Question

You have been assigned to determine whether more people prefer Coke or Pepsi. Assume that roughly half the population prefers Coke and half prefers Pepsi. How large a sample do you need to take to ensure that you can estimate, with 95% confidence, the proportion of people preferring Coke within 3% of the actual value? [Hint: proportion est. = 0.5] Round your answer to whole number

Answer 1

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11$  

And rounded up we have that n=1068

Step-by-step explanation:

For this case we have the following info given:

$ME=0.03$ the margin of error desired

$Conf= 0.95$ the level of confidence given

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)  

the critical value for 95% of confidence is $z=1.96$

We can use as estimator for the population of interest $\hat p=0.5$. And on this case we have that $ME =\pm 0.03$ 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:

$n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11$  

And rounded up we have that n=1068