In order to estimate the mean amount of time computer users spend on the intemet each month, how mary computer users must be surveyed in order to be $95 \%$ confident that your sample mean is in error by no more than 11 minutes? Assume that the standard deviation of the population of monthly time spent on the internet is 222 min. Use technology to find the estimated minimum required sample size. The minimum sample size required is $\square$ computer users (Round ip to the nearest whole number)

Step 1 ：We are given that the standard deviation of the population of monthly time spent on the internet is 222 minutes, the error margin is 11 minutes, and the Z-score for a 95% confidence level is 1.96.

Step 2 ：We can use the formula \(n = (Z*σ/E)^2\) to calculate the sample size, where \(Z\) is the Z-score, \(σ\) is the standard deviation, and \(E\) is the error margin.

Step 3 ：Substituting the given values into the formula, we get \(n = (1.96*222/11)^2\).

Step 4 ：Calculating the above expression, we get \(n = 1565\).

Step 5 ：Since the sample size cannot be a fraction, we round up to the nearest whole number, which gives us 1565.

Step 6 ：So, we need to survey at least 1565 computer users in order to be 95% confident that our sample mean is in error by no more than 11 minutes.

Step 7 ：Final Answer: The minimum sample size required is \(\boxed{1565}\) computer users.