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 90?% confident that his estimate is in error by no more than four percentage points? B) Assume that a recent survey suggests that about 91?% of computers use a new operating system

Answer 1

Let p be the true proportion of computers that use the new operating system.

First we find the z value corresponding to 90% confiedence interval. The value of total area under 2 tails is

The area under right tail is 0.1/2=0.05. The value of z is

Using standard normal table we get

We want that the estimate is in error by no more than four percentage points. That is the margin of error is +-0.04

If

is the standard error of sample proportion, the error can be written as

Since we do not know the value of p, we can get the mostconservative estimate of the sample size (largest sample size) when p=0.5

the value of n is

Hence 421 computers must be surveyed in order to be 90?% confident that his estimate is in error by no more than four percentage points

B) Suppose we have an estimate of the population proportion as

Now the estimate of n is

138 computers must be surveyed in order to be 90?% confident that his estimate is in error by no more than four percentage points