Question

A sample of 50 individuals are surveyed, and it is found that 10 of them regularly take zinc supplements. How many individuals should we survey so that, with 94% confidence, the true proportion taking zinc can be estimated to within 3 percentage points? (Note: 3 percentage points is 0.03 since we write a proportion as lying between 0 and 1.)

Answer 1

Solution :

Given that,

= 0.03

1 - = 1 - 0.03 = 0.97

margin of error = E = 3% = 0.03

At 94% confidence level the z is ,

Z/2 = 1.881 ( Using z table )

Sample size = n = (Z/2 / E)2 * * (1 - )

= (1.881 / 0.03)2 * 0.03 * 0.97

= 144.4

Sample size =144