Question

For any Hypothesis Test make sure to state Ho, Ha, Test statistic, p-value, whether you reject Ho, and your conclusion in the words of the claim. For any confidence interval make sure that you interpret the interval in context, in addition to using it for inference.

Round to the thousandths place

A survey is given to 300 random SCSU students to determine their opinion of being a “Tobacco Free Campus.” Of the 300 students surveyed, 228 were in favor a tobacco free campus.

1. Find a 95% confidence interval for the proportion of all SCSU students in favor of a tobacco free campus.
2. Interpret the interval in part a.
3. Find the error bound of the interval in part a.
4. 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, = 0.76
sample size, n = 300
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.76 * (1 - 0.76)/300) = 0.02466

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 = (0.76 - 1.96 * 0.02466 , 0.76 + 1.96 * 0.02466)
CI = (0.7117 , 0.8083)

we are 95% confident that the proportion of all SCSU students in favor of a tobacco free campus is between 0.7117 and 0.8083

Margin of Error, ME = zc * SE
ME = 1.96 * 0.02466
ME = 0.048

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p = 0.7
Alternative Hypothesis, Ha: p > 0.7

Test statistic,
z = (pcap - p)/sqrt(p*(1-p)/n)
z = (0.76 - 0.7)/sqrt(0.7*(1-0.7)/300)
z = 2.27

P-value Approach
P-value = 0.0116
As P-value < 0.05, reject the null hypothesis.

There is sufficient evidence to conclude that at least 70% of all students are in favor of a tobacco free campus