Question

7. A sociologist wishes to conduct a poll to estimate the percentage of Americans who favor affirmative action program for women and minorities for admission to colleges and universities. What sample size should be obtained if he wishes the estimate to be within 0.032 with 98% confidence if; a. He uses a 2008 estimate of 63.5% ? b. He does not use any prior estimate?

Answer 1

Solution :

Given that,

= 0.635

1 - = 1 - 0.635 = 0.365

margin of error = E = 0.032

At 98% confidence level the z is,

= 1 - 98%

= 1 - 0.98 = 0.02

/2 = 0.01

Z/2 = 2.326 ( Using z table )

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

= (2.326 / 0.032)2 * 0.635 * 0.365

=1224.57

Sample size =1225

(B)

Solution :

Given that,

= 0.5

1 - = 1 - 0.5 = 0.5

margin of error = E = 0.032

At 98% confidence level the z is,

= 1 - 98%

= 1 - 0.98 = 0.02

/2 = 0.01

Z/2 = 2.326 ( Using z table )

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

= (2.326 / 0.032)2 * 0.5 * 0.5

=1320.86

Sample size =1321