如何计算P值（急！）

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NS(45,116) NNS(44,908) Raw Per 1000000 raw Per1000000

Possession 125 2770.64 141 3139.75
Relationship 5 110.83 1 22.27
Health 7 155.16 1 22.27
Total 137 3036.62 143 3184.29

表格复制不了，所以有点乱。括号中是语料库大小， 红色是原始频数，后面是标准频数，possession, relationship,health分别是“have"这个词的不同义项，请问怎么分别用原始频数和标准频数计算这四个P值？
请高手指教！
 
回复: 如何计算P值（急！）

1) Define three variables: WordSense (string type), Learner (string type), Frequency (numeral without decimal point);
2) Enter data (Raw frequencies can be used in this case as the corpora are similar in size);
3) Weight cases for Frequency;
4) Crosstabulate (Analyze - Descriptive statistics - Crosstabs...) variables WordSense and Learner (do not forget to press Statistics and select Chi-square; as some frequencies are very low, chi or LL may not be reliable, so also press Exact and and select Exact (no real time simulation is required in this case);
5) You are taken to the output page.

chi square: 8.004, 2 df, p=0.018
LL: 8.807, 2 df, p=0.012
Fisher's Exact test: 7.724, p=0.017

In this case, the fisher's Exact test should be used. The differences between NS and NNS across the three wordsenses of "have" are statistically significant according to your data.

NOTE:
Instead of waiting for two days, you might have read the SPSS manual or googled the Web for an answer to such a question - answers are readily available from such sources. I have waited for two days in the hope of you can find an answer for yourself.
 
回复: 如何计算P值（急！）

I really tried to find the answer in a book or a website, but always ended up disappointed. I’m not interested in LL, ‘cause the result is hard to explain and most people seem to use chi-square more frequently. There is a lot of introduction about chi-square, but no example is provided related to corpus. So I had to turn to other sources for help.
Thanks a lot!! Your help means a lot to me!!
 
回复: 如何计算P值（急！）

wordsense learner frequency
total NS 137
NNS 143


I guess I can only compare the senses one by one? So I inserted the data as shown above and followed DR. Xiao’s instructions, and I got the following. How should I interpret the last three columns? Why is the result totally different from Dr. Xiao’s(he gave above)? And whatever group of frequency I put in, data in the last four columns barely changed, why? (df is forever 1, three kinds of significance is forever .000?) Shouldn't there be something like P=...?






Value df Asymp. Sig. (2-sided) Exact Sig. (2-sided) Exact Sig. (1-sided)
Pear Chi-Square 280.000(b) 1 .000 .000 .000
Conti Correction(a) 276.012 1 .000
Likeli Ratio 388.034 1 .000 .000 .000
F's Exact Test .000 .000
 
回复: 如何计算P值（急！）

1) Define three variables: WordSense (string type), Learner (string type), Frequency (numeral without decimal point);
2) Enter data (Raw frequencies can be used in this case as the corpora are similar in size);
3) Weight cases for Frequency;
4) Crosstabulate (Analyze - Descriptive statistics - Crosstabs...) variables WordSense and Learner (do not forget to press Statistics and select Chi-square; as some frequencies are very low, chi or LL may not be reliable, so also press Exact and and select Exact (no real time simulation is required in this case);

In this case, no info of corpus size is considered. So the assumption is that the two corpora are of equal size.
Following this, I think so long as we have normalized (proportional) frequencies of the search term, we can use this chi-square test method.

Alternatively, 2x2 crosstab chi-sqaure method (both freqs and corpus sizes) can be used instead.
 
回复: 如何计算P值（急！）

Dr.Xu.
How to set the varibles in spss when using 2x2 crosstab chi-sqaure method ?

thanks in advance.
 
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