+ Reply to Thread
Results 1 to 13 of 13

Thread: Margin of Error - Confidence Level

  1. #1
    Points: 1,795, Level: 24
    Level completed: 95%, Points required for next Level: 5

    Posts
    17
    Thanks
    0
    Thanked 0 Times in 0 Posts

    Question Margin of Error - Confidence Level




    I sent out a questionnaire to hundreds of people and recieved 154 responses.
    Some of the questions have a yes/no answer. So for one of them I received 69 yes and 85 no.
    What I want to add to my answers/graph is a margin of error for those that misinterpreted the question.
    This is where I am getting stumped. In excel, for confidence level, it asks for a standard deviation. I don't know how to calculate a standard deviation on yes/no responses.

    help/suggestions???

  2. #2
    Super Moderator
    Points: 10,097, Level: 67
    Level completed: 12%, Points required for next Level: 353
    Dragan's Avatar
    Location
    Illinois, US
    Posts
    1,820
    Thanks
    0
    Thanked 153 Times in 136 Posts
    Quote Originally Posted by psrs0810 View Post
    I sent out a questionnaire to hundreds of people and recieved 154 responses.
    Some of the questions have a yes/no answer. So for one of them I received 69 yes and 85 no.
    What I want to add to my answers/graph is a margin of error for those that misinterpreted the question.
    This is where I am getting stumped. In excel, for confidence level, it asks for a standard deviation. I don't know how to calculate a standard deviation on yes/no responses.

    help/suggestions???
    The distribution is Bernoulli (i.e. yes = 1 or no = 0).

    Thus, the Standard Deviation is Sqrt[p*q] = Sqrt[(69/154)*(85/154)].
    Last edited by Dragan; 09-01-2010 at 02:07 PM. Reason: Leave the N out

  3. #3
    Points: 1,795, Level: 24
    Level completed: 95%, Points required for next Level: 5

    Posts
    17
    Thanks
    0
    Thanked 0 Times in 0 Posts
    Dragon - thanks for the quick response!

    would you check me?
    so if the standard deviation is: = Sqrt[154*(69/154)*(85/154)]
    stdev= 6.171257

    and the confidence level = (.05, 6.171257, 154)
    confidence = .9747

  4. #4
    Super Moderator
    Points: 10,097, Level: 67
    Level completed: 12%, Points required for next Level: 353
    Dragan's Avatar
    Location
    Illinois, US
    Posts
    1,820
    Thanks
    0
    Thanked 153 Times in 136 Posts
    Quote Originally Posted by psrs0810 View Post
    Dragon - thanks for the quick response!

    would you check me?
    so if the standard deviation is: = Sqrt[154*(69/154)*(85/154)]
    stdev= 6.171257

    and the confidence level = (.05, 6.171257, 154)
    confidence = .9747

    I adjusted my post ( I thought too quickly ).

    Leave the 154 out like I did above.

  5. #5
    Points: 1,795, Level: 24
    Level completed: 95%, Points required for next Level: 5

    Posts
    17
    Thanks
    0
    Thanked 0 Times in 0 Posts
    I now have a confidence level of .0785

    does that sound right?

  6. #6
    Super Moderator
    Points: 10,097, Level: 67
    Level completed: 12%, Points required for next Level: 353
    Dragan's Avatar
    Location
    Illinois, US
    Posts
    1,820
    Thanks
    0
    Thanked 153 Times in 136 Posts
    Quote Originally Posted by psrs0810 View Post
    I now have a confidence level of .0785

    does that sound right?

    Yes, it looks good i.e. the standard error X 1.96.

  7. #7
    Points: 1,795, Level: 24
    Level completed: 95%, Points required for next Level: 5

    Posts
    17
    Thanks
    0
    Thanked 0 Times in 0 Posts
    I need to make another post about this.


    I applied the stdev of Sqrt[(A1/A3)*(A2/A3)] to all of my Yes/No questions
    A1 = Total Yes
    A2 = Total No
    A3 = Total of Yes and No

    My concern is that my standard deviation is so low that my confidence level (Margin of error) is below 1. Therefore when I apply my +- (error bar) to my Total Yes and Total No there is no fluctuation. It's basically saying that the data is very confident and no potential adjustments.

    I tried ajusting my Alpha in the confidence level and the end result is minute.

    HELP!!!

  8. #8
    TS Contributor
    Points: 18,686, Level: 86
    Level completed: 68%, Points required for next Level: 164

    Posts
    2,802
    Thanks
    9
    Thanked 514 Times in 489 Posts
    You need to know which parameter you are interested in to construct the
    confidence interval. For example, if you want to construct for the
    proportion of Yes, then the parameter of interest is p \in [0, 1]
    and the approximate confidence interval is
    \left[\hat{p} \pm z_{\alpha}\sqrt{\frac{\hat{p}(1 - \hat{p})} {n} } \right], where \hat{p} = \frac {X} {n}, X \sim Binomial(n, p),
    i.e. X is the number of observed Yes.

    However, if you want to construct for the number of expected Yes,
    then the parameter of interest is np
    The point estimate is n\hat{p} = X
    and thus the estimated variance is n\hat{p}(1 - \hat{p})
    So the approximate confidence interval is
    \left[n\hat{p} \pm z_{\alpha}\sqrt{n\hat{p}(1 - \hat{p})} \right]

  9. #9
    Super Moderator
    Points: 10,097, Level: 67
    Level completed: 12%, Points required for next Level: 353
    Dragan's Avatar
    Location
    Illinois, US
    Posts
    1,820
    Thanks
    0
    Thanked 153 Times in 136 Posts
    Quote Originally Posted by psrs0810 View Post
    ...What I want to add to my answers/graph is a margin of error for those that misinterpreted the question....
    The question I have -- going back to the original post -- is how do you know who the people are (and how many) that misinterpreted the question??

  10. #10
    Points: 1,795, Level: 24
    Level completed: 95%, Points required for next Level: 5

    Posts
    17
    Thanks
    0
    Thanked 0 Times in 0 Posts
    BGM and Dragon - thank you for your replys.

    These people are from other organizations that are in the same position that I am in. I am trying to get a feel for how they are handling their processes. That is how this questionnaire derived.

    The total number of responders was 154.
    As for the number of who misinterpreted the question(s), I am not sure. but I was going with a high confidence level for most of the questions.

    I was trying to structure the confidence level around how polling results work, if that makes sense.

    You guys have been a big help thus far - thank you.

    Oh, BGM - I REALLY appreciate the statistical formulas, but I need to see the numbers working to understand the formula. I am a visual person, so I need to see the numbers being applied.

    Thanks

  11. #11
    Points: 1,795, Level: 24
    Level completed: 95%, Points required for next Level: 5

    Posts
    17
    Thanks
    0
    Thanked 0 Times in 0 Posts
    If I apply 1.96 * SQRT((p * (1-p))/N)
    would this be accurate?
    P=69 (number that answered yes) would that be correct?
    N=154 (total number who returned the survery)

    (69*(-68))/154
    (-4692)/154
    -30.4673
    SQRT (30.4673)
    5.51974
    *1.96
    Margin of Error = 10.81869

  12. #12
    Beep
    Points: 63,426, Level: 100
    Level completed: 0%, Points required for next Level: 0
    Awards:
    Discussion EnderPosting AwardCommunity AwardFrequent PosterActivity Award
    Dason's Avatar
    Location
    Ames, IA
    Posts
    11,334
    Thanks
    266
    Thanked 2,204 Times in 1,883 Posts
    Quote Originally Posted by psrs0810 View Post
    If I apply 1.96 * SQRT((p * (1-p))/N)
    would this be accurate?
    P=69 (number that answered yes) would that be correct?
    N=154 (total number who returned the survery)

    (69*(-68))/154
    (-4692)/154
    -30.4673
    SQRT (30.4673)
    5.51974
    *1.96
    Margin of Error = 10.81869
    No. \hat{p} is the proportion of the sample that answered yes (ie. 69/154)

  13. #13
    Points: 1,795, Level: 24
    Level completed: 95%, Points required for next Level: 5

    Posts
    17
    Thanks
    0
    Thanked 0 Times in 0 Posts

    OK - if I correct P to be 69/154 or .44805

    I am getting a margin of error of 7.85%

    (1-.44805) = .551948

    .44805*.551948 = .247301

    .247301/154 = .001606

    Sqrt .001606 = .040073

    .040073 * 1.96 = .078543 or 7.85%
    ???
    Last edited by psrs0810; 09-03-2010 at 07:33 AM. Reason: miss calculation

+ Reply to Thread

           




Similar Threads

  1. Margin of error in a confidence interval.
    By JasonStats in forum Statistics
    Replies: 0
    Last Post: 03-28-2010, 10:03 PM
  2. Margin of Error/Confidence Interval Question
    By girugamesh in forum Statistics
    Replies: 3
    Last Post: 05-06-2009, 08:11 PM
  3. Margin of Error
    By chopkins in forum Statistics
    Replies: 0
    Last Post: 04-23-2008, 01:38 AM
  4. Confidence level of combined error margins
    By etenebris in forum Psychology Statistics
    Replies: 0
    Last Post: 01-15-2008, 03:43 AM
  5. confidence and margin of error
    By tejay in forum Statistics
    Replies: 2
    Last Post: 03-12-2006, 07:29 PM

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts






Advertise on Talk Stats