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Math Help - sample size

  1. #1
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    sample size

    I just got a strange result...

    1. I have the standard deviation = 200
    2. I know the mean of the sample is at most 100 more than the mean of the population.

    I must find the size of the sample with 95% confidence.

    I thought that #2 meant the standard error for the mean was 100;

    s_{\bar{x}} = \frac{s}{\sqrt{n}}

    100 = \frac{200}{\sqrt{n}}

    But it leads to;

    n = 4

    Also, with this method I can't get any confidence interval.
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  2. #2
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    Quote Originally Posted by AnnM View Post
    I just got a strange result...

    1. I have the standard deviation = 200
    2. I know the mean of the sample is at most 100 more than the mean of the population.

    I must find the size of the sample with 95% confidence.

    I thought that #2 meant the standard error for the mean was 100;

    s_{\bar{x}} = \frac{s}{\sqrt{n}}

    100 = \frac{200}{\sqrt{n}}

    But it leads to;

    n = 4

    Also, with this method I can't get any confidence interval.
    Suppose the samplw size is n, the assume large sample statistics so:

    m\sim N(\mu,\sigma^2/n)

    Now our confidence interval [n_1,n_2] is chosen so that for any n \in [n_1,n_2]

    p(m>\mu+100|n)<0.05

    That is for any n \in [n_1,n_2] the probability of getting a result worse than that actually observed is less than 5\%.

    RonL
    Last edited by CaptainBlack; May 8th 2008 at 11:46 PM.
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  3. #3
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    Thank you for your answer, but I don't really get your notation.

    I have the standard deviation of the sample, s = 200

    The mean of the sample is at most 100 more than the mean of the population. I interpret that as = the standard error of the mean is 100.

    The equation for the standard deviation for the mean is

    <br />
s_{\bar{x}} = \frac{s}{\sqrt{n}}<br />

    ...but it seems that I can't extract n from there (well, I can with a little algebra but the answer makes no sense). Why ? Beside, even if I could get 'n' from the equation, I would have no interval of confidence.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by AnnM View Post
    Thank you for your answer, but I don't really get your notation.

    I have the standard deviation of the sample, s = 200

    The mean of the sample is at most 100 more than the mean of the population. I interpret that as = the standard error of the mean is 100.

    The equation for the standard deviation for the mean is

    <br />
s_{\bar{x}} = \frac{s}{\sqrt{n}}<br />

    ...but it seems that I can't extract n from there (well, I can with a little algebra but the answer makes no sense). Why ? Beside, even if I could get 'n' from the equation, I would have no interval of confidence.
    Sorry I had misread the question

    <br />
p(m>\sigma+100|n)<0.05

    should be:

    <br />
p(m>\mu+100|n)<0.05

    Which gives us a nicer condition:

    <br />
p(m-\mu >100|n)<0.05

    <br />
p\left(\frac{m-\mu}{\sqrt{n} \sigma} >\frac{1}{2\sqrt{n}}\right)<0.05

    RonL
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  5. #5
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    I don't understand why my reasoning doesn't work.

    I have no idea how you got to your equation.

    And even if I did, we don't have the information to solve it.

    In short, the equation for the interval should be;

    <br />
 \left[\bar{x} - 1.96\frac{200}{\sqrt{n}};\bar{x} + 1.96\frac{200}{\sqrt{n}}\right]<br />

    But, as I don't have the mean, I can't get the sample size.
    Last edited by AnnM; May 9th 2008 at 09:20 AM.
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  6. #6
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    In fact, I got;

    <br />
 n =  \left[\frac{Z_{a/2}s}{s_x}\right]^2 = \left[\frac{1.96(200)}{100}\right]^2 = 15.37<br />

    It doesn't seem to make sense either...
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  7. #7
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    d = 100

    Why doesn't n = 16 make sense?

    If only we had the entire problem statement...
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  8. #8
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    Quote Originally Posted by TKHunny View Post
    Why doesn't n = 16 make sense?
    Because;

    <br />
    " alt="s_{\bar{x}} = \frac{s}{\sqrt{n}}
    " />
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