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.

614

B.

612

C.

611                          

D.

613

E.

Insufficient information to answer the question

Answer 1

Solution :

Given that,

= 0.5

1 - = 1 - 0.5 = 0.5

margin of error = E = 4.15% = 0.0415

At 96% confidence level the z is ,

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

= 610.03

Sample size =611