Solved – Mean comparison for statistical significance

meanmultiple-comparisonsstandard deviation

I was told to come here by a friend to ask a stats question.

I took Social Stats and Methods of Social Research about 4 years ago and to be honest, I don't remember much about it. Now I'm writing a research paper and I can't find any basic tutorials to assist with a problem that would probably be pretty simple to a Stats major.
So I'm running tests on the lunar effect, and I guess the qualitative details aren't important.

So In "Group F" I have a sample size of 44 Days, and when measuring incidence within the days, the total was Σ=1593, which generates a mean of 36.20 per day on average.
In "Group J" I have a sample size of 27 Days, and when measuring incidence within the days, the total was Σ=997, which generates a mean of 36.93 per day on average.
I'm trying to see if there is a statistically significant difference between the means comparing Group F and Group J.

I guess CL=95% or α=0.05 (I'm not sure) and I don't remember anything about standard deviation.

So my question is, what is my step by step process to figure this out (becuase i have to do it 9 more times) and how do I find out my standard deviation and CL/α?

Edit) Greg Snow, I'll answer the questions first.

Yes, I have each individual number for each day. So tell me if this is how you find the standard deviation. 1) Subtract the mean from each numerical result from each day. 2) Square all of those values. 3) Add the results 4) Take the square root of that total… That's correct, right?

There where certainly no 0's. lowest number I ever saw was something in the high 20's (I don't have the numbers in front of me). Highest was around low 70's (those maxes an mins are considering both groups)…

No. Each day is independent and isn't dependent on days before or after.

The days don't overlap. The number of days is more related to their frequency of occurrence (full moon nights vs. non-full moon nights). I guess you could call full moon days predetermined lol. But the non-full moon days are the 5th, 10th, 15th, 20th, and 25th day after the full moon (using a 3 full moon day method).

That should answer your questions

Ok I see. thank you for your help! However, I have a question. What's the difference between dividing the variance by N or N-1? I know it's Population vs. Sample, but how do you know which to use?

Best Answer

If both of those samples look normal then you could use a 2 sample t-test. This tests the null hypothesis that there is no difference between the means of your two samples. It seems reasonable to assume the variance of each population is the same, but the sample sizes are slightly different.

$$t = \frac{\overline{X}_1 - \overline{X}_2}{S_{X_1X_2} \cdot \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$$


$$S_{X_1X_2} = \sqrt{\frac{(n_1-1)S_{X_1}^2+(n_2-1)S_{X_2}^2}{n_1+n_2-2}}$$

$\overline{X}_1$ and $\overline{X}_2$ are the sample means for each group $S_{X_1,X_2}$ is the pooled standard deviation, $n_1$ and $n_2$ are the sample sizes of each group, and $S_{X_1}$ and $S_{X_2}$ are the sample standard deviations of each group.

You can do this in R using the function t.test, which has a number of options, including the ability to return confidence intervals and such.