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Thread: slight discrepancy between my answer and textbook...

  1. #1
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    slight discrepancy between my answer and textbook...

    OK, problem has sample n = 37, $\displaystyle \sum x = 277.875$, $\displaystyle \sum x^2 = 2093.109375$. Estimate with 99% confidence the mean.

    I get $\displaystyle \mu = \frac{\sum x}{n} \Rightarrow \frac{277.875}{37} = 7.510$

    and $\displaystyle \sigma = \sqrt{\sum x^2P(x) - \mu^2} = \sqrt {\frac{2093.109375}{37} - 7.51^2} \approx .413$

    For a 99% confidence rating, $\displaystyle z_{.005} = 2.576$. Therefore, the confidence interval should be
    $\displaystyle \mu \pm 2.576\frac {\sigma}{\sqrt n} \Rightarrow 7.51 \pm 2.576 \frac {.413}{\sqrt 37} \Rightarrow 7.51 \pm .175$.

    The book gives answer $\displaystyle 7.51 \pm .176$. Not much of a difference, I know; I'm just wondering whether the difference might be due to rounding assumptions or whether I've made a more fundamental mistake.
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  2. #2
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    Quote Originally Posted by earachefl View Post
    OK, problem has sample n = 37, $\displaystyle \sum x = 277.875$, $\displaystyle \sum x^2 = 2093.109375$. Estimate with 99% confidence the mean.

    I get $\displaystyle \mu = \frac{\sum x}{n} \Rightarrow \frac{277.875}{37} = 7.510$

    and $\displaystyle \sigma = \sqrt{\sum x^2P(x) - \mu^2} = \sqrt {\frac{2093.109375}{37} - 7.51^2} \approx .413$

    For a 99% confidence rating, $\displaystyle z_{.005} = 2.576$. Therefore, the confidence interval should be
    $\displaystyle \mu \pm 2.576\frac {\sigma}{\sqrt n} \Rightarrow 7.51 \pm 2.576 \frac {.413}{\sqrt 37} \Rightarrow 7.51 \pm .175$.

    The book gives answer $\displaystyle 7.51 \pm .176$. Not much of a difference, I know; I'm just wondering whether the difference might be due to rounding assumptions or whether I've made a more fundamental mistake.
    I don't know that you are doing this but I suspect it is the case. The variance you should be using is the unbiased estimator of the population variance based on the sample, which is:

    $\displaystyle
    V=\frac{1}{n-1}\sum(x- \bar{x})^2 = \frac{1}{n-1}\sum x^2 - \frac{n}{n-1}\bar{x}^2
    $

    RonL
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  3. #3
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    Thanks - that's not even mentioned in my book.... I'll check it out.
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  4. #4
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    Actually, when I use that formula, I get the answer 7.51 $\displaystyle \pm .17723903$, which is also off from the book's answer but a bit higher. Still confused.
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  5. #5
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    D'oh!! I have realized the error of my ways - I forgot how to do the standard deviation (hey, that was Chapter 2, I'm in Chapter 9 now )

    so if I use $\displaystyle s = \sqrt
    { \frac {\sum x^2 - \frac { (\sum x)^2}{n}}{n-1}}$

    then s = .416

    and the answer magically becomes $\displaystyle 7.51 \pm .176$ which is what the book says. Hoorah!

    It's amazing how easy it is to become confused between the standard deviation of a population and the standard deviation of a sample... at least for me
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by earachefl View Post
    Actually, when I use that formula, I get the answer 7.51 $\displaystyle \pm .17723903$, which is also off from the book's answer but a bit higher. Still confused.
    Quote Originally Posted by earachefl View Post
    D'oh!! I have realized the error of my ways - I forgot how to do the standard deviation (hey, that was Chapter 2, I'm in Chapter 9 now )

    so if I use $\displaystyle s = \sqrt
    { \frac {\sum x^2 - \frac { (\sum x)^2}{n}}{n-1}}$

    then s = .416

    and the answer magically becomes $\displaystyle 7.51 \pm .176$ which is what the book says. Hoorah!

    It's amazing how easy it is to become confused between the standard deviation of a population and the standard deviation of a sample... at least for me

    These should give the same result as what I gave you was a rewrite (if I did it correctly) of:

    $\displaystyle
    V = \frac {\sum x^2 - \frac { (\sum x)^2}{n}}{n-1}
    $

    RonL
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