A fast-food restaurant has an automated beverage machine that dispenses a set amount of liquid based on a size setting. Suppose that on the medium setting, the standard deviation of the amounts dispensed is Ï=17 mL . The manager of the restaurant plans on taking a sample of n medium drinks to construct 99\% percent confidence interval for the mean amount of liquid dispensed. They want the margin of error to be no more 10 mL

So the answer for this case would be n=20 rounded up to the nearest integer would be the sample required to obatin a margin of erros or 10 ml at 99% of confidence

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

represent the sample mean for the sample  

population mean (variable of interest)

represent the population standard deviation

n represent the sample size

Solution to the problem

The margin of error is given by this formula:

   (a)

And on this case we have that ME =10 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

  (b)

The critical value for 99% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.005;0;1)", and we got , replacing into formula (b) we got:

So the answer for this case would be n=20 rounded up to the nearest integer would be the sample required to obatin a margin of erros or 10 ml at 99% of confidence

d) 48

step-by-step explanation:

16,32,48,64,80,96,112,128,144,160,176,192,208,224,240,256,272,288,304,320,336