@ddd5119,
You have a binomial distribution (broken or not broken) with a sample size of 40 and a success (not broken) rate of 37. 37/40= =.925 or 92.5% success rate. You'll need to decide what level of confidence you want around that rate before you make a judgment on the entire set of boxes. There are numerous ways to calculate the CI, but the most common one uses the normal approximation to the binomial distribution. If you want the 95% ci around .925 using a normal approximation you would use the formula:
p0 +/- (z of 1-a/2)( sqrt ((p0(1-p0)/n)) where p0 = 0.925 a=1.96 and n=40
The 95% confidence interval of 37/40 is 0.796 - 0.984. This means that you are 95% confident that the number of unbroken items in all of the boxes is between 80% and 98% of the total, or (160)(.8) to (160)(.98) = 128 to 157 unbroken items. You can convert that to broken items by subtraction if you'd rather take that approach.
Without using the CI, what do you think the overall number of broken items would be?