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 one percentage point?

a. Assume that nothing is known about the percentage of computers with new operating systems. n = ?
b. Assume that a recent survey suggests that 99% of computers use a new operating system. n = ?

Respuesta :

Answer:

a) [tex]n = 9604[/tex]

b) n = 381

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

The margin of error is:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

90% confidence level

So [tex]\alpha = 0.1[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].

a. Assume that nothing is known about the percentage of computers with new operating systems. n = ?

When we do not know the proportion, we use [tex]\pi = 0.5[/tex], which is when we are going to need the largest sample size.

The sample size is n when M = 0.01.

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

[tex]0.01 = 1.96\sqrt{\frac{0.5*0.5}{n}}[/tex]

[tex]0.01\sqrt{n} = 1.96*0.5[/tex]

[tex]\sqrt{n} = \frac{1.96*0.5}{0.01}[/tex]

[tex](\sqrt{n})^{2} = (\frac{1.96*0.5}{0.01})^{2}[/tex]

[tex]n = 9604[/tex]

b. Assume that a recent survey suggests that 99% of computers use a new operating system. n = ?

Now we have that [tex]\pi = 0.99[/tex]. So

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

[tex]0.01 = 1.96\sqrt{\frac{0.99*0.01}{n}}[/tex]

[tex]0.01\sqrt{n} = 1.96*\sqrt{0.99*0.01}[/tex]

[tex]\sqrt{n} = \frac{1.96*\sqrt{0.99*0.01}}{0.01}[/tex]

[tex](\sqrt{n})^{2} = (\frac{1.96*\sqrt{0.99*0.01}}{0.01})^{2}[/tex]

[tex]n = 380.3[/tex]

Rouding up

n = 381

Based on the information given, the number of computers that must be surveyed in order to be 90% confident assume that nothing is known about the percentage of computers will be 423.

When nothing is known about the percentage of computers with new operating systems, the number of computers will be:

= [1.645² × 0.5 × (1 - 0.5)] / 0.04²

= 423

Assume that a recent survey suggests that 99% of computers use a new operating system, the number of computers will be:

= [1.645² × 0.90 × (1 - 0.90)] / 0.04²

= 152

Learn more about percentage on:

https://brainly.com/question/24304697

ACCESS MORE
EDU ACCESS