The owner of touchdown sports bar wants to develop a time standard for the task of mixing a specialty cocktail. in a preliminary study, he observed one of his bartenders perform this task seven times with an average of 90 seconds and a standard deviation of five seconds. how many observations should be made if he wants to be 95.44 percent confident that the maximum error in the observed time is one second?

100 observations needed for desired accuracy and confidence. The formula for the confidence of a sampling is: ME = z*d/sqrt(n) where ME = Margin of error z = z score for desired level of confidence d = standard deviation n = number of samples The z score desired is calculated as follows. If you want a 95% confidence, you calculate 1 - 0.95 = 0.05, then you divide the result by 2, getting 0.025, and finally you use a standard normal table to get the z score for the desired probability. So for this problem of 95.44% we get (1 - 0.9544)/2 = 0.0456/2 = 0.0228 Looking up a standard normal table, the value of 0.0228 is found to have a z-score of 2.0, a.k.a. 2 standard deviations from the normal. So let's substitute the known values into the formula and solve for n. ME = z*d/sqrt(n) 1 = 2*5/sqrt(n) sqrt(n) = 2*5 sqrt(n) = 10 n = 100 So the owner needs at least 100 samples to be 95.44% certain that his measurement error is within 1 second of the correct time.