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 95​% confident that his estimate is in error by no more than three percentage points question mark?
​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

Answer:  a. n= 1068

b. n= 164

Step-by-step explanation:

The formula to find the sample size :

$n=p(1-p)(\dfrac{z^{*}}{E})^2$

, where p=prior population proportion , z* = critical z-value and E = Margin of error.

Here , let p=proportion of computers that use a new operating system.

Given : Confidence level = 95%

i.e. z* = 1.96  [by z-table]

Margin of error : E = 3% =0.03

a. If p is unknown , then we assume p=0.5

Then, $n=(0.5)(1-0.5)(\dfrac{1.96}{0.03})^2=1067.11111\approx1068$

i.e. n= 1068

b. p=0.96

Then, $n=(0.96)(1-0.96)(\dfrac{1.96}{0.03})^2=163.908266667\approx164$

i.e. n= 164.