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What I am trying to do?

I have a data set which consists of only one undergraduate student's all courses scores. Let's assume, he has completed about 70 courses where 40 courses are related to science and remaining 30 courses are related to arts. Score range of each course is 0 to 100. I am interested to find whether there exists any difference in scores of science and arts related courses. Therefore, I have divided the data; in one group (arts) of data, there are 30 values and in another group (science), there are 40 values.

What did stop me to find the difference?

  1. As there is only two groups of data, I could use Student's T Test. However, as student's t test has assumption of independence [1], I can not use that formula.
  2. William M Connelly answered a question of RG where he remarked when should we use Paired T Test

    Moreover, it is basically only applicable when you have a "before" and "after" value recorded from a single "subject" (a subject could be a cell, a piece of tissue, or a human etc). Really, what it is asking is "is there a systematic difference between the before and after?"

    Therefore, I can not use Paired T Test also.

  3. The same problems occurred when I wanted to use Non parametric tests like Mann Whitney U Test or Wilcoxon Signed-Rank Test.

My Question

How can I find the score difference of arts related courses (30 courses) and science related courses (40 courses) when there is only one student's data and data are not paired?

Note: I have followed repeated measures related different questions of SE including this one and this one. However, I feel sorry to say you that I did not find the answer of my question.


Update

Here is a test data set which is relevant to the described data set in my asked question. I prepared this using Python.

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The independence requirement for the t-test is not really relevant here, as you have only one students's data (If you had data for more students, that would be more of an issue) ... but there is also an assumption of normal distribution and that is also doubtful here. You cannot use a paired test as this is not paired data.

I would here use a permutation test. Under the assumption that the scoring is the same for science and non-science courses, the labels science, non-science is just like they were attached arbitrarily to the courses. So you can simulate the permutation distribution of the difference of the means, say, by permuting the labels, say, $R=9999$ times, and each time compute the differences of the means. Draw the histogram, and over that, draw the observed difference as a line.


After questions in comments: Why is the independence assumption not relevant in this case with data from only one person? Because of exchangeability. Under the null hypothesis of no difference between groups, the data is exchangeable, see for instance Wilcoxon signed rank test independence assumption and search this site.

So with your example data: First, one should always do some visualizing of the data. For your data, parallel boxplots with data values overlaid is a good choice:

Parallel boxplots with overlaid data points

Then we do the permutation test, based on difference in means of the two groups.

meandiff_obs <- diff(with(score_data, tapply(Score, Subject, mean))) # Science - Arts

# Permute 9999 times

perm_dist <- replicate(9999, {tags <- score_data$Subject
    scores <- sample(score_data$Score, length(score_data$Score))
    diff(tapply(scores, tags, mean))
} )

perm_dist <- data.frame(meandiff=c(perm_dist))  

And we can show the results graphically:

Permutation distribution

with the observed meandiff as a red line.


Another question in comments:

Can you please provide an example where under the null hypothesis of no difference between two groups of data, data are not exchangeable. Basically, I suspect that Under the null hypothesis of no difference between groups, all data are exchangeable. Then, permutation test will be applicable everywhere.

This is opening a can of worms ... the paper (a review Joan F Box' biography of her father) contains:

There is a difficulty---not described by Box---with randomization and permutation tests as they are usually worked out when the experimental layout is at all complex: for example, if it is a two-way layout. The permutation test typically is based on what might be called the null null hypothesis3 of identical treatments, or at least identical distributions for the treatments. Yet another of Fisher's great contributions is the idea of factorial design with its associated analysis of vari- ance, in which various kinds of treatments (e.g., row and column treatments) may be looked at separately. (See, for example, the discussion on analysis of variance on p. 110, or of factorial designs on pp. 164-166.) So here we have a secondary paradox within the larger one over long-run support for randomization. Indeed, the obscurity about no treatment difference as the basis for a permutation tests and the untangling of treatment differences by different factors, interactions, and so on, lay at the core of the first major confrontation between Fisher and Jerzy Neyman (Neyman 1935). Neyman in effect pointed out the paradox, and Fisher retaliated ferociously.

You might consider asking a separate question about this!


For the record, the code for the plots:

library(ggplot2)
ggplot(score_data, aes(Subject, Score)) +
    geom_boxplot(notch=TRUE) +
    geom_jitter(position=position_jitter(0.03), color="blue") +
    ggtitle("Distribution of scores for two kinds of subjects") 

ggplot(perm_dist, aes(meandiff)) + geom_histogram() +
    geom_vline(xintercept=meandiff_obs, color="red") +
    ggtitle("Permutation distribution")

... and for reading the data:

score_data_text <-
  "   Arts  Science
88  20
85  13
84  58
74  66
66  49
53  46
63  42
15  46
28  39
65  15
27  78
75  66
19  81
37  67
85  21
80  55
69  94
20  18
73  24
87  27
58  67
86  71
66  13
19  29
19  18
15  85
52  80
54  17
14  98
59  91
0   24
0   82
0   67
0   49
0   39
0   34
0   92
0   10
0   15
0   77"  # Added zeros to simplify reading,  to be removed!

score_data <- read.table(textConnection(score_data_text), header=TRUE)
score_data <- as.list(score_data)
score_data$Arts <- with(score_data, Arts[Arts>0])
score_data <- with(score_data, data.frame(Subject=factor(c(rep("Arts", 30), rep("Science", 40))), Score=c(Arts, Science)) )
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  • $\begingroup$ Dear @kjetil_b_halvorsen, thanks for your answer. I am really aware regarding all of the assumptions to use t test, including normality. Can you please refer a paper or book or any reliable article which says that The independence requirement for the t-test is not really relevant here, as you have only one students's data $\endgroup$ – Md. Sabbir Ahmed Mar 27 at 13:56
  • $\begingroup$ kjetil b halvorsen, can I use Independent T test in this case? Or please, show me how can I use permutation test in this case? $\endgroup$ – Md. Sabbir Ahmed Mar 28 at 8:55
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    $\begingroup$ Thanks a lot for your updated answer. I need to understand "Under the null hypothesis of no difference between groups, the data is exchangeable" Can you please provide an example where under the null hypothesis of no difference between two groups of data, data are not exchangeable. Basically, I suspect that Under the null hypothesis of no difference between groups, all data are exchangeable. Then, permutation test will be applicable everywhere. $\endgroup$ – Md. Sabbir Ahmed Mar 31 at 16:43
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    $\begingroup$ You really know a lot. Thanks a lot for helping me a lot and I am sorry for disturbing you again and again. I apology for that. Can you please answer this question also with description rather than just quoting what other have said. $\endgroup$ – Md. Sabbir Ahmed Apr 1 at 17:35

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