Question

In preparing a report on the​ economy, we need to estimate the percentage of businesses that plan to hire additional employees in the next 60 days. ​a) How many randomly selected employers must we contact in order to create an estimate in which we are 98​% confident with a margin of error of 5​%? ​b) Suppose we want to reduce the margin of error to 3​%. What sample size will​ suffice? ​c) Why might it not be worth the effort to try to get an interval with a margin of error of 1​%?

Answer 1

Answer:

a) $n=\frac{0.5(1-0.5)}{(\frac{0.05}{2.33})^2}=542.89$  

And rounded up we have that n=543

b)  $n=\frac{0.5(1-0.5)}{(\frac{0.03}{2.33})^2}=1508.03$  

And rounded up we have that n=1509

c)  $n=\frac{0.5(1-0.5)}{(\frac{0.01}{2.33})^2}=13572.25$  

And rounded up we have that n=13573

Step-by-step explanation:

For this case since we don't have previous info we can assume that the best estimator for the true proportion is $\hat p =0.5$

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)  

Part a

The significance level is $\alpha=1-0.98 =0.02$The critical value would be for this case:

$z_{\alpha/2}= 2.33$

And on this case we have that $ME =\pm 0.05$ 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)  

And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.05}{2.33})^2}=542.89$  

And rounded up we have that n=543

Part b

$n=\frac{0.5(1-0.5)}{(\frac{0.03}{2.33})^2}=1508.03$  

And rounded up we have that n=1509

Part c

$n=\frac{0.5(1-0.5)}{(\frac{0.01}{2.33})^2}=13572.25$  

And rounded up we have that n=13573