# chi squared with proportions

We want to know whether there are any differences in the location of mole hills that occur in two fields.

We have been looking at the distance from a fence that mole hills occur. We have separated the distance into bins, 0-2 m = bin 1, 2-4 = bin 2, 4-6 = Bin 3, 6-8 is bin 4 etc to bin 7 which is >12 m.

In each bin we have counted the frequency of mole hills that occur in each bin. Field one -- 1=118 2=158 3=89 4=84 5=95 6=84 7=22

This was then repeated for a second field. We want to know if there is any difference in the frequency of mole hills that occur in each bin a) within the same field and b) when we compare frequencies from both fields.

Ho- there is no difference in frequency of mole hills observed in each bin in field one OR field 2 (there is no difference in location in field one/field two)

Ho - there is no difference between the frequencies observed in each bin between field 1 and field 2. (There is no difference in mole hill location between field one and 2)

I thought just a contingency table as we have the frequency for mole hill observed however, for each bin we do not have a number for 'No mole hill' to compare it to.

Can you do cross tabs with just proportions?

    BIN * Field Crosstabulation
Field          Total
1      2
0.00-2.00   Count   118     60       178
Expected Count  104.0   74.0    178.0
% within BIN    66.3%   33.7%   100.0%

2.02-4.00   Count   158     121     279
Expected Count  163.1   115.9   279.0
% within BIN    56.6%   43.4%   100.0%

4.01-6.00   Count   89      99      188
Expected Count  109.9   78.1    188.0
% within BIN    47.3%   52.7%   100.0%

6.01-8.00   Count   84      85      169
Expected Count  98.8    70.2    169.0
% within BIN    49.7%   50.3%   100.0%

8.01-10.00  Count   95      40      135
Expected Count  78.9    56.1    135.0
% within BIN    70.4%   29.6%   100.0%

10.01-12.00 Count    84     22      106
Expected Count  62.0    44.0    106.0
% within BIN    79.2%   20.8%   100.0%

>12.00  Count       22      35      57
Expected Count  33.3    23.7    57.0
% within BIN    38.6%   61.4%   100.0%

Total       Count   650 462     1112
Expected Count  650.0   462.0   1112.0
% within BIN    58.5%   41.5%   100.0%

Chi-Square Tests
Value    df  Asymptotic Significance (2-sided)
Pearson Chi-Square 55.790a 6   .000
Linear-by-Linear   .283c   1   .595
N of Valid Cases   1112


But if I wanted to look at where the significant differences were and did multiple chi squared tests would this inflate the p-value similar to performing multiple t-tests?

• Relax. We'd be happy to help you. There's no need to make a mountain out of this. Feb 8, 2016 at 0:14
• Your null hypothesis shouldn't be "there is no difference in frequency of mole hills observed in each bin in Field 1"; you already know there's a difference in the observed frequencies. It should be that there's an equal chance of a molehill's occurring in any bin. (Which seems a little odd given that the last bin's any distance over 12m. How far from the fence did you look for molehills?) And you're probably assuming that each molehill's made independently of the others. Does that seem reasonable? - did you notice any clustering of molehills? Feb 8, 2016 at 10:26
• Anyway, given that in Field 1 you've counted 650 molehills, what would the expected counts be if the chances of a molehill's occurring in each bin were equal? Feb 8, 2016 at 10:27
• Hi, thanks for your help, I have put in my crosstabs in the question above. And yes all the observations were independent
– Beth
Feb 8, 2016 at 10:46

This is about a comment and a half, but I have a graph to offer.

The chi-square test is clear-cut but scientifically not very helpful. The two fields do look very different. Don't forget to plot the data, even for a simple table like this.

For example (bar heights are proportional to percent within each field, not the raw counts; the reverse may make more sense scientifically):

To go further, plot (e.g.) Pearson residuals, (observed $$-$$ expected) / square root of expected.

• Ok that's great thanks, Ill have a look into that now. Is there no way to perform post-hoc tests on chi squared to see where the differences are? ive read bits about a tukey test being performed however, I am just starting to do learn statistics and am not very mathematical therefore lots of lines of syntax don't make a lot of sense! @NickCox
– Beth
Feb 8, 2016 at 13:22
• I don't think there is a widely practised and easily defensible way of doing what you want. Not being mathematical need not be a barrier here; I have not studied mathematics formally since age 17. Feb 8, 2016 at 13:54
• @Beth: Note the Pearson residuals should be approximately normally distributed under the null - looking at their pattern across the bins & informally discounting "noise" is probably more useful than performing more tests. Feb 11, 2016 at 13:48