Question

g 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 99​% confident that his estimate is in error by no more than four percentage points question mark Complete parts​ (a) through​ (c) below. ​a) Assume that nothing is known about the percentage of computers with new operating systems. equals nothing ​(Round up to the nearest​ integer.) ​b) Assume that a recent survey suggests that about 88​% of computers use a new operating system. equals nothing ​(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. ​Yes, using the additional survey information from part​ (b) dramatically increases the sample size. C. ​No, using the additional survey information from part​ (b) does not change the sample size. D. ​No, using the additional survey information from part​ (b) only slightly increases the sample size.

Answer 1

Answer:

a) n= 1045 computers

b) n= 442 computers

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

Step-by-step explanation:

Hello!

The variable of interest is

X: Number of computers that use the new operating system.

You need to find the best sample size to take so that the proportion of computers that use the new operating system can be estimated with a 99% CI and a margin of error no greater than 4%.

The confidence interval for the population proportion is:

p' ± $Z_{1-\alpha /2}$ * $\sqrt{\frac{p'(1-p')}{n} }$

$Z_{1-\alpha /2}= Z_{0.995}= 2.586$

a) In this item there is no known value for the sample proportion (p') when something like this happens, you have to assume the "worst-case scenario" that is, that the proportion of success and failure of the trial are the same, i.e. p'=q'=0.5

The margin of error of the interval is:

d= $Z_{1-\alpha /2}$ * $\sqrt{\frac{p'(1-p')}{n} }$

$\frac{d}{Z_{1-\alpha /2}} = \sqrt{\frac{p'(1-p)}{n} }$

$(\frac{d}{Z_{1-\alpha /2}})^2 = \frac{p'(1-p')}{n}$

$n * (\frac{d}{Z_{1-\alpha /2}})^2 = p'(1-p')$

$n= [p'(1-p')]*(\frac{Z_{1-\alpha /2}}{d} )^2$

$n=[0.5(1-0.5)]*(\frac{2.586}{0.04} )^2= 1044.9056$

n= 1045 computers

b) This time there is a known value for the sample proportion: p'= 0.88, using the same confidence level and required margin of error:

$n= [p'(1-p')]*(\frac{Z_{1-\alpha /2}}{d} )^2$

$n= [0.88*0.12]*(\frac{{2.586}}{0.04})^2= 441.3681$

n= 442 computers

c) The additional information in part b affected the required sample size, it was drastically decreased in comparison with the sample size calculated in a).

I hope it helps!