = 2*P(T < -abs(t)). So we can write 2*T.DIST(-ABS(t),df,TRUE), which is what T.DIST.2T(ABS(t),df) calculates.

TDIST(t,df,1) returns P(t < T) for t > 0. So we can write 2*TDIST(ABS(t),df,1), which is what TDIST(ABS(t),df,2) calculates.

Thank you Joe. Will run some tests then get back here.

I really appreciate you taking the time to take me through this.

Would you be kind enough to display how you would state the null and alternative hypothesis for μ1 > μ2 and μ1 < μ2. Do i compute another set of degrees of freedom and t when testing for μ1 > μ2 and μ1 < μ2, or do i use the same values I used for for μ1 = μ2?

Originally Posted by Onditi

Would you be kind enough to display how you would state the null and alternative hypothesis for μ1 > μ2 and μ1 < μ2. Do i compute another set of degrees of freedom and t when testing for μ1 > μ2 and μ1 < μ2, or do i use the same values I used for for μ1 = μ2?

As I understand it (not an expert), yes: referring to the stattrek.com tutorial, SE, df and t would be the same because d is simply zero.

I provided the null hypotheses in post #15. Note that I corrected one mathematical expression.

If algebra is not your strength, here is something more complete.

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According to one or two webpages, we do not always accept the alternate hypothesis when we reject the null hypothesis.

I am not citing the webpages because their authority and descriptions are unclear. You might do your own research, if it matters to you.

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The expressions for #3 assume that t is calculated the same way as for #2 (my preference).

Alternatively, calculate t as follows:

t = (d - (u1-u2)) / SE

and reject the null hypothesis when:

T.DIST(t,df,TRUE) < significance level