Question

Tom Nook would like to launch an investigation to see what proportion of fish caught are horse mackerals. What sample size would be required to do this with 95% confidence to within 0.03?

Answer 1

Solution :

Given that,

= 0.5

1 - = 1 - 0.5 = 0.5

margin of error = E =0.03

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96 ( Using z table )

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

= (1.96 / 0.03)2 * 0.5* 0.5

= 1067.111

Sample size =1067