The UNIVARIATE Procedure

Example 5: Examining the Data Distribution and Saving Percentiles

Procedure features:

PROC UNIVARIATE statement options:

	ALPHA=
	CIBASIC
	CIPCTLNORMAL
	MU0=
	NORMAL
	PLOTS
	PLOTSIZE=

OUTPUT statement

Other features:

PRINT procedure

Data set:

SCORE

This example

specifies the confidence level for the confidence limits
computes a lower confidence limit for the parameters
computes two-sided confidence limits for the quantiles based on the assumption of normal data
specifies the null hypothesis mean for the tests for locations
tests the hypothesis that the data are normally distributed
produces a stem-and-leaf plot, box plot, and normal probability plot and increase the plot size
computes additional percentiles
creates an output data set with percentiles
prints the output data set.

Program

options nodate pageno=1 linesize=64 pagesize=58;

proc univariate data=score mu0=80 alpha=.1 cibasic(type=lower)
                cipctlnormal normal plots plotsize=26;

   var final;

   output out=pctscore median=Median pctlpts=98 50 20 70
          pctlpre=Pctl_ pctlname=Top Mid Low;
   title 'Examining the Distribution of Final Exam Scores';
run;

proc print data=pctscore noobs;
   title1 'Quantile Statistics for Final Exam Scores';
   title2 'Output Data Set from PROC UNIVARIATE';
run;

Output

The estimate of the mean test score is 82.4, with a standard deviation of 8.6. The 90 percent lower confidence limit for the mean is 79.
The Tests for Location table includes three hypothesis tests. To determine whether the Student's t statistic is appropriate, you must determine if the data are approximately normally distributed.
PROC UNIVARIATE calculates the Shapiro-Wilk W statistic because the sample size is below 2000. All p-values from the tests for normality are >0.15, which provides insufficient evidence to reject the assumption of normality. The probability plot also supports the assumption that the data are normal. Therefore, the t statistic appears appropriate. The p-value of .35 for this test provides insufficient evidence to reject the null hypothesis that the mean test score is 80.
Examination of the box plot, which is nonsymmetric, and the small sample size, which causes low power, make the sign test a more appropriate test of location. The p-value of .75 for this test provides insufficient evidence to reject the null hypothesis that the mean test score is 80.
The three plots display the data distribution. The PLOTSIZE= option enlarges the plots so that you can easily see if the data are approximately normal.

The PCTSCORE data set contains one observation. The median value in Median is equivalent to the 50th percentile in PCTL_MID.

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The PCTSCORE data set contains one observation. The median value in Median is equivalent to the 50th percentile in PCTL_MID.