Results 1 to 10 of 10

Math Help - Standard deviation

  1. #1
    Newbie
    Joined
    Sep 2007
    Posts
    5

    Standard deviation

    I'm interested in calculating standard deviation and have read your web site.
    I hope to get your help:
    If I have three groups of data A, B and C,
    the mean values for A =a, B =b, and C =c,
    the standard deviation(SD) for A =SD1, B =SD2, and C =SD3,
    then what is the standard deviation for the average value of A, B and C, namely (A+B+C/3)?
    Could you please tell me a formula to calculate their SD. If there are more groups of data?

    Thank you very much.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,211
    Thanks
    419
    Awards
    1
    Quote Originally Posted by frkchen1438 View Post
    I'm interested in calculating standard deviation and have read your web site.
    I hope to get your help:
    If I have three groups of data A, B and C,
    the mean values for A =a, B =b, and C =c,
    the standard deviation(SD) for A =SD1, B =SD2, and C =SD3,
    then what is the standard deviation for the average value of A, B and C, namely (A+B+C/3)?
    Could you please tell me a formula to calculate their SD. If there are more groups of data?

    Thank you very much.
    If you have data sets A, B, C, ... of sizes N_A, N_B, N_C, ... and standard deviations of \sigma_A, \sigma_B, \sigma_C, ..., where sets A, B, C, ... all measure the same property, then
    \bar{\sigma} = \frac{\sqrt{N_A \sigma_A^2 + N_B \sigma_B^2 + N_C \sigma_C^2 + ...}}{N_A + N_B + N_C + ...}

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2007
    Posts
    5

    Red face

    Quote Originally Posted by topsquark View Post
    If you have data sets A, B, C, ... of sizes N_A, N_B, N_C, ... and standard deviations of \sigma_A, \sigma_B, \sigma_C, ..., where sets A, B, C, ... all measure the same property, then
    \bar{\sigma} = \frac{\sqrt{N_A \sigma_A^2 + N_B \sigma_B^2 + N_C \sigma_C^2 + ...}}{N_A + N_B + N_C + ...}

    -Dan
    Thank you very much.
    I have used your method to calculate my data, but the final standard deviations are much smaller than any standard deviation from every data sets. Is it ok? It is not in the range between max and min values?
    New Question:
    If I have two data sets A and B, with means a and b, and standard deviations SDa and SDb, respectively.
    Now I need to calculate the ratio:
    a/b, then what is the final standard deviation for the ratio?
    Thank you again.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by topsquark View Post
    If you have data sets A, B, C, ... of sizes N_A, N_B, N_C, ... and standard deviations of \sigma_A, \sigma_B, \sigma_C, ..., where sets A, B, C, ... all measure the same property, then
    \bar{\sigma} = \frac{\sqrt{N_A \sigma_A^2 + N_B \sigma_B^2 + N_C \sigma_C^2 + ...}}{N_A + N_B + N_C + ...}

    -Dan
    Typo? The pooled variance formula gives:

    s_{pooled} = \sqrt{ \frac{(N_A-1) s_A^2 + (N_B-1) s_B^2 + (N_C-1) s_C^2 + ...}{(N_A-1) + (N_B-1) + (N_C-1) + ...}}

    RonL
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,211
    Thanks
    419
    Awards
    1
    Quote Originally Posted by CaptainBlack View Post
    Typo? The pooled variance formula gives:

    s_{pooled} = \sqrt{ \frac{(N_A-1) s_A^2 + (N_B-1) s_B^2 + (N_C-1) s_C^2 + ...}{(N_A-1) + (N_B-1) + (N_C-1) + ...}}

    RonL
    I used the wrong standard deviation formula again. (Sigh) Thanks for the spot.

    -Dan
    Last edited by ThePerfectHacker; September 11th 2007 at 09:03 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Sep 2007
    Posts
    5

    Thumbs up thank you

    Quote Originally Posted by CaptainBlack View Post
    Typo? The pooled variance formula gives:

    s_{pooled} = \sqrt{ \frac{(N_A-1) s_A^2 + (N_B-1) s_B^2 + (N_C-1) s_C^2 + ...}{(N_A-1) + (N_B-1) + (N_C-1) + ...}}

    RonL
    Could you tell me where these formula come from? Thank you very much!
    How about the standard deviation of a ratio of two mean values?
    Thank you.
    Frank
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by frkchen1438 View Post
    Could you tell me where these formula come from? Thank you very much!
    How about the standard deviation of a ratio of two mean values?
    Thank you.
    Frank
    The pooled variance formula you can find on wikipedia under pooled variance, and more hits than you will want if you type pooled variance into Google.

    RonL
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Sep 2007
    Posts
    5

    Yes, thanks

    Quote Originally Posted by CaptainBlack View Post
    The pooled variance formula you can find on wikipedia under pooled variance, and more hits than you will want if you type pooled variance into Google.

    RonL
    Yes, I have found a lot!
    But I also hope to find the standard deviation of a ratio between two average values(in the case I have the standard deviation of each average)
    Thanks.

    Frank
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by frkchen1438 View Post
    Yes, I have found a lot!
    But I also hope to find the standard deviation of a ratio between two average values(in the case I have the standard deviation of each average)
    Thanks.

    Frank
    You will almost certainly end up with an approximation for this.

    RonL
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Sep 2007
    Posts
    5

    Question you mean use the same formula?

    Quote Originally Posted by CaptainBlack View Post
    You will almost certainly end up with an approximation for this.

    RonL
    Right? Why?
    Thank you.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Standard Deviation
    Posted in the Statistics Forum
    Replies: 4
    Last Post: September 8th 2010, 04:05 AM
  2. Replies: 5
    Last Post: May 18th 2009, 05:14 AM
  3. standard deviation and mean deviation
    Posted in the Statistics Forum
    Replies: 1
    Last Post: March 16th 2008, 06:09 AM
  4. i need standard deviation help
    Posted in the Statistics Forum
    Replies: 2
    Last Post: February 22nd 2008, 06:03 PM
  5. The mean and standard deviation
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: February 17th 2008, 09:15 PM

Search Tags


/mathhelpforum @mathhelpforum