Question

12. If a manufacturer conducted a survey among randomly selected target market households and wanted to be 95% confident that the difference between the sample estimate and the actual market share for its new product was no more than 3%, what sample size would be needed?\

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:

We have the following info given:

$Confidence= 0.95$ the confidence level desired

$ME =0.03$ represent the margin of error desired

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 confidence level is 95% or 0.95, the significance is $\alpha=0.05$ and the critical value for this case using the normal standard distribution would be $z_{\alpha/2}=1.96$

Since we don't have prior information we can use $\hat p= 0.5$ as an unbiased estimator

Also we know 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