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find the 90% confidence interval for the variance and standard deviation of a sample of 16 and a standard deviation of 2.1 Assume the variable is normally distributed.

  • Statistics -

    There are different ways you can do this kind of problem, but the formula below might be one of the easier ways:

    s/[1 + (1.645/√2n)]
    s/[1 - (1.645/√2n)]

    ...where s = standard deviation, 1.645 represents the 90% confidence interval using a z-table, and n = sample size.

    Substituting your values:
    (2.1)/[1 + (1.645/√2*16)]
    (2.1)/[1 - (1.645/√2*16)]

    Note: After you finish the above calculations, square the standard deviation values to find the variance values.

    I hope this will help.

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