Hi

I would consider an alternative approach.  For each ITEM, calculate the 
percentage of students who passed that item. Then do a paired-difference test 
of significance for pre vs post with items as the random factor (i.e., 
"subjects").  This will tell you whether there was an overall change.  Then you 
might calculate a change score and a confidence interval about the mean change 
score to identify items that scored unexpectedly high or low.

Or you could do some analysis just of post-test scores for items.  What follows 
would depend on the nature of the items.  If objective questions (TF, 
MultChoice), then determine the chance probability of correct and confidence 
intervals about that value to see which items fall outside the CI.  If not 
objective but just scored 0 and 1, then perhaps determine one-sided CI about 0 
(assuming such a statistic exists)?

Or if you could a priori (i.e., without looking at results or using factor 
analysis or something like that) group some of the items together (e.g., from 
same chapter, same concept, ...), you could produce a score rather than a 0/1 
value and do some more standard statistical tests, such as within-s factorial 
(time, scale) or simple effects of time within scale.

Take care
Jim


James M. Clark
Professor & Chair of Psychology
[email protected]
Room 4L41A
204-786-9757
204-774-4134 Fax
Dept of Psychology, U of Winnipeg
515 Portage Ave, Winnipeg, MB
R3B 0R4  CANADA


>>> John Kulig <[email protected]> 16-Jan-13 9:24 AM >>>
Hi Annette 

Perhaps McNemar's Test for significance of changes, for dichotomous data. For 
each item, set up a table that looks like a 2*2 chi square but has "pretest" 
and "post-test" as variables (in texts its usually labelled "before" and 
"after") . 
Posttest 
- + 
+ A B 
Pretest 
- C D 

So every S appears as one count in the table and A+B+C+D = total N. But the key 
cells are A and D. Since (A+D) equals the people who changed, we expect half of 
(A+D) in A and D if the Null is true. The frequency expected for both A and D 
is (A+D)/2 ... Then you just do a chi square on those two cells. The formula 
simplifies to 

chi square = (A - D)squared / (A + D) with df = 1. 

The Wikipedia link is http://en.wikipedia.org/wiki/McNemar%27s_test, and I just 
noticed it reorders the rows and columns so that B and C are the "change" 
variables. It also reminds us about Yates correction .. Unfortunately, I don't 
use SPSS much these days so I don't know how to find it or code the variables 
for the chi square! Not sure about the chance issue. The p values may be all 
you need ,.. but you may want a correction for chance?? Others will know more 
.... 




========================== 
John W. Kulig, Ph.D. 
Professor of Psychology 
Coordinator, University Honors 
Plymouth State University 
Plymouth NH 03264 
========================== 

----- Original Message -----

From: "Annette Taylor" <[email protected]> 
To: "Teaching in the Psychological Sciences (TIPS)" 
<[email protected]> 
Sent: Tuesday, January 15, 2013 6:21:42 PM 
Subject: [tips] my crummy knowledge of stats 

I know this is a basic question but here goes: 

I have categorical data, 0,1 which stands for incorrect (0) or correct (1) on a 
test item. 

I have 25 items and I have a pretest and a posttest and I want to know on which 
items students improved significantly, and not just by chance. Just eyeballing 
the data I can tell that there are some on which the improved quite a bit, some 
not at all and some are someplace in the middle and I can't make a guess at 
all. That is why we have statistics. Yeah! .... hmmmm....bleh..... 

As far as I know, the best thing to do is a chi-square test for each of 25 
items; but of course that will mean that with a .05 sig level I will have at 
least one false positive, maybe more, but most assuredly at least one. This 
seems to be a risk. At any rate I can use SPSS and the crosstabs command allow 
for calculation of the chi-square. 

I know that when I do planned comparisons with multiple t-tests, I can do a 
Simes' correction in which I can rank order my final, obtained alphas, and 
adjust for the number of comparisons and reject from the point from which the 
obtained alpha failed to exceed the corrected-for-number-of-comps alpha. But as 
far as I know, I cannot do that with 25 chi square tests. There is probably 
some reason why I can no more do that, that relates to the reason for why I 
cannot do 25 t-tests in this situation with categorical data. 

Is there a better way to answer my research question? I need a major professor! 
Oh wait, that's me... drat! I need to hire a statistician. Oh wait, I'd need $$ 
for that and I don't have any. So I hope tipsters can stand in as a 
quasi-hired-statistician and help me out. 

Oh, I get the digest. I don't mind waiting until tomorrow or the next day for a 
response, but a backchannel is fine. [email protected] 

I will be at APS this year. Any other tipsters planning to be there? Let's have 
a party! I'd love to put personalities to names. 

Thanks 

Annette 

Annette Kujawski Taylor, Ph. D. 
Professor, Psychological Sciences 
University of San Diego 
5998 Alcala Park 
San Diego, CA 92110 
[email protected] 
--- 
You are currently subscribed to tips as: [email protected]. 
To unsubscribe click here: 
http://fsulist.frostburg.edu/u?id=13338.f659d005276678c0696b7f6beda66454&n=T&l=tips&o=23044
 
or send a blank email to 
leave-23044-13338.f659d005276678c0696b7f6beda66...@fsulist.frostburg.edu 


---
You are currently subscribed to tips as: [email protected].
To unsubscribe click here: 
http://fsulist.frostburg.edu/u?id=13251.645f86b5cec4da0a56ffea7a891720c9&n=T&l=tips&o=23065
 
or send a blank email to 
leave-23065-13251.645f86b5cec4da0a56ffea7a89172...@fsulist.frostburg.edu

---
You are currently subscribed to tips as: [email protected].
To unsubscribe click here: 
http://fsulist.frostburg.edu/u?id=13090.68da6e6e5325aa33287ff385b70df5d5&n=T&l=tips&o=23069
or send a blank email to 
leave-23069-13090.68da6e6e5325aa33287ff385b70df...@fsulist.frostburg.edu

<<attachment: Jim_Clark.vcf>>

Reply via email to