Question

help!

1. In a recent survey conducted by a professor of UM, 200 students were asked whether or not they have a satisfying experience with the e-learning approach adopted by the school in the current semester. Among the 200 students interviewed, 121 said they have a satisfying experience. What is the 99% confidence interval for the proportion of all UM students who have a satisfying experience with the e-learning approach?

Select one:

a. 0.537 to 0.673

b. 0.524 to 0.686

c. 0.516 to 0.694

d. 0.548 to 0.662

2.

A university bookstore claims that 40% of its customers are satisfied with the service and prices. If this claim is true, what is the probability that in a random sample of 500 customers less than 37% are satisfied?

Select one:

a. 0.9147

b. 0.0853

c. 0.0823

d. 0.9177

Answer 1

Answer Date: 02/06/2020 1) The 99% confidence interval for the proportion of all UM students who have a satisfying experience with the e-learning approach is, First, compute proportion, the z-critical value then find confidence interval. The proportion is, = n 121 200 = 0.605 The Z-critical value is, From the standard normal table, the z-critical value at 99% confidence is 2.576.

P(1-P) 99%C.I.= zcx n = 0.605 + 2.576x 10.605(1 -0.605) 200 = 0.516 to 0.694 The 99% confidence interval for the proportion of all UM students who have a satisfying experience with the e-learning approach is 0.516 and 0.694. The answer is b) The probability that in a random sample of 500 customers less than 37% are satisfied is, First, compute the mean and standard deviation of sampling distribution of sample proportion.

The mean of the sampling distribution of sample proportion is, 4₂ = = = 0.40 The standard deviation of the sampling distribution of sample proportion is, P(1-P) O n (0.40) (1 -0.40) 500 = 0.0216 The probability that in a random sample of 500 customers less than 37% are satisfiedis, First, compute the z score then find probability based on standard normal table.

For p = 0.37 converts to р- из Z = 0.37 -0.40 0.0216 = -1.39 From the standard normal distribution table, the associated probability for the area to the left is shown below, P(z<-1.39) = 0.0823 The probability that in a random sample of 500 customers less than 37% are satisfied is 0.0823. The answer is C.