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18 July, 10:44

A sample of 2,000 union members was selected and asked about their opinions regarding a proposed contract. A total of 1,600 members were in favor of it. A 95% confidence interval estimated that the population proportion was between 0.78 and 0.82. This indicates that about 80 out of 100 similarly constructed intervals would include the population proportion.

True or False

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  1. 18 July, 13:09
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    So Equation (1) and Equation (2) both are not same C. I hence the given statement is false.

    Step-by-step explanation:

    Given:

    total member = 2000

    Favor member=1600

    Mean=1600/2000=0.8

    To Find:

    C. I of 95 % and it is true for given statement

    Solution:

    Now to calculate the

    C. I=0.8±Z*Sqrt[p (1-p) / n]

    C. I=0.8±1.96Sqrt[0.8 (0.2) / 2000]

    C. I=0.8±0.0175

    C. I=0.7825 to 0. 8175 ... Equation (1)

    Now

    Above C. I should be same when n = 100 and x = 80 So p=x/n=80/100=0.8

    Hence

    C. I.=p±Z*Sqrt[p (1-p) / n]

    C. I=0.8±1.96*sqrt[0.8*0.2/100]

    =0.8±1.96*0.04

    =0.8±0.0784

    Hence C. I.=0.7216 to 0.8784 ... Equation (2)

    By comparing

    So Equation (1) and Equation (2) both are not same C. I hence the given statement is false.
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