In order to estimate the mean amount of time computer users spend on the internet each month. how many computer users must be surveyed in order to be 95% confident that your sample mean is within 15 minutes of the population mean? assume that the standard deviation of the population of monthly time spent on the internet is 210 min. what is a major obstable to getting a good estimate of the population mean? use technology to find the estimated minimum required sample size. the minimum sample size required is computer users. g

So the answer for this case would be n=753 rounded up to the nearest integer

The minimum sample size required is 753 computer users

Step-by-step explanation:

1) 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 (variable of interest)

Me=15, margin of error required

Confidence =0.95 or 95%

2) Calculate the sample size

The margin of error is given by this formula:

   (a)

And on this case we have that ME =15 min 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% 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=753 rounded up to the nearest integer

The minimum sample size required is 753 computer users