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

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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

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Maya 2 weeks 2021-09-15T04:41:19+00:00 2 Answers 0

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  1. Charlotte
    0
    2021-09-15T04:42:25+00:00

    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

  2. Charlotte
    0
    2021-09-15T04:42:58+00:00

    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

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