Question

A programmer plans to develop a new software system. In planning for the operating system that he will​ use, he needs to estimate the percentage of computers that use a new operating system. How many computers must be surveyed in order to be 99​% confident that his estimate is in error by no more than two percentage points question mark s?​a) Assume that nothing is known about the percentage of computers with new operating systems. n=_______.(Round up to the nearest​ integer.)b) Assume that a recent survey suggests that about 96​% of computers use a new operating system. n=_______.(Round up to the nearest​ integer.)

Answer 1

a)
`n=(0.5(1-0.5))/(((0.02)/(2.58))^2)=4160.25`

And rounded up we have that n=4161

b)
`n=(0.96(1-0.96))/(((0.02)/(2.58))^2)=639.01`

And rounded up we have that n=640

Explanation:

Part a

The confidence level is 99% , our significance level would be given by
`\alpha=1-0.99=0.01` and
`\alpha/2 =0.005`. And the critical value would be given by:

`z_(\alpha/2)=-2.58, z_(1-\alpha/2)=2.58`

The margin of error for the proportion interval is given by this formula:

`ME=z_(\alpha/2)\sqrt{(\hat p (1-\hat p))/(n)}` (a)

And on this case we have that
`ME =\pm 0.02` and we are interested in order to find the value of n, if we solve n from equation (a) we got:

`n=(\hat p (1-\hat p))/(((ME)/(z))^2)` (b)

Since we don't have prior info for the true proportion we can use
`\hat p=0.5`. And replacing into equation (b) the values from part a we got:

`n=(0.5(1-0.5))/(((0.02)/(2.58))^2)=4160.25`

And rounded up we have that n=4161

Part b

For this case we have a prior estimation ofr the proportion:

`\hat p=0.96`

And replacing we got:

`n=(0.96(1-0.96))/(((0.02)/(2.58))^2)=639.01`

And rounded up we have that n=640