Final answer:
a) The probability that fewer than 120 students in the sample were in favor of a new cafe is 0.9999 or 99.99%. b) i. The 95% confidence interval for the true proportion of students who are in favor of a new cafe is (0.618, 0.782). b) ii. We cannot conclude that more than 70% of the students are in favor of a new cafe based on the 95% confidence interval.
Step-by-step explanation:
a) To answer this question, we will use the binomial distribution and the cumulative probability function. Given that the administration believes that 70% of the students are in favor of a new cafe, we can assume that the population proportion, p, is 0.7. We want to find the probability that fewer than 120 students in the sample were in favor of a new cafe. Using the binomial distribution formula, P(X
b) i. To construct a 95% confidence interval for the true proportion of students who are in favor of a new cafe, we can use the formula: CI = p ± z * √(p(1-p)/n), where CI is the confidence interval, p is the sample proportion, z is the z-score corresponding to the desired confidence level (in this case, 95%), and n is the sample size. Plugging in the values, we have CI = 130/200 ± 1.96 * √((130/200)(1-130/200)/200). Calculating this, we find that the lower bound of the confidence interval is approximately 0.618 and the upper bound is approximately 0.782. Therefore, the 95% confidence interval for the true proportion of students who are in favor of a new cafe is (0.618, 0.782).
b) ii. Based on the 95% confidence interval, we can determine whether we can conclude that more than 70% of the students are in favour of a new cafe. If the lower bound of the confidence interval is greater than 70%, then we can conclude that more than 70% of the students are in favour of a new cafe. In this case, the lower bound of the confidence interval is 0.618, which is less than 70%. Therefore, we cannot conclude that more than 70% of the students are in favour of a new cafe.