Question

A survey is given to 300 random SCSU students to determine their opinion of being a “Tobacco Free Campus.” Of the 300 students surveyed, 230 were in favor a tobacco free campus. Find a 95% confidence interval for the proportion of all SCSU students in favor of a tobacco free campus. Interpret the interval in part a.

Find the error bound of the interval in part a. The Dean claims that at least 70% of all students are in favor of a tobacco free campus.

Can you support the Dean’s claim at the 95% confidence level? Justify!

Answer 1

sample proportion, = 0.7667
sample size, n = 300
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.7667 * (1 - 0.7667)/300) = 0.0244

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

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.7667 - 1.96 * 0.0244 , 0.7667 + 1.96 * 0.0244)
CI = (0.719 , 0.815)
Therefore, based on the data provided, the 95% confidence interval for the population proportion is 0.719 < p < 0.815 , which indicates that we are 95% confident that the true population proportion p is contained by the interval (0.719 , 0.815)

Margin of Error, ME = zc * SE
ME = 1.96 * 0.0244
ME = 0.05

yes, it support because confidence interval does not contain 70%