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Anonymous's avatar
Anonymous
Not applicable
6 years ago
Solved

Breaking down the normal distribution by attributes

Hey,   We have just conducted a survey and I'm looking to understand if the respondents who answered the survey are representitive of our population as a whole. Mainly, are we getting the same prop...
Need Help
  • v-lionel-msft's avatar
    v-lionel-msft
    6 years ago

    Hi Anonymous ,

     

    First of all, your model relationship may be a bit problematic, it should be like this:

    Secondly, the results returned by your two percentage measures may have problems, please refer to the following formula:

    Percent of respondents = 
VAR x = 
CALCULATE(
    DISTINCTCOUNT('Sample'[External Data Reference]),
    ALLEXCEPT(
        'Sample',
        'Sample'[Sector]
    )
)
VAR y = 
CALCULATE(
    DISTINCTCOUNT('Sample'[External Data Reference]),
    ALL('Sample')
)
RETURN
DIVIDE(
    x,
    y,
    BLANK()
)

Percent of total Population = 
VAR x = 
CALCULATE(
    DISTINCTCOUNT(Population[ID]),
    ALLEXCEPT(
        'Population',
        'Population'[Sector]
    )
)
VAR y = 
CALCULATE(
    DISTINCTCOUNT(Population[ID]),
    ALLSELECTED('Sample'[Sector])
)
RETURN
 DIVIDE(
    x,
    y,
    BLANK()
) 
    Left_Tailed_Hypothesis = 
NORM.DIST(
    [Percent of respondents], 
    [Percent of total Population],
    1,
    TRUE()
)

    The value of COUNTROWS(ALLSELECTED('Sample')) is too big, which is not suitable for Standard_dev parameters.
    And its value is equal to a fixed value 899.
     
    You can refer to https://docs.microsoft.com/en-us/dax/norm-dist-dax to learn NORM.DIST() function.
     

    Best regards,
    Lionel Chen

     

    If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.

     

     

sturlaws's avatar
sturlaws
Icon for Resident Rockstar rankResident Rockstar
6 years ago

Great.

 

If you don't have it already in your report, create a section-table/dimension, and change you measures to this:

Percentage of respondents =
DIVIDE (
    [Number respondents];
    CALCULATE ( [Number respondents]; ALL ( 'dimSector' ) )
)

 

And as v-lionel-msft mentions, the value you are using for standard deviation does not make any sense. I can't really see how you should calculate the standard deviation the way you are trying to do. I am not that experienced in statistics, but is perhaps the chi-square test is more appropriate?

Cheers,
Sturla