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Math Help - Unbiased Mean and Variance problem

  1. #1
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    Unbiased Mean and Variance problem

    Hello~!! I'm new to this forum

    Please show me how to do this little problem (from my A-level CIE stats book)-

    Question:

    Unbiased estimates of the mean and variance of a population, based on a random sample of 24 observations, are 5.5 and 2.42 respectively. Another random observation of 8.0 is obtained. Find new unbiased estimates of the mean and variance with this new information.

    (the textbook says "8.0", but I think it should just be 8, but not too sure)

    (the answer is Mean: 5.6 and Unbiased Variance: 2.569)
    Thanks in advance!!
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  2. #2
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    Quote Originally Posted by agbrianlee355 View Post
    Hello~!! I'm new to this forum

    Please show me how to do this little problem (from my A-level CIE stats book)-

    Question:

    Unbiased estimates of the mean and variance of a population, based on a random sample of 24 observations, are 5.5 and 2.42 respectively. Another random observation of 8.0 is obtained. Find new unbiased estimates of the mean and variance with this new information.

    (the textbook says "8.0", but I think it should just be 8, but not too sure)

    (the answer is Mean: 5.6 and Unbiased Variance: 2.569)
    Thanks in advance!!
    Use \bar{X} = \frac{\sum_{i=1}^n x_i}{n} as the unbiassed estimator of the mean.

    When n = 24: 5.5 = \frac{\sum_{i=1}^{24} x_i}{24} \Rightarrow \sum_{i=1}^{24} x_i = (24)(5.5) = 132.

    Therefore the sum of the 25 observations is 132 + 8 = 140.

    Therefore the new unbiassed estimate of the mean is 140/25 = 5.6.


    Starting with the unbiassed estimator of the variance, you proceed in an analogous way to get the new unbiassed estimate of the variance.
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