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 99​% confident that his estimate is in error by no more than two percentage points question mark s?​a) Assume that nothing is known about the percentage of computers with new operating systems. n=_______.(Round up to the nearest​ integer.)b) Assume that a recent survey suggests that about 96​% of computers use a new operating system. n=_______.(Round up to the nearest​ integer.)

Answer 1

Answer:

a) $n=\frac{0.5(1-0.5)}{(\frac{0.02}{2.58})^2}=4160.25$  

And rounded up we have that n=4161

b) $n=\frac{0.96(1-0.96)}{(\frac{0.02}{2.58})^2}=639.01$  

And rounded up we have that n=640

Step-by-step explanation:

Part a

The confidence level is 99% , our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$. And the critical value would be given by:

$z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58$

The margin of error for the proportion interval is given by this formula:  

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$    (a)  

And on this case we have that $ME =\pm 0.02$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$   (b)  

Since we don't have prior info for the true proportion we can use $\hat p=0.5$. And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.02}{2.58})^2}=4160.25$  

And rounded up we have that n=4161

Part b

For this case we have a prior estimation ofr the proportion:

$\hat p=0.96$

And replacing we got:

$n=\frac{0.96(1-0.96)}{(\frac{0.02}{2.58})^2}=639.01$  

And rounded up we have that n=640