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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 ​% confident that his estimate is in error by no more than percentage point Complete parts​ (a) through​ (c) below.


A) Assume nothing is known about the percentage of computers with new operating systems

n =
round up to the nearest integer

b) Assume that the recent survey suggest that about 96% of computers use a 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?

A.

​Yes, using the additional survey information from part​ (b) dramatically reduces the sample size.

B.

​No, using the additional survey information from part​ (b) does not change the sample size.

C.

​Yes, using the additional survey information from part​ (b) dramatically increases the sample size.

D.

​No, using the additional survey information from part​ (b) only slightly increases the sample size.

Sagot :

Using the z-distribution, we have that:

a) A sample of 601 is needed.

b) A sample of 93 is needed.

c) A.  ​Yes, using the additional survey information from part​ (b) dramatically reduces the sample size.

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]\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 z-score that has a p-value of [tex]\frac{1+\alpha}{2}[/tex].

The margin of error is:

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

95% confidence level, hence[tex]\alpha = 0.95[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], so [tex]z = 1.96[/tex].

For this problem, we consider that we want it to be within 4%.

Item a:

  • The sample size is n for which M = 0.04.
  • There is no estimate, hence [tex]\pi = 0.5[/tex]

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

[tex]0.04 = 1.96\sqrt{\frac{0.5(0.5)}{n}}[/tex]

[tex]0.04\sqrt{n} = 1.96\sqrt{0.5(0.5)}[/tex]

[tex]\sqrt{n} = \frac{1.96\sqrt{0.5(0.5)}}{0.04}[/tex]

[tex](\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.5(0.5)}}{0.04}\right)^2[/tex]

[tex]n = 600.25[/tex]

Rounding up:

A sample of 601 is needed.

Item b:

The estimate is [tex]\pi = 0.96[/tex], hence:

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

[tex]0.04 = 1.96\sqrt{\frac{0.96(0.04)}{n}}[/tex]

[tex]0.04\sqrt{n} = 1.96\sqrt{0.96(0.04)}[/tex]

[tex]\sqrt{n} = \frac{1.96\sqrt{0.96(0.04)}}{0.04}[/tex]

[tex](\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.96(0.04)}}{0.04}\right)^2[/tex]

[tex]n = 92.2[/tex]

Rounding up:

A sample of 93 is needed.

Item c:

The closer the estimate is to [tex]\pi = 0.5[/tex], the larger the sample size needed, hence, the correct option is A.

For more on the z-distribution, you can check brainly.com/question/25404151  

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