Question

2. Starting salaries. You are planning a survey of starting salaries for recent computer science majors. In the latest survey by the National Association of Colleges and Employers, the average starting salary was reported to be $61,287. If you assume that the standard deviation is $3850, what sample size do you need to have a margin of error equal to $500 with 95% confidence? (report as an integer)

3. Suppose that in the setting of question 2 you have resources to contact 300 recent graduates. If all respond, what would be your margin of error? (report to nearest integer).

4. Suppose that in the setting of question 2 you have resources to contact 300 recent graduates. If only 50% respond, what would be your margin of error? (report to nearest integer).

Answer 1

Solution :

Given that,

Population standard deviation = = 3850

Margin of error = E = 500

2) At 95% confidence level

= 1-0.95% =1-0.95 =0.05

/2 =0.05/ 2= 0.025

Z/2 = Z_{0.025 = 1.960}

Z/2 = 1.960  

sample size = n = [Z_/2* / E] ²

n = [1.96  * 3850 / 500 ]²

n = 228

Sample size = n = 228

3) n = 300

Margin of error = E = Z/2 * ( /n)

=1.96 * ( 3850 /300 )

=435

4) n =300

= 0.50

1 - = 0-050 = 0.50

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.960 * ((0.50*(0.50) /300 )

= 0.05658 = 0