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 three percentage points question mark Complete parts​ (a) through​ (c) below. ​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 97​% of computers use a new operating system. n= ​(Round up to the nearest​ integer.) ​c) Does the additional survey information from part​ (b) have much of an effect on the sample size that is​ required?

Answer 1

Answer to the question)

Given :

Confidence level = 99% for which Z = 2.58

Margin of error ( e ) = 3% ~0.03

.

Part a)

Percent ( P) is unknown

thus we consider the default value of P = 0.05

.

The formula of sample size is :

n = p*(1-p) * (Z/e)^2

.

on plugging the values we get:

n = 0.5*0.5*(2.58/0.03)^2

n = 1849

.

Part b)
given : p = 0.97

n = 0.97*0.03*(2.58/0.03)^2

n = 215.22 ~ 216

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Part c)

On comparing when there is no information about the percent one needs to get a sample of 1849 units, which is very large sample needed

But once there is information about the percent p = 0.97, a sample of 216 units is sufficient to maintain the error up to 3%.

Thus prior information, helps to go for a smaller sample and maintain the error as well