Question

Suppose a seed distributor is interested in estimating the proportion of a certain type of beet seeds that will germinate. How large of a sample size is required if they wanted to estimate this proportion with a 88% confidence interval and a margin of error of 0. 03? Assume that the distributor does not have any prior knowledge of this proportion.

Answer 1

Solution :

Given that,

= 0.5 ( use 0.5)

1 - = 1 - 0.5 = 0.5

margin of error = E = 0.03

At 88% confidence level the z is ,

= 1 - 88% = 1 - 0.88= 0.12

/ 2 = 0.06

Z/2 = Z0.06 = 1.55( Using z table )

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

= (1.55 / 0.03)2 * 0.5 * 0.5

= 667.36

Sample size =668