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 ​% 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.

Answer 1

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 $\pi$, and a confidence level of $\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which z is the z-score that has a p-value of $\frac{1+\alpha}{2}$.

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

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

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 $\pi = 0.5$

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.04 = 1.96\sqrt{\frac{0.5(0.5)}{n}}$

$0.04\sqrt{n} = 1.96\sqrt{0.5(0.5)}$

$\sqrt{n} = \frac{1.96\sqrt{0.5(0.5)}}{0.04}$

$(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.5(0.5)}}{0.04}\right)^2$

$n = 600.25$

Rounding up:

A sample of 601 is needed.

Item b:

The estimate is $\pi = 0.96$, hence:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.04 = 1.96\sqrt{\frac{0.96(0.04)}{n}}$

$0.04\sqrt{n} = 1.96\sqrt{0.96(0.04)}$

$\sqrt{n} = \frac{1.96\sqrt{0.96(0.04)}}{0.04}$

$(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.96(0.04)}}{0.04}\right)^2$

$n = 92.2$

Rounding up:

A sample of 93 is needed.

Item c:

The closer the estimate is to $\pi = 0.5$, 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