Kenny's Condition A #2
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Hello I got a 6x3 table with
this descriptive result:
The chi-square-test gives
a significant result:
If I’m right,
chi-square test is a sort of omnibus-test. How can I perform with PASW 18 an a
posteriori test in order to compare the relative amount of Yes within the three
categories of the variable “function” ? Or, in other words: which
one are significant: 40:60, 24:76 or/and 54:46? I appreciate any help. Tom |
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Thank you for your email. I will be out of the
office beginning Friday, August 20th, returning on Monday, August 30th and will
respond to your message when I return. Stephen Salbod Data Analyst Pace University Psychology Department, NYC |
Re: Chi Square test and a posteriori multiple comparison
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In reply to this post by Tom
The percentages for the Yes response are already in the table,
as row percentages: Funktion=1 à %Yes=12/20=60% Funktion=2 à %Yes=129/170=75.9% Funktion=3 à %Yes=456/980=46.53% Total à% Yes=597/1170=51.0% The other percentages in the table refer to column totals and
table total, and are irrelevant for your purposes. Now for significance. Are these percentages significant? It
depends on what you mean. Significantly different from zero? Probably yes
because they are quite large, and sample sizes are large except for Funktion=1,
but even in that case it may be sufficient. In general, use a t-test to do
that comparison, in SPSS (comparing two independent samples). Significantly different
from the overall percentage, which is 51%? Use the same t-test procedure in
SPSS, comparing percentages to the fixed value 51%. In both cases: For the
first category of funktion, for which the sample is exceedingly small (20
cases) the normality assumption of the t-test may not be met; so you may want
to use nonparametric tests instead. Hector De: SPSSX(r) Discussion
[mailto:[hidden email]] En nombre de Balmer Thomas Hello I got a 6x3 table with this descriptive result:
The chi-square-test gives a significant result:
If I’m right, chi-square test is a sort of omnibus-test. How can I
perform with PASW 18 an a posteriori test in order to compare the relative
amount of Yes within the three categories of the variable “function” ? Or, in
other words: which one are significant: 40:60, 24:76 or/and 54:46? I appreciate any help. Tom Se certificó
que el correo entrante no contiene virus. |
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Re: Chi Square test and a posteriori multiple comparison
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In reply to this post by Tom
Thomas:
I would do another procedure rather than the one Hector suggests. The reason is that the data are di- and tri-chotomous, and t-tests require that the raw data be continuous for the test variable. Instead: run a logistic regression with the Yes/No as the dependent, dichotomous outcome variable. Put in one independent variable, the tri-chotomous (1,2,3) variable. If you have some a priori hypotheses as to which of the three rows are different, consider using them instead. The a priori comparisons number k -1 = 3 - 1 = 2, so that you are only allowed 2 of the 3 comparisons you mentioned. This allows for orthogonal comparisons, with no need to adjust the p-level such as in the 3 post-hoc tests you suggested. There are two basic patterns as to how to compare the 3 cells this way: A. Compare the hypothesized most different cell to the other two combined, then contrast the two that were first combined against each other. B. Pick a "control" cell and do direct comparisons of it with EACH of the other 2, respectively. Procedure A uses all the data on the first test, but is robust on that first test. Procedure B uses 2/3 of the data on each test, but is more specific. the trick is judging whether you want more robustness or more specificity. The names of the orthogonal a priori test is logistic regression are the same as in general linear models. Note that some of them are NOT orthogonal. Joe Burleson ________________________________ From: SPSSX(r) Discussion [[hidden email]] On Behalf Of Balmer Thomas [[hidden email]] Sent: 20 August 2010 10:56 To: [hidden email] Subject: Chi Square test and a posteriori multiple comparison Hello I got a 6x3 table with this descriptive result: Funktion * G4 Kreuztabelle G4 Gesamt No Yes Funktion 1 Anzahl 8 12 20 % innerhalb von Funktion 40.0% 60.0% 100.0% % innerhalb von G4 1.4% 2.0% 1.7% % der Gesamtzahl .7% 1.0% 1.7% 2 Anzahl 41 129 170 % innerhalb von Funktion 24.1% 75.9% 100.0% % innerhalb von G4 7.2% 21.6% 14.5% % der Gesamtzahl 3.5% 11.0% 14.5% 3 Anzahl 524 456 980 % innerhalb von Funktion 53.5% 46.5% 100.0% % innerhalb von G4 91.4% 76.4% 83.8% % der Gesamtzahl 44.8% 39.0% 83.8% Gesamt Anzahl 573 597 1170 % innerhalb von Funktion 49.0% 51.0% 100.0% % innerhalb von G4 100.0% 100.0% 100.0% % der Gesamtzahl 49.0% 51.0% 100.0% The chi-square-test gives a significant result: Chi-Quadrat-Tests Wert df Asymptotische Signifikanz (2-seitig) Chi-Quadrat nach Pearson 50.600a 2 .000 Likelihood-Quotient 52.878 2 .000 Zusammenhang linear-mit-linear 39.583 1 .000 Anzahl der gültigen Fälle 1170 a. 0 Zellen (.0%) haben eine erwartete Häufigkeit kleiner 5. Die minimale erwartete Häufigkeit ist 9.79. If I’m right, chi-square test is a sort of omnibus-test. How can I perform with PASW 18 an a posteriori test in order to compare the relative amount of Yes within the three categories of the variable “function” ? Or, in other words: which one are significant: 40:60, 24:76 or/and 54:46? I appreciate any help. Tom ===================== To manage your subscription to SPSSX-L, send a message to [hidden email] (not to SPSSX-L), with no body text except the command. To leave the list, send the command SIGNOFF SPSSX-L For a list of commands to manage subscriptions, send the command INFO REFCARD |
Re: Chi Square test and a posteriori multiple comparison
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What Joseph Burleson proposes is right, but it requires entering the much
more complicated field of logistic regression. For the case at hand I tilt to the view that one can use t-test nonetheless. For one thing, dichotomous yes/no variables can be treated as interval measures: the percentage of cases with the value 1 (if the other value is 0) equals the mean of the variable considered as an interval scale. T-tests are, besides, frequently used to assess percentage differences. However, CROSSTABS provides other tests for contingency tables that may be more adequate, such as those based on chi square. Hector -----Mensaje original----- De: SPSSX(r) Discussion [mailto:[hidden email]] En nombre de Burleson,Joseph A. Enviado el: Friday, August 20, 2010 2:22 PM Para: [hidden email] Asunto: Re: Chi Square test and a posteriori multiple comparison Thomas: I would do another procedure rather than the one Hector suggests. The reason is that the data are di- and tri-chotomous, and t-tests require that the raw data be continuous for the test variable. Instead: run a logistic regression with the Yes/No as the dependent, dichotomous outcome variable. Put in one independent variable, the tri-chotomous (1,2,3) variable. If you have some a priori hypotheses as to which of the three rows are different, consider using them instead. The a priori comparisons number k -1 = 3 - 1 = 2, so that you are only allowed 2 of the 3 comparisons you mentioned. This allows for orthogonal comparisons, with no need to adjust the p-level such as in the 3 post-hoc tests you suggested. There are two basic patterns as to how to compare the 3 cells this way: A. Compare the hypothesized most different cell to the other two combined, then contrast the two that were first combined against each other. B. Pick a "control" cell and do direct comparisons of it with EACH of the other 2, respectively. Procedure A uses all the data on the first test, but is robust on that first test. Procedure B uses 2/3 of the data on each test, but is more specific. the trick is judging whether you want more robustness or more specificity. The names of the orthogonal a priori test is logistic regression are the same as in general linear models. Note that some of them are NOT orthogonal. Joe Burleson ________________________________ From: SPSSX(r) Discussion [[hidden email]] On Behalf Of Balmer Thomas [[hidden email]] Sent: 20 August 2010 10:56 To: [hidden email] Subject: Chi Square test and a posteriori multiple comparison Hello I got a 6x3 table with this descriptive result: Funktion * G4 Kreuztabelle G4 Gesamt No Yes Funktion 1 Anzahl 8 12 20 % innerhalb von Funktion 40.0% 60.0% 100.0% % innerhalb von G4 1.4% 2.0% 1.7% % der Gesamtzahl .7% 1.0% 1.7% 2 Anzahl 41 129 170 % innerhalb von Funktion 24.1% 75.9% 100.0% % innerhalb von G4 7.2% 21.6% 14.5% % der Gesamtzahl 3.5% 11.0% 14.5% 3 Anzahl 524 456 980 % innerhalb von Funktion 53.5% 46.5% 100.0% % innerhalb von G4 91.4% 76.4% 83.8% % der Gesamtzahl 44.8% 39.0% 83.8% Gesamt Anzahl 573 597 1170 % innerhalb von Funktion 49.0% 51.0% 100.0% % innerhalb von G4 100.0% 100.0% 100.0% % der Gesamtzahl 49.0% 51.0% 100.0% The chi-square-test gives a significant result: Chi-Quadrat-Tests Wert df Asymptotische Signifikanz (2-seitig) Chi-Quadrat nach Pearson 50.600a 2 .000 Likelihood-Quotient 52.878 2 .000 Zusammenhang linear-mit-linear 39.583 1 .000 Anzahl der gültigen Fälle 1170 a. 0 Zellen (.0%) haben eine erwartete Häufigkeit kleiner 5. Die minimale erwartete Häufigkeit ist 9.79. If Im right, chi-square test is a sort of omnibus-test. How can I perform with PASW 18 an a posteriori test in order to compare the relative amount of Yes within the three categories of the variable function ? Or, in other words: which one are significant: 40:60, 24:76 or/and 54:46? I appreciate any help. Tom ===================== To manage your subscription to SPSSX-L, send a message to [hidden email] (not to SPSSX-L), with no body text except the command. To leave the list, send the command SIGNOFF SPSSX-L For a list of commands to manage subscriptions, send the command INFO REFCARD Se certificó que el correo entrante no contiene virus. Comprobada por AVG - www.avg.es Versión: 8.5.441 / Base de datos de virus: 271.1.1/3081 - Fecha de la versión: 08/20/10 06:35:00 ===================== To manage your subscription to SPSSX-L, send a message to [hidden email] (not to SPSSX-L), with no body text except the command. To leave the list, send the command SIGNOFF SPSSX-L For a list of commands to manage subscriptions, send the command INFO REFCARD |
Re: Chi Square test and a posteriori multiple comparison
Administrator
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In reply to this post by Burleson,Joseph A.
My notes on categorical data (http://www.angelfire.com/wv/bwhomedir/notes/categorical.pdf, section 2.8) show how to partition the overall likelihood ratio chi-square into orthogonal components. I've not checked, but because I'm using the likelihood ratio chi-square, I expect the method shown in my notes will give identical results to two orthogonal contrasts carried out via logistic regression, as proposed by Joe.
By the way, the example in the notes also uses a 3x2 table, so it should be fairly easy to adapt to Thomas's problem. HTH.
--
Bruce Weaver bweaver@lakeheadu.ca http://sites.google.com/a/lakeheadu.ca/bweaver/ "When all else fails, RTFM." PLEASE NOTE THE FOLLOWING: 1. My Hotmail account is not monitored regularly. To send me an e-mail, please use the address shown above. 2. The SPSSX Discussion forum on Nabble is no longer linked to the SPSSX-L listserv administered by UGA (https://listserv.uga.edu/). |
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In reply to this post by Tom
Ok - a lot of informations, thank you all! |
Re: AW: Re: Chi Square test and a posteriori multiple comparison
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Administrator
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It's still not clear to me what question, or questions, you are trying to answer.
The overall chi-square test is statistically significant. This tells you that the row and column variables are not independent of each other -- i.e., the proportion of Yes's is not the same at each of the 3 levels. The SPLIT FILE analysis you show asks if the proportion of Yes's is significantly different from .5 at each level of the other variable, but it makes no comparisons across the 3 levels. The orthogonal contrast method picks two rows from the table, and asks if the proportion of Yes's differs across those two rows. Then it collapses those two rows together, and asks if the proportion of Yes's in those two rows combined differs from the proportion of Yes's in the 3rd row. If the overall chi-square test is of interest, it seems to me that the questions addressed by the orthogonal contrast method are probably more interesting than the SPLIT FILE questions. Finally, now that you've shown us the counts, it's possible to do the analyses using the WEIGHT command. E.g., * Read in the data. data list list / r c kount (3f5.0). begin data 1 1 8 1 2 12 2 1 129 2 2 41 3 1 456 3 2 524 end data. weight by kount. crosstabs r by c / stat = chisq /cells = count row. * The likelihood ratio chi-square for the 3x2 table = 53.152 . * Row 2 is the one that looks a lot different from the other two. * So for the orthogonal contrasts, I'd like to compare rows 3 and 1, * and then 3 and 1 pooled vs 2. * Row 1 vs Row 3 . temporary. select if any(r,1,3). /* Omit row 2 . crosstabs r by c / stat = chisq /cells = count row. * Rows 1 & 3 pooled vs Row 2 . temporary. recode r (3=1). /* Temporarily recode 3 to 1. crosstabs r by c / stat = chisq /cells = count row. * Likelihood ratio chi-square for first table = .339 . * Likelihood ratio chi-square for second table = 52.813 . * The sum of the two = 53.152, which is the value for * the overall table. So the likelihood ratio chi-square * values for the two orthogonal components add up to the * total, as they should do. * The SPLIT FILE method, which IMO, is probably not what you want. split file by r. NPAR TESTS /CHISQUARE=c /EXPECTED=EQUAL /MISSING ANALYSIS. split file off. HTH.
--
Bruce Weaver bweaver@lakeheadu.ca http://sites.google.com/a/lakeheadu.ca/bweaver/ "When all else fails, RTFM." PLEASE NOTE THE FOLLOWING: 1. My Hotmail account is not monitored regularly. To send me an e-mail, please use the address shown above. 2. The SPSSX Discussion forum on Nabble is no longer linked to the SPSSX-L listserv administered by UGA (https://listserv.uga.edu/). |
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Thanks, Bruce, yes, these orthogonal contrast were what I've been looking for.
-----Ursprüngliche Nachricht----- Von: SPSSX(r) Discussion [mailto:[hidden email]] Im Auftrag von Bruce Weaver Gesendet: Montag, 23. August 2010 23:58 An: [hidden email] Betreff: Re: AW: Re: Chi Square test and a posteriori multiple comparison It's still not clear to me what question, or questions, you are trying to answer. The overall chi-square test is statistically significant. This tells you that the row and column variables are not independent of each other -- i.e., the proportion of Yes's is not the same at each of the 3 levels. The SPLIT FILE analysis you show asks if the proportion of Yes's is significantly different from .5 at each level of the other variable, but it makes no comparisons across the 3 levels. The orthogonal contrast method picks two rows from the table, and asks if the proportion of Yes's differs across those two rows. Then it collapses those two rows together, and asks if the proportion of Yes's in those two rows combined differs from the proportion of Yes's in the 3rd row. If the overall chi-square test is of interest, it seems to me that the questions addressed by the orthogonal contrast method are probably more interesting than the SPLIT FILE questions. Finally, now that you've shown us the counts, it's possible to do the analyses using the WEIGHT command. E.g., * Read in the data. data list list / r c kount (3f5.0). begin data 1 1 8 1 2 12 2 1 129 2 2 41 3 1 456 3 2 524 end data. weight by kount. crosstabs r by c / stat = chisq /cells = count row. * The likelihood ratio chi-square for the 3x2 table = 53.152 . * Row 2 is the one that looks a lot different from the other two. * So for the orthogonal contrasts, I'd like to compare rows 3 and 1, * and then 3 and 1 pooled vs 2. * Row 1 vs Row 3 . temporary. select if any(r,1,3). /* Omit row 2 . crosstabs r by c / stat = chisq /cells = count row. * Rows 1 & 3 pooled vs Row 2 . temporary. recode r (3=1). /* Temporarily recode 3 to 1. crosstabs r by c / stat = chisq /cells = count row. * Likelihood ratio chi-square for first table = .339 . * Likelihood ratio chi-square for second table = 52.813 . * The sum of the two = 53.152, which is the value for * the overall table. So the likelihood ratio chi-square * values for the two orthogonal components add up to the * total, as they should do. * The SPLIT FILE method, which IMO, is probably not what you want. split file by r. NPAR TESTS /CHISQUARE=c /EXPECTED=EQUAL /MISSING ANALYSIS. split file off. HTH. Balmer Thomas wrote: > > Ok - a lot of informations, thank you all! > > So then, I did what Martin suggested (splitting the file), which gave me: > > Funktion G4 > 1 Chi-Quadrat .800a > df 1 > Asymptotische Signifikanz .371 > Exakte Signifikanz .503 > Punkt-Wahrscheinlichkeit .240 > 2 Chi-Quadrat 45.553b > df 1 > Asymptotische Signifikanz .000 > Exakte Signifikanz .000 > Punkt-Wahrscheinlichkeit .000 > 3 Chi-Quadrat 4.718c > df 1 > Asymptotische Signifikanz .030 > Exakte Signifikanz .032 > Punkt-Wahrscheinlichkeit .005 > a. Bei 0 Zellen (.0%) werden weniger als 5 Häufigkeiten erwartet. Die > kleinste erwartete Zellenhäufigkeit ist 10.0. > b. Bei 0 Zellen (.0%) werden weniger als 5 Häufigkeiten erwartet. Die > kleinste erwartete Zellenhäufigkeit ist 85.0. > c. Bei 0 Zellen (.0%) werden weniger als 5 Häufigkeiten erwartet. Die > kleinste erwartete Zellenhäufigkeit ist 490.0. > > This means, if Im right, that the proportions of function 2 and 3 are > significantly different from equal (50:50), which is just a comparison > within the category/group, not of the PROPORTION of the Yes/No between the > groups. Or: does these results let me say, that group/ function 1 (with > 60% Yes) does not answer significantly less Yes than group/ function 2 > (with 75%)? I suppose, rather no > > If I got Bruce right, the Maximum Likelihood chi-square would be a > possibility. Along with his paper I produced this: First, all the three > functions, then, second, a comparison of function 1 and 2 only: > L2 Observed E ln(0/E) O*ln(O/E) > 1yes 8 10 -0.22314355 -1.78514841 > 2 129 86 0.40546511 52.3049989 > 3 456 497 -0.08609722 -39.2603308 > 1no 12 10 0.18232156 2.18785868 > 2 41 84 -0.71724473 -29.407034 > 3 524 483 0.08147503 42.6929161 > 26.7332605 > L2 53.466521 > df=2 highly sign. > Which means, that the three groups differ significantly from each other. > > > O E ln(O/E) O*ln(O/E) > 1 yes 8 14 -0.55961579 -4.4769263 > 2 yes 129 123 0.04762805 6.14401832 > 1 no 12 6 0.69314718 8.31776617 > 2 no 41 47 -0.13657554 -5.59959694 > 4.38526125 > L2 8.7705225 > df=1 ** > > > Which means, that function 1 and 2 are signficantly different answering > Yes. > > Am I right, that this second solution gives me the desired comparison for > two of the three independent groups, which the first solution doesnt? > Is the expected frequency (funktion 1 no) of only 6 a problem? > And: How can I do this with PASW/SPSS? > > > > [hidden email] > http://www.phbern.ch/weiterbildung > -----Ursprüngliche Nachricht----- > Von: SPSSX(r) Discussion [mailto:[hidden email]] Im Auftrag von > Bruce Weaver > Gesendet: Freitag, 20. August 2010 21:21 > An: [hidden email] > Betreff: Re: Chi Square test and a posteriori multiple comparison > > My notes on categorical data > (http://www.angelfire.com/wv/bwhomedir/notes/categorical.pdf, section 2.8) > show how to partition the overall likelihood ratio chi-square into > orthogonal components. I've not checked, but because I'm using the > likelihood ratio chi-square, I expect the method shown in my notes will > give > identical results to two orthogonal contrasts carried out via logistic > regression, as proposed by Joe. > > By the way, the example in the notes also uses a 3x2 table, so it should > be > fairly easy to adapt to Thomas's problem. > > HTH. > > > > Burleson,Joseph A. wrote: >> >> Thomas: >> >> I would do another procedure rather than the one Hector suggests. The >> reason is that the data are di- and tri-chotomous, and t-tests require >> that the raw data be continuous for the test variable. >> >> Instead: run a logistic regression with the Yes/No as the dependent, >> dichotomous outcome variable. Put in one independent variable, the >> tri-chotomous (1,2,3) variable. >> >> If you have some a priori hypotheses as to which of the three rows are >> different, consider using them instead. The a priori comparisons number k >> -1 = 3 - 1 = 2, so that you are only allowed 2 of the 3 comparisons you >> mentioned. This allows for orthogonal comparisons, with no need to adjust >> the p-level such as in the 3 post-hoc tests you suggested. >> >> There are two basic patterns as to how to compare the 3 cells this way: >> A. Compare the hypothesized most different cell to the other two >> combined, >> then contrast the two that were first combined against each other. >> B. Pick a "control" cell and do direct comparisons of it with EACH of the >> other 2, respectively. >> >> Procedure A uses all the data on the first test, but is robust on that >> first test. >> Procedure B uses 2/3 of the data on each test, but is more specific. >> >> the trick is judging whether you want more robustness or more >> specificity. >> >> The names of the orthogonal a priori test is logistic regression are the >> same as in general linear models. Note that some of them are NOT >> orthogonal. >> >> Joe Burleson >> >> ________________________________ >> From: SPSSX(r) Discussion [[hidden email]] On Behalf Of Balmer >> Thomas [[hidden email]] >> Sent: 20 August 2010 10:56 >> To: [hidden email] >> Subject: Chi Square test and a posteriori multiple comparison >> >> Hello >> >> I got a 6x3 table with this descriptive result: >> Funktion * G4 Kreuztabelle >> >> >> >> G4 >> >> Gesamt >> >> No >> >> Yes >> >> Funktion >> >> 1 >> >> Anzahl >> >> 8 >> >> 12 >> >> 20 >> >> % innerhalb von Funktion >> >> 40.0% >> >> 60.0% >> >> 100.0% >> >> % innerhalb von G4 >> >> 1.4% >> >> 2.0% >> >> 1.7% >> >> % der Gesamtzahl >> >> .7% >> >> 1.0% >> >> 1.7% >> >> 2 >> >> Anzahl >> >> 41 >> >> 129 >> >> 170 >> >> % innerhalb von Funktion >> >> 24.1% >> >> 75.9% >> >> 100.0% >> >> % innerhalb von G4 >> >> 7.2% >> >> 21.6% >> >> 14.5% >> >> % der Gesamtzahl >> >> 3.5% >> >> 11.0% >> >> 14.5% >> >> 3 >> >> Anzahl >> >> 524 >> >> 456 >> >> 980 >> >> % innerhalb von Funktion >> >> 53.5% >> >> 46.5% >> >> 100.0% >> >> % innerhalb von G4 >> >> 91.4% >> >> 76.4% >> >> 83.8% >> >> % der Gesamtzahl >> >> 44.8% >> >> 39.0% >> >> 83.8% >> >> Gesamt >> >> Anzahl >> >> 573 >> >> 597 >> >> 1170 >> >> % innerhalb von Funktion >> >> 49.0% >> >> 51.0% >> >> 100.0% >> >> % innerhalb von G4 >> >> 100.0% >> >> 100.0% >> >> 100.0% >> >> % der Gesamtzahl >> >> 49.0% >> >> 51.0% >> >> 100.0% >> >> >> >> The chi-square-test gives a significant result: >> >> Chi-Quadrat-Tests >> >> >> >> Wert >> >> df >> >> Asymptotische Signifikanz (2-seitig) >> >> Chi-Quadrat nach Pearson >> >> 50.600a >> >> 2 >> >> .000 >> >> Likelihood-Quotient >> >> 52.878 >> >> 2 >> >> .000 >> >> Zusammenhang linear-mit-linear >> >> 39.583 >> >> 1 >> >> .000 >> >> Anzahl der gültigen Fälle >> >> 1170 >> >> >> >> >> >> a. 0 Zellen (.0%) haben eine erwartete Häufigkeit kleiner 5. Die minimale >> erwartete Häufigkeit ist 9.79. >> >> >> If Im right, chi-square test is a sort of omnibus-test. How can I >> perform >> with PASW 18 an a posteriori test in order to compare the relative amount >> of Yes within the three categories of the variable function ? Or, in >> other words: which one are significant: 40:60, 24:76 or/and 54:46? >> >> I appreciate any help. >> Tom >> >> ===================== >> To manage your subscription to SPSSX-L, send a message to >> [hidden email] (not to SPSSX-L), with no body text except the >> command. To leave the list, send the command >> SIGNOFF SPSSX-L >> For a list of commands to manage subscriptions, send the command >> INFO REFCARD >> >> > > > ----- > -- > Bruce Weaver > [hidden email] > http://sites.google.com/a/lakeheadu.ca/bweaver/ > > "When all else fails, RTFM." > > NOTE: My Hotmail account is not monitored regularly. > To send me an e-mail, please use the address shown above. > > -- > View this message in context: > http://spssx-discussion.1045642.n5.nabble.com/Kenny-s-Condition-A-2-tp2641949p2642709.html > Sent from the SPSSX Discussion mailing list archive at Nabble.com. > > ===================== > To manage your subscription to SPSSX-L, send a message to > [hidden email] (not to SPSSX-L), with no body text except the > command. To leave the list, send the command > SIGNOFF SPSSX-L > For a list of commands to manage subscriptions, send the command > INFO REFCARD > > > ----- -- Bruce Weaver [hidden email] http://sites.google.com/a/lakeheadu.ca/bweaver/ "When all else fails, RTFM." NOTE: My Hotmail account is not monitored regularly. To send me an e-mail, please use the address shown above. -- View this message in context: http://spssx-discussion.1045642.n5.nabble.com/Kenny-s-Condition-A-2-tp2641949p2645397.html Sent from the SPSSX Discussion mailing list archive at Nabble.com. ===================== To manage your subscription to SPSSX-L, send a message to [hidden email] (not to SPSSX-L), with no body text except the command. To leave the list, send the command SIGNOFF SPSSX-L For a list of commands to manage subscriptions, send the command INFO REFCARD ===================== To manage your subscription to SPSSX-L, send a message to [hidden email] (not to SPSSX-L), with no body text except the command. To leave the list, send the command SIGNOFF SPSSX-L For a list of commands to manage subscriptions, send the command INFO REFCARD |
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