mean test within linear model

Question

Hi Folks,

I am hoping there is a test within the Fit Model that can return a “p-value” for observing the data in one’s sample if the population mean were a given value.

When I run a REML model on the data I am working with, I get the following least square means:

What I would like to do is test whether the population mean for each of these levels is zero. I can tell that because the 95% confidence interval does not include zero for all four levels that p > 0.05. However, I am hoping to figure out how much smaller. Obviously p is going to be larger for the “wrinkly NS” level than for the “Ancestor” level.

I have looked through the menus and help screens and can’t find a way to do this. My boss (who generally knows JMP better than I do) suggested that based on the Standard Error, the 95% Confidence Interval, and the Degrees of Freedom from the Fixed Effects Test that I could “Reverse Engineer” the confidence interval.

Essentially, by calculating the factor to multiply the Standard Error by to get a 95% confidence interval with these degrees of freedom, then I should be able to figure out what % confidence interval would just exclude zero, and that would give me the p value I am looking for.

This sounds feasable, but I am hoping there is an easier way. Does anyone have suggestions? Let me know if more information would help.

Thanks a bunch for any help!

Mark_Bailey · Accepted Answer

I think that you can use a t-test here. The numerator is the alternative hypothesis minus the null hypothesis and the denominator is the standard error of the estimate. The sample t ratio for Level = Ancestor is (2.3213706 - 0) / 0.25634312 = 9.0557164. Then determine the probability of a value at least this large from the t cumulative distribution function with the appropriate degrees of freedom, 4.157 in this case. You can compute all of this information by opening a script window (select File > New > Script) and typing in the following code:

p value = 1 - t Distribution( (2.3213706 - 0) / 0.25634312, 4.157 );

I also added the computation of the 95% confidence interval. I get this answer:

Note that the small differences are due to the rounded values in the JMP window. I do not have the full precision results to work with. It makes a big difference when you are out in the tails of the distribution.