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?\

And rounded up we have that n=1068

Step-by-step explanation:

We have the following info given:

the confidence level desired

represent the margin of error desired

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

   (a)  

The confidence level is 95% or 0.95, the significance is and the critical value for this case using the normal standard distribution would be

Since we don't have prior information we can use as an unbiased estimator

Also we know that and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

  (b)  

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

And rounded up we have that n=1068

im like 90% sure its a meme

step-by-step explanation:

good luck tho