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7 June, 20:14

Suppose a sample of 1475 floppy disks is drawn. Of these disks, 1328 were not defective. Using the data, construct the 80%confidence interval for the population proportion of disks which are defective. Round your answers to three decimal places.

Lower:

Upper:

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  1. 7 June, 20:41
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    The 80% confidence interval estimate of the true population proportion of disks which are defective is 0.100 + / - 0.010

    = (0.090, 0.110)

    Lower: 0.090

    Upper: 0.110

    Step-by-step explanation:

    Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

    The confidence interval of a statistical data can be written as.

    p+/-z√ (p (1-p) / n)

    Given that;

    Proportion of disks which are not defective p' = 1328/1475 = 0.900

    Proportion of disks which are defective p = 1 - p' = 1 - 0.900 = 0.100

    p = 0.100

    Number of samples n = 1475

    Confidence interval = 80%

    z value (at 80% confidence) = 1.28

    Substituting the values we have;

    0.100 + / - 1.28√ (0.100 (1-0.100) / 1475)

    0.100 + / - 1.28 (0.007811334658)

    0.100 + / - 0.009998508363

    0.100 + / - 0.010

    = (0.090, 0.110)

    The 80% confidence interval estimate of the true population proportion of disks which are defective is 0.100 + / - 0.010

    = (0.090, 0.110)

    Lower: 0.090

    Upper: 0.110
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