Question

Question 3 9 pts Academic advising: In 2014 the Community College Survey of Student Engagement reported that 32% of the students surveyed rarely or never use academic advising services. Suppose that in reality, 42% of community college students rarely or never use academic advising services at their college. In a simulation we select random samples from this population. For each sample we calculate the proportion who rarely or never use academic advising services What is the smallest sample in the list below that will satisfy the conditions for the use of a normal model to represent the sampling distribution? D 15 20 25 D30

Answer 1

Answer: The smallest sample in the list below that will satisfy the conditions for the use of a normal model to represent the sampling distribution is n=25 Explanation: From the given information, Let p=0.42 represents the proportion of community college students rarely or never use academic advising services. Following are the conditions that should be satisfied for the use of a normal model to represent the sampling distribution. In case of sampling distribution of sample proportions, the distribution is approximately normal if and only if, np > 10 ng > 10 For sample size 15: np=15x0.42 = 6.3 nq = n(1-P) =15(1-0.42) =15(0.58) = 8.7 Here, both the conditions np and nq were less than 10. For sample size 20: np=20x0.42 = 8.4

nq=n(1-2) = 20(1-0.42) = 200.58) =11.6 Here, np less than 10. For sample size 25: np = 25 x 0.42 = 10.5 nq=n(1-2) = 25(1-0.42) = 25(0.58) = 14.5 Here, both the conditions np and nq were greater than 10. For sample size 30: np = 30x0.42 = 12.6 ng=n(1-p) = 30(1-0.42) =30(0.58) = 17.4 Here, both the conditions np and ng were greater than 10. Therefore, the smallest sample that will satisfy the conditions for the use of a normal model to represent the sampling distribution is n=25.