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18 September, 07:18

A population is known to be normally distributed with a standard deviation of 2.8. (a) Compute the 95% confidence interval on the mean based on the following sample of nine: 8, 9, 10, 13, 14, 16, 17, 20, 21. (b) Now compute the 99% confidence interval using the same data.

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  1. 18 September, 07:27
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    Step-by-step explanation:

    Mean = sum of terms/number of terms

    The mean of the given sample is

    (8 + 9 + 10 + 13 + 14 + 16 + 17 + 20 + 21) / 9 = 14.2

    a)

    For a confidence level of 95%, the corresponding z value is 1.96.

    We will apply the formula

    Confidence interval

    = mean ± z * standard deviation/√n

    Where

    n represents the number of samples

    It becomes

    14.2 ± 1.96 * 2.8/√9

    = 14.2 ± 1.96 * 0.933

    = 14.2 ± 1.83

    The lower end of the confidence interval is 12.37

    The upper end of the confidence interval is 16.03

    b)

    For a confidence level of 99%, the corresponding z value is 2.58

    We will apply the formula

    Confidence interval

    = mean ± z * standard deviation/√n

    Where

    n represents the number of samples

    It becomes

    14.2 ± 2.58 * 2.8/√9

    = 14.2 ± 2.58 * 0.933

    = 14.2 ± 2.4

    The lower end of the confidence interval is 11.8

    The upper end of the confidence interval is 16.6
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