z****k
发帖数: 1057 | 1 投票选美，四个mm候选，按身高从高到矮依次编号1234
1号mm得票中，19票来自男性，17票来自女性
2号mm得票中，21票来自男性，14票来自女性
3号mm得票中，13票来自男性，17票来自女性
4号mm得票中，16票来自男性，21票来自女性
问：相对女人来讲，男人是否更加喜欢高个mm? 应该用什么检验方法？
（假设每一票都来自一个不同的人） |
z*******n
发帖数: 15481 | |
z*****9
发帖数: 669 | 3 你这confounding factor太多了，拿郑海霞和刘璇让我选，我肯定选刘璇；但如果是志
玲姐姐和凤姐，我肯定选志玲姐姐
【在 z****k 的大作中提到】
: 投票选美，四个mm候选，按身高从高到矮依次编号1234
: 1号mm得票中，19票来自男性，17票来自女性
: 2号mm得票中，21票来自男性，14票来自女性
: 3号mm得票中，13票来自男性，17票来自女性
: 4号mm得票中，16票来自男性，21票来自女性
: 问：相对女人来讲，男人是否更加喜欢高个mm? 应该用什么检验方法？
: （假设每一票都来自一个不同的人）
|
o****o
发帖数: 8077 | 4 每个mm的总票数都不一样，她们面对的投票男生都是一群人么？ |
a******n
发帖数: 11246 | 5 大牛就是大牛，问到了点子上：）
多正确答案。尊重自己的选择。努力为之奋斗。
【在 o****o 的大作中提到】
: 每个mm的总票数都不一样，她们面对的投票男生都是一群人么？
|
z****k
发帖数: 1057 | 6 是同一群人
而且必须四选一
【在 o****o 的大作中提到】
: 每个mm的总票数都不一样，她们面对的投票男生都是一群人么？
|
o****o
发帖数: 8077 | 7 sounds like a trend test on a 2X4 contingency table
you can use Cochran-Armitage trend test in PROC FREQ.
Or you can use Score Statistics from PROC LOGISTIC, which is the same as a 2-sided trend test.I treat Gender as response variable because I read the original post as ppl are given 4 different treatments which are MM of 4 different height categories and vote for their favorite.
***************************************************************;
data test;
input height_category gender $ count;
cards;
1 M 19
1 F 17
2 M 21
2 F 14
3 M 13
3 F 17
4 M 16
4 F 21
;
run;
ods graphics on;
proc freq data=test
table gender*height_category
/ trend plots=freqplot(twoway=stacked);
weight count;
run;
ods graphics off;
proc logistic data=test;
model gender = height_category;
freq count;
run;
note that both the Score Statistic and the two-sided trend Z statistic have a p-value of 0.2337
Looks like with MM's of higher height, it is not more likely, in statistical sense, that male will favor them comparing to females, even though there is weak tendency.
Hope my explanation is correct.
【在 z****k 的大作中提到】
: 是同一群人
: 而且必须四选一
|
z****k
发帖数: 1057 | 8 谢谢大牛
回答个问题连sas code都一起给了，实在太专业了
但是对于这个回答我仍然有一个小问题：
对于这样的分析方法，是不是暗含一个前提假设就是每一个数据点的count之间是相互
独立的？
我这个问题里每个人都必须投一票，不投给1,2,3,就必须投给4，因此count之间是相
关的
2-sided trend test.I treat Gender as response variable because I read the
original post as ppl are given 4 different treatments which are MM of 4
different height categories a
【在 o****o 的大作中提到】
: sounds like a trend test on a 2X4 contingency table
: you can use Cochran-Armitage trend test in PROC FREQ.
: Or you can use Score Statistics from PROC LOGISTIC, which is the same as a 2-sided trend test.I treat Gender as response variable because I read the original post as ppl are given 4 different treatments which are MM of 4 different height categories and vote for their favorite.
: ***************************************************************;
: data test;
: input height_category gender $ count;
: cards;
: 1 M 19
: 1 F 17
: 2 M 21
|
D*********2
发帖数: 535 | 9 they r iid, each record means each person's vote, not the aggregated count.
I was thinking should we use ordinal logistic regression? since 1, 2, 3, 4
here are ranked. do u have other variables? |
o****o
发帖数: 8077 | 10 of course you can use ordinal logistic regression
model height_category = gender /link=clogit;
you will find that the score statistic and its p-value are the same.
【在 D*********2 的大作中提到】
: they r iid, each record means each person's vote, not the aggregated count.
: I was thinking should we use ordinal logistic regression? since 1, 2, 3, 4
: here are ranked. do u have other variables?
|
|
|
j*****e
发帖数: 182 | 11 data one;
input choice gender $ count;
cards;
1 f 17
1 m 19
2 f 14
2 m 21
...;
run;
/*CMH test for norminalXordinal table
Use the chi-square test for "Row mean score differ". This is equivalent to
ANOVA */
proc freq data=one;
table gender*choice/cmh;
weight count;
run;
/*Cumulative logit model.
You have to understand the model before your interpretating the SAS output.
*/
proc genmod data=one;
class gender/param=ref;
model choice=gender/dist=multi link=clogit;
weight count;
run; |
B******y
发帖数: 9065 | 12 I think your question is reasonable.
I couldn't provide a solution here, but would agree with you. Each count are
dependent, while Cochran-Armitage trend test is based on independence
assumption so it shouldn't be applicable to this case.
【在 z****k 的大作中提到】
: 谢谢大牛
: 回答个问题连sas code都一起给了，实在太专业了
: 但是对于这个回答我仍然有一个小问题：
: 对于这样的分析方法，是不是暗含一个前提假设就是每一个数据点的count之间是相互
: 独立的？
: 我这个问题里每个人都必须投一票，不投给1,2,3,就必须投给4，因此count之间是相
: 关的
:
: 2-sided trend test.I treat Gender as response variable because I read the
: original post as ppl are given 4 different treatments which are MM of 4
|
o****o
发帖数: 8077 | 13 I guess his question only implies "reference category is redundant".
are
【在 B******y 的大作中提到】
: I think your question is reasonable.
: I couldn't provide a solution here, but would agree with you. Each count are
: dependent, while Cochran-Armitage trend test is based on independence
: assumption so it shouldn't be applicable to this case.
|
z****k
发帖数: 1057 | 14 这句话没懂
能麻烦稍微解释一下吗？
谢谢！
【在 o****o 的大作中提到】
: I guess his question only implies "reference category is redundant".
:
: are
|
p***r
发帖数: 920 | 15 I think the assumption is:
M: p_m1, p_m2, p_m3, 1-p_m1-p_pm2-p_m3
F: p_f1, p_f2, p_f3, 1-p_f1-p_pf2-p_f3
and LZ want to testing hypothesis:
H_a: p_m1>p_f1
2-sided trend test.I treat Gender as response variable because I read the
original post as ppl are given 4 different treatments which are MM of 4
different height categories and vote for their favorite.
【在 o****o 的大作中提到】
: sounds like a trend test on a 2X4 contingency table
: you can use Cochran-Armitage trend test in PROC FREQ.
: Or you can use Score Statistics from PROC LOGISTIC, which is the same as a 2-sided trend test.I treat Gender as response variable because I read the original post as ppl are given 4 different treatments which are MM of 4 different height categories and vote for their favorite.
: ***************************************************************;
: data test;
: input height_category gender $ count;
: cards;
: 1 M 19
: 1 F 17
: 2 M 21
|
