1
$\begingroup$

Hi I conducted the research where I measured germination of seeds in different ecological conditions. Each seed sample (one germination data) was exposed to different combination of ecological conditions - like growing on light or dark, on different temperatures (5,15,23°C), keeping seeds on warm or cold for 4, 8, 12 or 16 weeks.

Now I would like to do two-sample T-test to confirm that my plant definitively needs light to germinate. That is noticeable straight away from the data. Low germination in the dark and high in the light for any other combination of ecological conditions.

Now, I'm not sure which data can I put into the t-test since there are different combination of ecological parameters for each data point. If I limit the t-test only to the data set where I changed light or dark conditions and everything else is static, I don't get the wanted info from all of my data.

I don't have a broad knowledge of statistics, but am I trying to get the information from mixed data populations?

$\endgroup$
  • $\begingroup$ It seems unlikely that a $t$-test is going to be the best thing to do here. It looks like an analysis of variance (ANOVA) if I understand your study design. $\endgroup$ – mdewey Oct 13 '18 at 15:50
  • $\begingroup$ I am basically trying to prove that germination correlates more with light than with dark. So I have a continous variable with % germinations and binary variable with 1 (light) and 0 (dark). I can do Point-biserial correlation now, but I'm still not sure whether or not I should include all the data (where there are multiple variables changing) or just some of it (where the only variable changing is light/dark). $\endgroup$ – Alan_dj Oct 13 '18 at 16:09
  • 1
    $\begingroup$ It seems you have three factors: Illumination (two levels Light and Dark), Temperature (three levels 5, 15, 23), and Time (four levels). If data are normal and you have the same number of replications in each of the $2 \times 3 \times 4 = 24$ cells of your data table, then it's a simple three-way ANOVA design. You could test whether Illumination, Temperature, and Time make a difference; you could also test interactions such as Illumination * Temp, Illumination * Time, etc. $\endgroup$ – BruceET Oct 13 '18 at 17:37
  • 1
    $\begingroup$ How did you calculate the percentages? You might be better of with a binomial model if you know the total number of seeds and the number for seeds that germinated. Since percentages are often not normally distributed, ANOVA might not be the best choice. $\endgroup$ – Stefan Oct 13 '18 at 19:23
  • $\begingroup$ Hi Stefan! I used time-to event model described in this paper - researchgate.net/publication/… I used as the article suggest, drc package in R software to get the germination percentages and their standard errors. $\endgroup$ – Alan_dj Oct 13 '18 at 20:08
0
$\begingroup$

Here are fake data with the corresponding three-factor ANOVA. I have no idea what values the real data may take, so this is just to illustrate output from a three-factor ANOVA as done by Minitab.

The data. I have 24 germination counts. The following tables show the counts in the various data cells.

Results for Illum = Dark [Rows for 1,2,3 are the three temperatures; Columns for the four times.]

Rows: Temp   Columns: Time

           1      2      3      4    All

1      43.86  46.35  48.86  44.48  45.89
           1      1      1      1      4

2      40.33  47.31  56.62  49.52  48.44
           1      1      1      1      4

3      49.99  59.97  63.86  57.11  57.73
           1      1      1      1      4

All    44.73  51.21  56.45  50.37  50.69
           3      3      3      3     12

Cell Contents:  Germ  :  Value
                         Count

.

Results for Illum = Light 

Rows: Temp   Columns: Time

           1      2      3      4    All

1      63.85  64.94  62.98  73.14  66.23
           1      1      1      1      4

2      61.07  74.16  69.43  71.11  68.94
           1      1      1      1      4

3      81.52  71.18  71.51  73.53  74.43
           1      1      1      1      4

All    68.81  70.09  67.97  72.59  69.87
           3      3      3      3     12

Cell Contents:  Germ  :  Value
                         Count

Here is the ANOVA table

ANOVA: Germ versus Illum, Temp, Time 

Factor  Type   Levels  Values
Illum   fixed       2  1, 2
Temp    fixed       3  1, 2, 3
Time    fixed       4  1, 2, 3, 4


Analysis of Variance for Germ

Source      DF       SS       MS      F      P
Illum        1  2207.32  2207.32  98.88  0.000
Temp         2   432.19   216.10   9.68  0.013
Time         3   105.81    35.27   1.58  0.290
Illum*Temp   2    18.47     9.24   0.41  0.679
Illum*Time   3   138.02    46.01   2.06  0.207
Temp*Time    6   105.29    17.55   0.79  0.611
Error        6   133.94    22.32
Total       23  3141.05

S = 4.72477   R-Sq = 95.74%   R-Sq(adj) = 83.65%

The Illumination effect is highly significant (P-value < 0.0005), and the Temperature effect is significant (P-value = 0.013). The Time effect is not significant and neither are the three two-way interactions. [The three-way interaction of Illumination, Temperature, and Time is not distinguishable from random error in this design with only one observation in each of the 24 cells.]

The table of expected mean squares shown below shows that each of the main effects Illumination, Temperature, and Time, and all three of the interactions are 'tested against' Error. That is, the Mean Square for Error is the denominator of each of the relevant F-ratios. [The notation (7) stands for the population error variance $\sigma^2$; expressions Q[] stand for quadratic forms involving the various factors. These notations are unique to Minitab as far as I know.]

                                 Expected Mean
                                 Square for Each
                                 Term (using
                Variance  Error  restricted
   Source      component   term  model)
1  Illum                      7  (7) + 12 Q[1]
2  Temp                       7  (7) + 8 Q[2]
3  Time                       7  (7) + 6 Q[3]
4  Illum*Temp                 7  (7) + 4 Q[4]
5  Illum*Time                 7  (7) + 3 Q[5]
6  Temp*Time                  7  (7) + 2 Q[6]
7  Error           22.32         (7)

You could do ad hoc tests to see the pattern of differences of the three Temperature means.

This is hardly a complete exposition of the three-factor ANOVA or of Minitab's analysis. But perhaps it is enough to get you started. Please find an intermediate level applied statistics textbook that explains three-factor ANOVA's.

| cite | improve this answer | |
$\endgroup$

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.