Question

A professor wishes to estimate the proportion of college students that have never cheated

during a course. She wishes to estimate the proportion to be within​ 4.15% with​ 96%

confidence. How many teenagers are needed for the​ sample?

A. 613

B. 611                          

C. 612

D. 614

E. Insufficient information to answer the question

Answer 1

Solution :

Given that,

= 0.5 ( assume )

1 - = 1 - 0.5 = 0.5

margin of error = E = 4.15% = 0.0415

At 96% confidence level the z is ,

= 1 - 96% = 1 - 0.96 = 0.04

/ 2 =0.02

Z/2 = Z0.02= 2.05 ( Using z table )

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

= (2.05 / 0.0415)2 * 0.5 * 0.5

= 611

Sample size =611