How many observations should a time study analyst plan for in an operation that has a standard deviation of .9 minutes per piece if the goal is to estimate the mean time per piece to within .1 minute with a confidence of 95.5 percent

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)

s represent the sample 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 =0.1 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 95.5% of confidence interval now can be founded using the normal distribution. And in excel we can use this formula to find it:"=-NORM.INV(0.023;0;1)", and we got , replacing into formula (b) we got:

So the answer for this case would be n=4 rounded up to the nearest integer

answer: 2 litters juice bottle

half of a gallon is 50% of a full gallon 2 litters of juice is 52% of a full gallon

hope this ^_^