Suppose the investigators had made a rough guess of 330 for the value of s before collecting data.
What sample size would be necessary to obtain an interval width of 50 ml for a confidence level of 95%?

So the answer for this case would be n=670 rounded up to the nearest integer

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 sicne we want a confidence interval with a wisth of 50 ml then we have that ME =50/2=25 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

  (b)

We can use an estimator of the population deviation the sample deviation

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got , replacing into formula (b) we got:

So the answer for this case would be n=670 rounded up to the nearest integer

is there more to this question?