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3 years ago

11

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.

2 answers:

lawyer [7]3 years ago

8 0

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

Digiron [165]3 years ago

4 0

<em>Answer:</em>

<em>n = 1067</em>

<em>Step-by-step explanation:</em>

<em />

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

<em>So, standard error = 0.5/SQRT(n) </em>

<em>Z-value for a 95% CI = 1.9596 </em>

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

<em>So, n = 1067 (approx.)</em>

<em />

<em>This will be your approximate answer : n = 1067</em>

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