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I'm doing some research about fertilizer and the effect on mitosis. I got the following results(simplified):

╔═════════════╦═══════════════╦════════════╗
║    phase    ║ without fert. ║ with fert. ║
╠═════════════╬═══════════════╬════════════╣
║ interphase  ║ 400           ║        200 ║
║ prophase    ║ 30            ║        200 ║
║ metaphase   ║ 60            ║         40 ║
║ anaphase    ║ 20            ║         50 ║
╚═════════════╩═══════════════╩════════════╝

However which test (chi square or t test) should be used for this data? I'm only familiar with chi square to compare expected ratios e.g. 1:2:3:2 with some observed numbers and not to compare different "samples". So please explain why the test you recommend should be used.

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  • $\begingroup$ Is this a homework question? If so, you should add the self-study tag, and read the details under it. $\endgroup$ Dec 18 '16 at 18:47
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    $\begingroup$ No I just removed and simplified a lot of the data, this would make the table huge @JeremyMiles $\endgroup$ Dec 18 '16 at 19:23
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You don't say what the results are precisely or what hypothesis you are testing.

A chi-square test is easy to identify here as comparing whether columns of a two-way table of frequencies have equal probability distributions, or equivalently its row frequencies are in proportion to the total column frequencies. Whether it's what you want is the important question.

If it is, the test is otiose. What a quick look at the table suggests is amply confirmed: the two columns are quite different and $P$-values are minute on any assumption.

Here are some results from Stata, but any decent software allows this test:

. tabchii  400 200 \ 30 200 \ 60 40 \ 20 50 , pearson

          observed frequency
          expected frequency
          Pearson residual

----------------------------
          |       col       
      row |       1        2
----------+-----------------
        1 |     400      200
          | 306.000  294.000
          |   5.374   -5.482
          | 
        2 |      30      200
          | 117.300  112.700
          |  -8.061    8.223
          | 
        3 |      60       40
          |  51.000   49.000
          |   1.260   -1.286
          | 
        4 |      20       50
          |  35.700   34.300
          |  -2.628    2.681
----------------------------

         Pearson chi2(3) = 208.8595   Pr = 0.000
likelihood-ratio chi2(3) = 225.5996   Pr = 0.000

. ret li

scalars:
                  r(N) =  1000
                  r(r) =  4
                  r(c) =  2
               r(chi2) =  208.8595272477521
                  r(p) =  5.13643443208e-45
            r(chi2_lr) =  225.5995641056678
               r(p_lr) =  1.23651307520e-48

The Pearson residual (observed $-$ expected) / root of expected quantifies how far individual cells depart from a null expectation row total $\times$ column total / table total.

The P-value is here $5 \times 10^{-45}$ or $10^{-48}$, or minute either way.

If your results are something else, then this bet is off. I can't imagine where a t test would be used here.

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  • $\begingroup$ Thankyou for the explanation!, I thought the t test wouldn't be a good option (but searching on the internet gave we a lot of question about the difference, so I got confused). $\endgroup$ Dec 18 '16 at 19:22

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