@estherw1,
Not sure why you would use 90% instead of the typical 95% that is routinely used in science but here is how you answer your question. Let's say you take 10 measurements and get a mean of 9 and a standard deviation of 1. The confidence interval around that value assuming a normal distribution is:
9 +/- (Z)(SD)/sqrt(N)
Where SD is the standard deviation (1 in this case), N is the number of samples (10 in this case) and Z is a value depending on the confidence interval you want. 90% is 1.645, 95% is 1.96. You can find these numbers on the Internet by Googling something like "
90% z score"
So in this case you get 9+/- 1.645/sqrt(10) = 9 +/- 0.52.
If you were expecting 10, you have shown that the answer is not 10. If you were expecting 9.3, you have not disproved your hypothesis. Note that you have not proved the average is 9.3, after all you measured 9, but you have not disproved the true average is 9.3 (or 8.7 or anything within .52 of 9).