In: Statistics and Probability
Referring back to the scenario in the previous few questions, suppose the app company wanted to make a 99% confidence interval for the true satisfaction rate with a margin of error of 2%.
How big a sample would they need to take?
4147.36
3382
3381.42
4148
Solution :
Given that,
= 0.5
1 - = 1 - 0.5 = 0.5
margin of error = E = 2% = 0.02
At 99% confidence level the z is,
= 1 - 99%
= 1 - 0.99 = 0.01
/2 = 0.005
Z/2 = 2.576 ( Using z table )
Sample size = n = (Z/2 / E)2 * * (1 - )
= (2.576 / 0.02)2 * 0.5 * 0.5
=4147.36
Sample size = 4148 rounded