Statistical analysis for two questionnaires

Question

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I am working on a research paper and I need to organize the statistical analysis part. I have two questionnaires with the same number of participants (15 cases). Basically, we asked the participants to play two different games with two different conditions.

• Imagine: Game A (under conditions 1 and 2) (questionnaire with 5 questions). Game B (under conditions 1 and 2) (questionnaire with 7 questions). And we asked them to fill out the questionnaire to estimate the overall satisfaction for each condition. I am wondering that whats could be the best statistical method here to analyze the results for each condition. Is (one/two way) ANOVA a good solution? If yes is it a good way to do it on Python? (unfortunately, I am not familiar with R and I'm kinda in rush) Here is an example of the result of the questionnaires for only one person (1 out of 15).

Answer 1

Thanks for the revised question.

For Game A. The two conditions were scored very similarly, and you have only 7 scores.

a1 = c(4, 3.5, 3, 4, 4.5, 4, 4)
a2 = c(4,   4, 4, 3,   4, 4, 4)

A paired Wilcoxon signed rank test finds no difference between conditions. The P-value is not exact because of the ties in the data. Tests from R.

wilcox.test(a1, a2, pair=T)

    Wilcoxon signed rank test 
    with continuity correction

data:  a1 and a2
V = 5, p-value = 1
alternative hypothesis: 
 true location shift is not equal to 0

Treating the scores as numerical (even though I wonder if they may be ordinal categorical choices), we find no difference (high P-value), again because the scores are so nearly the same for the two conditions.

t.test(a1, a2)$p.val
[1] 1

For Game B. In all five categories, Condition 2 was preferred. The Wilcoxon signed rank test will not work because of ties in the differences. However, regarding data as numerical, a paired t test is significant at the 5% level (P-value $0.012 < 0.05 = 5\\%.)$

b1 = c(4, 3, 3, 3, 4)
b2 = c(5, 4, 5, 5, 4.5)
t.test(b1, b2, pair=T)

        Paired t-test

data:  b1 and b2
t = -4.3333, df = 4, p-value = 0.01232
alternative hypothesis: 
 true difference in means is not equal to 0
95 percent confidence interval:
 -2.1329335 -0.4670665
sample estimates:
mean of the differences 
                   -1.3