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karenktq
Calcite | Level 5

Based on the summary statistics using, the mode is 0. In that case, shouldn't the histogram reflect 0 as the highest frequency? 

 

*summary stats;
proc means data= libref.data2 mean std var min max n median mode;
run;

*histogram;
proc univariate data= libref.data2;
histogram;
run;

 

WhatsApp Image 2021-06-09 at 00.24.34.jpegWhatsApp Image 2021-06-09 at 00.25.20.jpeg

1 ACCEPTED SOLUTION

Accepted Solutions
FreelanceReinh
Jade | Level 19

Hello @karenktq,

 

These kind of "hairy" histograms typically occur when discrete data are binned inappropriately. In your example the default binning was inappropriate. I assume that your popularity values are integers. Note that apparently six consecutive values went into five bins. Thus, one of the five bins necessarily contains two values, not one as the other four, and therefore "stands out" in the histogram, easily rising above the true mode.

 

Here's an example demonstrating the issue and showing how the MIDPOINTS= option can be used to obtain appropriate (or, if misused, inappropriate) binning:

data test;
call streaminit(314159);
do _n_=1 to 50000;
  x=rand('poisson',160)-107;
  output;
end;
run;

ods graphics on;

proc univariate data=test;
histogram / midpoints=0 to 100 by 1.2 odstitle='Inappropriate Binning';
histogram / midpoints=0 to 100 odstitle='Appropriate Binning';
run;

Results:

Inappropriate_Binning.pngAppropriate_Binning.png

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4 REPLIES 4
PaigeMiller
Diamond | Level 26

It may be (in fact I'm pretty sure) that 0 is the mode, but because of the way the bins are grouped together for plotting it doesn't look like 0 is the mode in the plot. In other words, the plot is misleading you. However, there are options in the HISTOGRAM statement that let you change the way the bins are grouped.

https://documentation.sas.com/doc/en/pgmsascdc/9.4_3.4/procstat/procstat_univariate_syntax09.htm

--
Paige Miller
Reeza
Super User
The histogram bins data. So your binned data is more frequent than the mode but those are not the same measures so you're not comparing apples to apples.

Ie Mode is single most frequent value but each bar is showing values within a range, ie 0 to 0.9

The Bar @ 50 for example could be 50-50.9 which likely includes more values than just 50.0
FreelanceReinh
Jade | Level 19

Hello @karenktq,

 

These kind of "hairy" histograms typically occur when discrete data are binned inappropriately. In your example the default binning was inappropriate. I assume that your popularity values are integers. Note that apparently six consecutive values went into five bins. Thus, one of the five bins necessarily contains two values, not one as the other four, and therefore "stands out" in the histogram, easily rising above the true mode.

 

Here's an example demonstrating the issue and showing how the MIDPOINTS= option can be used to obtain appropriate (or, if misused, inappropriate) binning:

data test;
call streaminit(314159);
do _n_=1 to 50000;
  x=rand('poisson',160)-107;
  output;
end;
run;

ods graphics on;

proc univariate data=test;
histogram / midpoints=0 to 100 by 1.2 odstitle='Inappropriate Binning';
histogram / midpoints=0 to 100 odstitle='Appropriate Binning';
run;

Results:

Inappropriate_Binning.pngAppropriate_Binning.png

Ksharp
Super User

You ignore the important element : the width of BIN .

Change it by ENDPOINT= option , you would get different result.

 

ods select histogram;
proc univariate data=score_card ;
var total_score;
histogram total_score/ kernel endpoints=(490 to 650 by 10)  ;
run;

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