``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56``` ```SPSS AnswerNet: Result Solution ID: 100000632 Product: SPSS Base Title: Using date variables in TABLES Description: Q. I have a date variable where I'd like to get the mean and standard deviation in days. When I run the table, I get the mean, but the standard deviation isn't coming out correctly. I get either all asterisks, or some really high number. How can I get the standard deviation out of the date variable? A. The following shows how to get a table that has the mean date and standard deviation in days from the mean date. * Here is a sample data file. Suppose we want to get * the mean date and standard deviation in days of the * DT variable by the VARA variable. DATA LIST fixed / dt 1-8 (adate) vara 10. BEGIN DATA 06/11/50 1 11/17/53 1 01/19/56 2 08/26/59 2 END DATA. * The first thing we do is to use AGGREGATE to compute * a variable that represents the standard deviation of * the DT variable. We can then incorporate that value * back to our original data file. AGGREGATE outfile=temp.sav / break=vara /sddt=sd(dt). MATCH FILES file=* / table=temp.sav / by vara. * Since the new variable (SDDT) is stored as * of seconds, we can use a COMPUTE command to translate * this value into days by using the CTIME.DAYS function. COMPUTE sddt=trunc(ctime.days(sddt)). EXECUTE. FORMATS sddt (f5.0). * Some cosmetic work here. VARIABLE LABELS dt 'Mean Date' sddt 'Standard Deviation in Days' . * Now we will compute the table. We will use DT and * SDDT. Although SDDT represents a standard deviation, * we will use the MEAN statistics to represent it. * This is because the standard deviation is already * computed in the dataset, and we just want to display * the value in the dataset. We can use the MEAN statistic * to do that. TABLES observation=dt sddt /table=dt+sddt by vara /statistics mean(dt'') mean(sddt''). ```
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