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Thread: % Change taking into account standard deviation.

  1. #1
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    % Change taking into account standard deviation.

    Hi,

    I am a Masters student at the University of Exeter in England currently writing an essay comparing the rehabilitative efficacy of different exercise modes in treating Chronic Obstructive Pulmonary Disease (COPD). To compare treatment conditions I need to calculate the percentage change of various exercise measures from pre- to post-training. Whilst this is simple enough, I do not know how to calculate percentage change taking into account standard deviation values.

    For example:
    137(38) to 99(26) (values expressed as mean(SD))

    I calculate the above change as 27.7% ((38/137) x 100 = 27.7). However, apparently the correct percentage change is 25%. Can anyone help?

    Thanks very much,

    Paul
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  2. #2
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    Quote Originally Posted by pc269 View Post
    Hi,

    I am a Masters student at the University of Exeter in England currently writing an essay comparing the rehabilitative efficacy of different exercise modes in treating Chronic Obstructive Pulmonary Disease (COPD). To compare treatment conditions I need to calculate the percentage change of various exercise measures from pre- to post-training. Whilst this is simple enough, I do not know how to calculate percentage change taking into account standard deviation values.

    For example:
    137(38) to 99(26) (values expressed as mean(SD))

    I calculate the above change as 27.7% ((38/137) x 100 = 27.7). However, apparently the correct percentage change is 25%. Can anyone help?

    Thanks very much,

    Paul
    Let x = pre-measure with standard deviation $\displaystyle \delta x$.
    Let y = post-measure with standard deviation $\displaystyle \delta y$.

    These standard deviations can be used as the uncertainty in the measured value.

    Let percentage change be $\displaystyle z = \frac{100(x-y)}{x} = 100 - \frac{100 y}{x}$.

    What you want is the uncertainty $\displaystyle \delta z$ in z. It's found using $\displaystyle \frac{\delta z}{z} = \frac{\delta y}{y} + \frac{\delta x}{x}$.


    Therefore $\displaystyle \delta z = z \left(\frac{\delta y}{y} + \frac{\delta x}{x} \right) = \frac{100(x-y)}{x} \left(\frac{\delta y}{y} + \frac{\delta x}{x} \right)$.

    Then the percentage change is equal to $\displaystyle z \pm \delta z$.
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