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.

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mtosi17 mtosi17 · Answer 1 · 2019-12-15T15:06:34+01:00

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:

[tex]UL-LL=12[/tex]

[tex](\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[/tex]

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

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

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