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 two percentage points? a)Assume that nothing is known about the percentage of computers with new operating systems.

Answer 1

Answer: 2401

Step-by-step explanation:

Formula to find the sample size is given by :-

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

, where p = prior population proportion.

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

E= Margin of error.

As per given , we have

Confidence level : $1-\alpha=0.95$

⇒$\alpha=1-0.95=0.05$

Two -tailed z-value for $\alpha=0.05 : z_{\alpha/2}=1.96$

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 prior estimate of proportion is unknown.]

Required sample size will be :-

$n= 0.5(1-0.5)(\dfrac{1.96}{0.02})^2\\\\ 0.25(98)^2=2401$

Hence, the number of computer must be surveyed = 2401

Answer 2

Answer:

n = 1067

Step-by-step explanation:

Since nothing is known, we would assume that 50% of the computers use the new operating system.

So, standard error = 0.5/SQRT(n)

Z-value for a 95% CI = 1.9596

So, margin of error = 1.9596 x 0.5 / SQRT(n) = 0.03

So, n = 1067 (approx.)

This will be your approximate answer : n = 1067