A recent colleges dri association the student said that they drink alcoholic bev
ID: 3316308 • Letter: A
Question
A recent colleges dri association the student said that they drink alcoholic beverage at least once-a-week. article in a national newspaper claimed that 40% of all students in US nk some alcoholic beverages at least once a week. A students argued that actually fewer students drink alcohol. In support of its claim association has sampled at random 500 students of which 180 a) What is the estimated proportion of students who drink (at least once-a-week)? b) Construct an approximate 95% confidence interval for the true proportion p of An approximate 95% CI for p is from LCL- to UCL= . 4. An experiment was conducted to determine if individuals could be taught how to make decisions rationally and if such training would improve the quality of their career decisions. A group of 40 students were randomly divided into two groups. The experimental group, with n, = 23 students, received instructions in rational decision-making while the control group,n 17 did not. At the end of the instruction period, all subjects completed a multiple-choice test designed to assess the extent to which an individual knows how to apply rational principle in job-decision situations. High score on the test is indicative of application of rational principle in decision- making. The study found that for the experimental group, the average test- score was -65.1 with a sample standard deviation of Si-6.7, whereas for the control group the average was -61.3 with a sample standard deviation of S.-7.1 . Let and denote the mean test-scores for students with training and for students without training (in rational decision-making), respectively. It was assumed that the test-scores distributions of the two groups are approximately normal with the respective means but with equal standard deviations, so that .-o, ,,o a) Formally state the: Null Hypothesis, H,Explanation / Answer
(a) Estimated proportion = p = 180/500 = 0.36
(b) 95% confidence interval = 0.36+- 1.96((0.36)(0.64)/500)0.5
95% CI = (0.3179, 0.4021)