Question

Exhibit 18-2 Students in statistics classes were asked whether they preferred a 10-minute break or to get out of class 10 minutes early. In a sample of 150 students, 40 preferred a 10-minute break, 80 preferred to get out 10 minutes early, and 30 had no preference. We want to determine if there is a difference in students' preferences. Refer to Exhibit 18-2. The mean and the standard deviation of the sampling distribution of the number of students who preferred to get out early are

Answer 1

Answer:

The mean and the standard deviation of the sampling distribution of the number of students who preferred to get out early are 0.533 and 0.82

Step-by-step explanation:

According to the given data we have the following:

Total sample of students= 150

80 students preferred to get out 10 minutes early

Therefore, the mean of the sampling distribution of the number of students who preferred to get out early is = 80/150 = 0.533

Therefore,  standard deviation of the sampling distribution of the number of students who preferred to get out early= phat - p0/sqrt(p0(1-p)/)

= 0.533-0.5/sqrt(0.5*0.5/15))

= 0.816 = 0.82

Answer 2

Answer:

The mean and standard are 0.533 and 0.82 respectively

Step-by-step explanation:

We are given the following data

Total number of students = 150

Students preferred to get out of 10mins = 80

Our mean of the sampling distribution of the number of students who preferred to get out early will be

= 80÷150

= 0.533

The formula for the standard deviation is as follows

= mean - pⁿ/√(pⁿ(1-p)/)

Substituting the values we have

= 0.533-0.5/√(0.50×0.5/15))

=0.82