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?
Complete below parts: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) $n=\frac{0.5(1-0.5)}{(\frac{0.02}{2.05})^2}=2627$  

And rounded up we have that n=2627

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, 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)  

The estimated proportion would be 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

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 96% of confidence, our significance level would be given by $\alpha=1-0.96=0.04$ and $\alpha/2 =0.02$. And the critical value would be given by:

$z_{\alpha/2}=-2.05, z_{1-\alpha/2}=2.05$

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

The estimated proportion would be 0.5. And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.02}{2.05})^2}=2627$  

And rounded up we have that n=2627