Question

A survey of shoppers is planned to determine what percentage use credit cards. prior surveys suggest​ 63% of shoppers use credit cards. how many randomly selected shoppers must we survey in order to estimate the proportion of shoppers who use credit cards to within​ 4% with​ 95% confidence?

Answer 1

We would need a sample size of 560.

We first calculate the z-score associated. with this level of confidence:
Convert 95% to a decimal:  95% = 95/100 = 0.95
Subtract from 1:  1-0.95 = 0.05
Divide by 2:  0.05/2 = 0.025
Subtract from 1:  1-0.025 = 0.975

Using a z-table (http://www.z-table.com) we see that this is associated with a z-score of 1.96.

The margin of error, ME, is given by:
$ME=z*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

We want ME to be 4%; 4% = 4/100 = 0.04.  Substituting this into our equation, as well as our proportion and z-score,
$0.04=1.96\sqrt{\frac{0.63(1-0.63)}{n}} \\ \\\text{Dividing both sides by 1.96,} \\ \\\frac{0.04}{1.96}=\sqrt{\frac{0.63(0.37)}{n}} \\ \\\text{Squaring both sides,} \\ \\(\frac{0.04}{1.96})^2=\frac{0.63(0.37)}{n} \\ \\\text{Multiplying both sides by n,} \\ \\n(\frac{0.04}{1.96})^2=0.63(0.37) \\ \\n(\frac{0.04}{1.96})^2=0.2331 \\ \\\text{Isolating n,} \\ \\n=\frac{0.2331}{(\frac{0.04}{1.96})^2}=559.67\approx560$