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

`n=(0.5(1-0.5))/(((0.03)/(1.96))^2)=1067.11`

And rounded up we have that n=1068

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{(\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=(\hat p (1-\hat p))/(((ME)/(z))^2)` (b)

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

`n=(0.5(1-0.5))/(((0.03)/(1.96))^2)=1067.11`

And rounded up we have that n=1068