Question 935140
since you bought a box of 10 mints, then your sample size is 10.


your population mean is 10.


your population standard deviation is 3.


your sample mean is 95/10 = 9.5.


your standard error is equal to 3/sqrt(10) = .94868 rounded to 5 decimal places.


the definition of standard error is that it is the standard deviation of the distribution of sample means.


your z-score is equal to (x-m)/s, where x = 9.5 and m = 10 and s = .94868.


x is the mean of your sample, m is the mean of the population, s is the stnadard error.


this makes your z-score equal to -.53 rounded to 2 decimal places.


the area to the left of a z-score of -.53 is equal to .29805.


this is well above alpha at .95 confidence level or .90 confidence level.


the 2 tail alphas for those are:


for .95 confidence level, alpha = (1-.95)/2 = .025


for .90 confidence level, alpha = (1-.90)/2 = .05


your alpha is much higher than either of those, so your confidence level that the box came from the factory is much higher than .95.


this means that your confidence level that the box did not come from the same factory is much lower than .05.  in other words, you have very little confidence that the box did not come from the factory.


you can safely assume that the box of chocolates did come from the same factory.


the difference between the mean of the sample and the mean of the population is well within the normal confidence interval limits.


at 90% confidence limit (more restrictive, i.e. smaller than 95% confidence limit), the confidence interval would be calculated as follows:


critical z-factor = 1.645


margin of error = standard error * critical z factor = .94868 * 1.645 = 1.56 rounded to 2 decimal places.


your confidence interval is equal to population mean plus or minus margin of error which is equal to 10 plus or minus 1.56 which is equal to between 8.44 and 11.56 grams.


the mean of your box of mints at 9.5 grams is well within those limits.