```Date: Thu, 4 Mar 2010 17:41:45 -0500 Reply-To: Francois van der Walt Sender: "SAS(r) Discussion" From: Francois van der Walt Subject: Re: Distribution Comments: To: Peter Flom Hi Mohsen, Your question is trivial in some sense, but to answer it with integrity is more complex than anticipated. One can do various things including perform a T-test. I decided to use a Shewhart or XmR chart as made famous by Donald Wheeler in Making sense of Data. See page 142 and surrounding pages if you would like to know more. (Or send me an e-mail and I will send you an article I wrote in our internal news at GJI (www.gji.com.au) regarding this) It is also difficult to show the results here as the result is a chart in preferably hi-resolution graphics. Running the code will show that I and J is outside the limits that confirms that one could probably see looking at the data. Surprisingly G is also outside the limits that one would not easily identified! In conclusion using this analysis allows one to give an objective opinion and view. If you need any more background please let me know. Kind Regards Francois www.gji.com.au Full code: data dist; input Group\$ Freq; datalines; A 311 B 336 C 254 E 318 F 325 G 370 H 317 I 21 J 6 ; /*proc gchart data=dist; vbar group/freq=freq; run;*/ /* Make sure they are random in order */ data dist2; set dist; rand=uniform(0); /* proc sort data=dist2 out=dist2; by rand; run; */ /* Calculate range */ data dist3; set dist2; range=abs(freq-lag(freq)); /* Calculate summaries */ proc means data=dist3; var freq range; output out=summ median=medfreq medrange mean=meanfreq meanrange; run; data dist4 (keep= group indicator freq lnpl unpl url RangeInd range meanfreq); if _n_=1 then set summ; set dist3; unpl=meanfreq+3.14*medrange; lnpl=meanfreq-3.14*medrange; url=medrange*3.87; indicator="Within prediction"; if freq>unpl or frequrl then RangeInd="Over range limit"; run; symbol1 interpol=none v=star; symbol2 interpol=join; symbol3 interpol=join; /* For some reason the symbols and does not work as expected but one can still make out the result*/ proc gplot; plot freq*group= 1 meanfreq*group= 2 unpl*group= 2 lnpl*group= 2 /overlay; quit; run; On Wed, 24 Feb 2010 18:47:12 -0500, Peter Flom wrote: >>> >Hello All: >>> > >>> >I have the following data and I would like to know if the frequencies >>> >are evenly distributed across the plots. How I can prove that the >>> >frequency of plots are significantly different from each other. >>> > >>> >Plots Frequency >>> >A 311 >>> >B 336 >>> >C 254 >>> >E 318 >>> >F 325 >>> >G 370 >>> >H 317 >>> >I 21 >>> >J 6 >>> > >>> >Thanks, >>> >Mohsen >> > >I think this data passes the IOTT. > >The InterOcular Trauma Test > >It hits you between the eyes. > >:-) > >Peter > >Peter L. Flom, PhD >Statistical Consultant >Website: http://www DOT statisticalanalysisconsulting DOT com/ >Writing; http://www.associatedcontent.com/user/582880/peter_flom.html >Twitter: @peterflom ```

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