its not possible with the information you have provided.
I have these two sets of frequency data for 4 groups. I am trying to show that they are from the same population.
My null hypothesis is that they are the same. My alternate hypothesis is that they are different.
First I did a chi square goodness of fit test where the expected value was the average of the two. The Chi squared value for the data was 0.639 and at for 3 degrees of freedom the critical chi squared value is 7.82. Because the Chi squared value for the data is not above 7.82 I cannot say that they are different.
Although I cannot say that they are different I don't want to conclude that they are the same without knowing the probability of a type 2 error.
Now to show that they are the same I flip my null hypothesis around.
My null hypothesis is that they are different. My alternate hypothesis is that they are the same.
The data will surely support the alternative hypothesis. I do not know of a test which can show the chance of incorrectly rejecting the null hypothesis (which in my original case will be the chance of incorrectly failing to reject it).
Could someone tell me a test which can show this?