final code2

# final code2 - Final Exam 960:390 November 8 2010 Jiyoon...

This preview shows pages 1–11. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Final Exam 960:390 November 8, 2010 Jiyoon Jung *Problem 1 1) Provide summary statistics and appropriate graphic displays for each of the two age groups (over 35 age and under 35) Summary statistics is on output. Also, through histograms for each group, I can figure out which group has better scores. 2) Test whether the mean score for the over 35 age group exceed the mean score for the below 35 age group. Using PROC MEANS, I can figure out that over 35 age group mean is 27.588, and below 35 age group mean is 24.125. So, the result proves the statement ‘the mean score for the over 35 age group exceed the mean score for the below 35 age group.’ filename scores '~/scores.txt'; data scores1; infile scores firstobs=2 obs=5; input var1 @@; run; data scores2; infile scores firstobs=8 obs=11; input var2 @@; run; PROC MEANS DATA =scores1; run ; PROC MEANS DATA =scores2; run ; PROC MEANS data =scores1 min max nmiss var t ; var var1; title 'Summary Statistics from under 35 data' ; run ; PROC MEANS data =scores2 min max nmiss var t ; var var2; title1 'Summary Statistics from over 35 data' ; run ; proc chart data =scores1; vbar var1; title1 'Histogram of scores for under 35' ; run ; proc chart data =scores2; vbar var2; title1 'Histogram of scores for over 35' ; run ; *Output Final problem 3 16:59 Monday, November 8, 2010 8 The MEANS Procedure Analysis Variable : var1 N Mean Std Dev Minimum Maximum------------------------------------------------------------------ 32 24.1250000 10.3198837 8.0000000 54.0000000------------------------------------------------------------------ Final problem 3 16:59 Monday, November 8, 2010 9 The MEANS Procedure Analysis Variable : var2 N Mean Std Dev Minimum Maximum------------------------------------------------------------------ 34 27.5882353 14.0434772 7.0000000 68.0000000------------------------------------------------------------------ Summary Statistics from under 35 data 10 16:59 Monday, November 8, 2010 The MEANS Procedure Analysis Variable : var1 N Minimum Maximum Miss Variance t Value--------------------------------------------------------------- 8.0000000 54.0000000 0 106.5000000 13.22--------------------------------------------------------------- Summary Statistics from over 35 data 11 16:59 Monday, November 8, 2010 The MEANS Procedure Analysis Variable : var2 N Minimum Maximum Miss Variance t Value--------------------------------------------------------------- 7.0000000 68.0000000 0 197.2192513 11.45--------------------------------------------------------------- Histogram of scores for under 35 16:59 Monday, November 8, 2010 12 Frequency 12 + ***** | ***** 11 + ***** | ***** 10 + ***** ***** | ***** ***** 9 + ***** ***** | ***** ***** 8 + ***** ***** | ***** ***** 7 + ***** ***** | ***** ***** 6 + ***** ***** | ***** ***** 5 + ***** ***** ***** | ***** ***** ***** 4 + ***** ***** ***** | ***** ***** ***** 3 + ***** ***** ***** ***** | ***** ***** ***** ***** 2 + ***** ***** ***** ***** ***** | *****...
View Full Document

{[ snackBarMessage ]}

### Page1 / 44

final code2 - Final Exam 960:390 November 8 2010 Jiyoon...

This preview shows document pages 1 - 11. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online