2
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Relax. We'd be happy to help you. There's no need to make a mountain out of this. $\endgroup$ – gung Feb 8 '16 at 0:14
  • $\begingroup$ 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? $\endgroup$ – Scortchi Feb 8 '16 at 10:26
  • $\begingroup$ 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? $\endgroup$ – Scortchi Feb 8 '16 at 10:27
  • $\begingroup$ Hi, thanks for your help, I have put in my crosstabs in the question above. And yes all the observations were independent $\endgroup$ – Beth Feb 8 '16 at 10:46
2
$\begingroup$

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):

enter image description here

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

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – Beth Feb 8 '16 at 13:22
  • $\begingroup$ 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. $\endgroup$ – Nick Cox Feb 8 '16 at 13:54
  • $\begingroup$ @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. $\endgroup$ – Scortchi Feb 11 '16 at 13:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.