Question

A random sample of 1200 Freshmen at UGA were asked if they had applied for student football tickets. The proportion who answered yes was 0.86. The standard error of this estimate is 0.010.

(a) Find the margin of error for a 95% confidence interval for the population proportion who applied for football tickets.
(3 decimal places)

(b) Construct the 95% confidence interval.

Lower Limit	(3 decimal places)
Upper Limit	(3 decimal places)

(c) Since the sample was randomly selected, will this confidence interval be valid?

Yes because the sample size is greater than 30.

Yes because at least one of (1200)*(0.86) and (1200)*(1 - 0.86) is greater than 15.    

Yes because at least one of (1200)*(0.86) and (1200)*(1 - 0.86) is greater than 30.

None of these.

Yes because both (1200)*(0.86) and (1200)*(1 - 0.86) are greater than 15.

(d) Using the above confidence interval, what can you say about the true population proportion?

If we were to take many samples like this, and calculate a 95% confidence interval for each one, approximately 95% of these intervals would contain the true population proportion.

We are 95% confident that the sample proportion is in this interval.    

Answer 1

a)
Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

Margin of error, ME = 1.96 * 0.01 = 0.0196

ME = 0.020

b)
CI = (pcap - z*SE, pcap + z*SE)
CI = (0.86 - 1.96 * 0.01 , 0.86 + 1.96 * 0.01)
CI = (0.84 , 0.88)

lower limit = 0.840
upper limit = 0.880

c)
Yes because both (1200)*(0.86) and (1200)*(1 - 0.86) are greater than 15.

d)
If we were to take many samples like this, and calculate a 95% confidence interval for each one, approximately 95% of these intervals would contain the true population proportio