Question

Suppose that a random sample 80 recent graduates of students is taken and it is found that 30 of them participated in an undergraduate research project while at school. a) Find a 99% confidence interval for the population proportion of students who participate in undergraduate research while at school. b) Using the fact that 30 out of the 80 graduates participated in undergraduate research, how large a sample of graduates is needed to arrive at a 99% confidence interval with a margin of error of .10?

Answer 1

Data given is:

Sample size n = 80

Sample proportion p = 30/80 = 0.375

So, Standard error is:

S = (p*(1-p)/n)^0.5 = (0.375*(1-0.375)/80)^0.5 = 0.054

(a)

The 99% Ci is:

p - (2.58*S) < < p + (2.58*S)

0.375 - (2.58*0.054) < < 0.375 + (2.58*0.054)

0.235  < < 0.514

(b)

Margin of error for a 99% CI = 2.58*S

Here, S = (p*(1-p)/n)^0.5

Here we are asked to use the value of p as 0.375

So,

Margin of error for a 99% CI = 0.10 = 2.58*(0.375*(1-0.375)/n)^0.5

Solving we get:

n = 156