Answer:
The formula for a confidence interval for a proportion is:
^p ± z*√(^p(1-^p)/n)
where ^p is the sample proportion, z is the z-score corresponding to the desired confidence level, n is the sample size, and √(^p(1-^p)/n) is the standard error of the sample proportion.
Using a 95% confidence level, the z-score is 1.96 (from the standard normal distribution). Substituting the given values, we get:
0.68 ± 1.96*√(0.68(1-0.68)/50)
which simplifies to:
(0.5549, 0.8151)
Therefore, the 95% confidence interval for the proportion of OSU students who love their math class is (0.5549, 0.8151).
To ensure the relevant confidence interval conditions are met, we need to verify that the sample size is large enough to assume the sampling distribution of the sample proportion is approximately normal. The condition for this is that both np and n(1-p) are greater than or equal to 10, where n is the sample size and p is the proportion of interest. In this case, np = 500.68 = 34 and n(1-p) = 500.32 = 16, so both conditions are met.