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Math Help - test n varience unkown

  1. #1
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    test n varience unkown

    Im not sure how to do this question. A chemical test on a compoun gave the following results(grams of a particular element in 1kg of compound):1.60, 1.56 ,1.59, 1.53, 1.58 , 1.52, 1.59 ,1.64, 1.55. From these figures make an estimate of the size of the sample required so that the standard error of the mean will be about 0.005. Explain what assumptions have to be made, and what sources of error there could be in the estimate.

    I dont quite understand the structure of the working required. I believe the population mean to be 1(unsure) sample variece to be 0.0122, sample mean 1.57 and from here whether i use a chi squared test or t test and how inco-operate an n value to estimate the standard error??? Somewhere along the line do i use a null hypothesis of mean = 1, alternative mean < 1 then use a type 1 error? The answer is 54 using the sample varience, pop varience unkown, small sample so sample varience may not be very good estimate of pop varience. If somebody could show me the working that for 54 id appreciate it very much thanks.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by oxrigby View Post
    Im not sure how to do this question. A chemical test on a compoun gave the following results(grams of a particular element in 1kg of compound):1.60, 1.56 ,1.59, 1.53, 1.58 , 1.52, 1.59 ,1.64, 1.55. From these figures make an estimate of the size of the sample required so that the standard error of the mean will be about 0.005. Explain what assumptions have to be made, and what sources of error there could be in the estimate.

    I dont quite understand the structure of the working required. I believe the population mean to be 1(unsure) sample variece to be 0.0122, sample mean 1.57 and from here whether i use a chi squared test or t test and how inco-operate an n value to estimate the standard error??? Somewhere along the line do i use a null hypothesis of mean = 1, alternative mean < 1 then use a type 1 error? The answer is 54 using the sample varience, pop varience unkown, small sample so sample varience may not be very good estimate of pop varience. If somebody could show me the working that for 54 id appreciate it very much thanks.
    The standard error of the mean of a sample of size n is:

    se=\frac{\sigma}{\sqrt{n}}

    so the required sample size to give a se of 0.005 would be:

    {n}=\frac{\sigma^2}{0.005^2}

    we will estimate this by:

    \widehat{n}=\frac{s^2}{0.005^2}

    where s^2 is the unbiased estimator of the population variance derived from the sample.

    The source of error in this estimate is the sampling error in the sample based variance estimate.

    (check your arithmetic)

    CB
    Last edited by CaptainBlack; March 19th 2009 at 10:47 PM.
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