Results 1 to 6 of 6

Math Help - slight discrepancy between my answer and textbook...

  1. #1
    Junior Member
    Joined
    Jul 2007
    Posts
    43

    slight discrepancy between my answer and textbook...

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

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

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

    For a 99% confidence rating, z_{.005} = 2.576. Therefore, the confidence interval should be
    \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 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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by earachefl View Post
    OK, problem has sample n = 37, \sum x = 277.875, \sum x^2 = 2093.109375. Estimate with 99% confidence the mean.

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

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

    For a 99% confidence rating, z_{.005} = 2.576. Therefore, the confidence interval should be
    \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 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:

    <br />
V=\frac{1}{n-1}\sum(x- \bar{x})^2 = \frac{1}{n-1}\sum x^2 - \frac{n}{n-1}\bar{x}^2<br />

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jul 2007
    Posts
    43
    Thanks - that's not even mentioned in my book.... I'll check it out.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jul 2007
    Posts
    43
    Actually, when I use that formula, I get the answer 7.51 \pm .17723903, which is also off from the book's answer but a bit higher. Still confused.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jul 2007
    Posts
    43
    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 s = \sqrt <br />
{ \frac {\sum x^2 - \frac { (\sum x)^2}{n}}{n-1}}

    then s = .416

    and the answer magically becomes 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
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by earachefl View Post
    Actually, when I use that formula, I get the answer 7.51 \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 s = \sqrt <br />
{ \frac {\sum x^2 - \frac { (\sum x)^2}{n}}{n-1}}

    then s = .416

    and the answer magically becomes 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:

    <br />
V = \frac {\sum x^2 - \frac { (\sum x)^2}{n}}{n-1}<br />

    RonL
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Weighted percentage discrepancy.
    Posted in the Business Math Forum
    Replies: 2
    Last Post: March 25th 2011, 05:47 PM
  2. Integration by Substitution Discrepancy
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 6th 2010, 07:26 AM
  3. MacClaurin series discrepancy
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 7th 2009, 02:06 PM
  4. Textbook discrepancy
    Posted in the Calculus Forum
    Replies: 9
    Last Post: May 14th 2008, 09:47 PM

Search Tags


/mathhelpforum @mathhelpforum