Question

A federal lobbyist wishes to survey residents of a certain congressional district to see what proportion of the electorate is aware of that congressperson’s position on using state funds to pay for non-college level courses at a state college. a) What sample size is necessary if the 95% confidence interval for p is to have a width of at most .10? b) If the lobbyist has a strong reason to believe that at least two thirds of the electorate know of the position, how large a sample size would you recommend?

Answer 1

Answer:

a. 385

b. 342

Step-by-step explanation:

To solve the problem we do the following:

The equation to calculate the sample size is:

$n = (p)(q)(\frac{z}{E})^2$

a)

We have the following data:

Since no estimate of proportion is given, we will assume: p = q = 0.5

We know that For 95% confidence, z = 1.96

Width = 0.10

Hence, the margin of error would be:

0.10 / 2 = 0.05

E = 0.05

And so we can calculate sample size:

$n = (0.5)(0.5)(\frac{1.96}{0.05})^2$

n = 385  

b)

We have the following data for this point:

p = 2/3

q = 1 - p

q = 1 - 2/3

q = 1/3

And so we can calculate sample size:

$n = (2/3)(1/3)(\frac{1.96}{0.05})^2$

n = 342