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Posted 02-23-2016 08:20 PM
(1648 views)

The SAS homework is as follows.

A researcher was interested in comparing three methods of dropout prevention in at-risk high school students: a structured work-study program, a peer-buddy program, and teacher mentoring program. Thirty at-risk students were randomly assigned to each of these three treatments at the beginning of their sophomore year. At the end of the year, the researcher noted how many students were still enrolled in school and how many appeared to have dropped out. These are the data that were obtained:

| End of Year Status | |

Program | Still Enrolled | Not Enrolled |

Work-Study | 11 | 19 |

Peer Buddy | 16 | 14 |

Teacher Mentor | 20 | 10 |

What am I doing wrong in SAS?

input program $ persist dropout $;

datalines;

study 11 19

buddy 16 14

mentor 20 10

;

run;

proc anova data=case1;

class program;

model persistence dropout = program;

means / tukey;

run;

1 ACCEPTED SOLUTION

Accepted Solutions

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It sounds you need Cochran-Armitage Trend Test of PROC FREQ .

7 REPLIES 7

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It sounds you need Cochran-Armitage Trend Test of PROC FREQ .

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SAS one way ANOVA is not completely solved.

The proc freq idea only deals with the procedure. It does not expain why the input codes are not being accepted by SAS.

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SAS one way ANOVA is not completely solved. The input data is providing many errors

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Your data should look like this :

data have;

input program $ persist dropout ;

datalines;

study 11 19

buddy 16 14

mentor 20 10

;

run;

And Trend Test is for the ordinal variable , but your PROGRAM is nominal variable, so change their position for different Trend Test.

But I think the original order in your data is looking good .

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Hi KSharp,

Thanks for your insights. Another question? Why do you think I am not getting the TTest and probability result in my syntax?

Josie

Thanks for your insights. Another question? Why do you think I am not getting the TTest and probability result in my syntax?

Josie

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Thank you very much KSharp. Much appreciated. J

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