Date: Thu, 1 Sep 2005 02:54:35 -0400
Reply-To: Gabor Toth <gabortoth@HOTMAIL.COM>
Sender: "SPSSX(r) Discussion" <SPSSX-L@LISTSERV.UGA.EDU>
From: Gabor Toth <gabortoth@HOTMAIL.COM>
Subject: repost chi-squared goodness of fit test SPSS 12
Yesterday I posted a help message and I got some answers...Thank you!
...but it was not enough for me to do what I need. Here is the problem
again with a longer explanation. Please give help me with it. I am an
absolute newby, I've never done any statistics before, I have the SPSS
12.0 installed in my PC.
If somebody can send me a syntax for each cases is would be a great help
(I think). Or if somebody could explain how to do it with the SPSS in the
way you teach it to a 10 years old :-). Sorry for that...
The data is related to speech instructors communication behaviour with
disabled people.
I observed 30 instructors for 6 month during their teaching sessions. I
observed one instructor at a time for an average of 7.5 minutes (each of
them 29 times, so it is all together about 109 hours observation time) and
recorded their interpersonal interventions (type of behaviour). I made a
six-category table (6 column) to collect their behaviour into these six
categories and 30 row for the 30 instructors. I marked the usage frequency
of each type of category during the observation into the adequate cell.
Now I have the data table (see below) with usage-frequency numbers of how
many time each instructor used each category (in my opinion as an
observer) during the 6 month.
From this data I made a usage-frequency order (or rank list) to make a
rank order from 1-6 to show which category from the 6 was the "most" and
which is the "least" utilized.
What I need to do is:
1) To test whether the distribution of behaviour of one instructor or all
instructors taken together (the Total row below in the table) is governed
by a given distribution (a multinomial one, for example, or that all
possible outcomes have the same probability of occurence).The null
hypothesis is that "the probabilities of occurence of all categories are
equal"
2) To test whether the distribution of one instructor is different from
the distribution of behaviour of a second instructor, or, in the general
case, whether (in the group of 30 instructors) there are differences at
all (i.e. there are at least two instructors behaving differently).
Yesterday I got an answer that this second case (differences between
instructors) uses chi-squares in crosstabs command and is slightly more
complicated than the first one.
For example from the raw data:
The first instructor has been observed 522 times to behave in the
prescriptive way and the second one was 202 times confrontative.
Here is the raw data:
VARIABLE ID
ID = Participant ID (N = 30)
pres = Prescriptive category
info = Informative category
conf =C onfronting category
cath = Cathartic category
cataly = Catalytic category
supp = Supportive category
ID pres info conf cath cataly supp
1 522 429 313 81 109 204
2 435 520 202 79 320 115
3 310 463 118 57 197 521
4 551 493 143 86 199 380
5 464 548 230 72 144 315
6 348 436 169 93 258 550
7 493 322 514 59 168 233
8 377 206 431 120 78 574
9 203 114 339 85 434 525
10 580 449 265 83 121 346
11 430 347 547 141 69 257
12 406 231 115 73 459 546
13 116 260 371 60 377 576
14 462 578 145 82 289 375
15 232 376 461 94 146 530
16 523 490 287 77 167 405
17 489 147 71 198 345 560
18 145 290 398 52 428 573
19 438 404 164 84 209 519
20 319 427 117 92 236 556
21 549 87 200 173 356 462
22 460 321 140 66 270 585
23 174 205 75 320 530 488
24 58 341 458 233 452 598
25 261 171 61 364 553 437
26 568 410 165 88 286 410
27 479 74 229 118 401 552
28 61 110 259 400 478 579
29 113 403 89 270 119 518
30 440 311 112 63 201 545
Thank you very much!
