Question

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 12 minutes of the population​ mean? Assume that the standard deviation of the population of monthly time spent on the internet is 227 min. What is a major obstacle 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 nothing computer users. ​(Round up to the nearest whole​ number.) What is a major obstacle to getting a good estimate of the population​ mean?
A. It is difficult to precisely measure the amount of time spent on the​ internet, invalidating some data values.
B. The data does not provide information on what the computer users did while on the internet.
C. There may not be 1 comma 3751,375 computer users to survey.
D. There are no obstacles to getting a good esitmate of the population mean.

Answer 1

Answer:

The minimum sample size required is 5,499 computer users.

The major obstacle is that it is a big sample size if it is a small organization wanting to perform it.

Step-by-step explanation:

We need to calculate a sample size in order to have 95% confidence interval with 12 minutes of witdth.

So, we have that the difference between the upper limit and the lower limit is 12:

$UL-LL=12$

$(\mu+z*\sigma/\sqrt{n})-(\mu-z*\sigma/\sqrt{n})=12\\\\2z\sigma/\sqrt{n}=12\\\\\sqrt{n}=(1/6)*z\sigma\\\\n=(1/36)z^2\sigma^2$

For a 95% CI, the value of z is 1.96. The standard deviation is 227.

$n=z^2\sigma^2/36=(1.96)^2*(227)^2/36=3.8416*51,529/36=5498.72\\\\n=5499$

The minimum sample size required is 5,499 computer users.