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 two percentage points

Answer 1

Answer:

The number of computers that must be surveyed is 1681.

Step-by-step explanation:

The formula to find the sample size is given by:-

$n=p(1-p)(\frac{\frac{z_{\alpha } }{2} }{E} )^{2}$

where p = prior population proportion,

$\frac{z_{\alpha }}{2}$= Two-tailed z-value for ∝

E= Margin of error.

As per the given, we have

Confidence level :

1 - ∝ = 0.90

⇒∝= 1 - 0.90= 0.10

Two -tailed z-value for ∝= 0.10 : $\frac{z_{\alpha }}{2}$ = 1.64

E= 2%=0.02

We assume that nothing is known about the percentage of computers with new operating systems.

Let us take p=0.5  [we take p= 0.5 if the prior estimate of proportion is unknown.]

The required sample size will be:-

$n= 0.5(1-0.5)(\frac{1.64}{0.02} )^{2}$

=> $0.25(82)^{2}$

=> 1681

Hence, the number of the computer must be surveyed = 1681

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