```Date: Thu, 1 Sep 2005 02:54:35 -0400 Reply-To: Gabor Toth Sender: "SPSSX(r) Discussion" From: Gabor Toth 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! ```

Back to: Top of message | Previous page | Main SPSSX-L page