Statistical analysis for two questionnaires 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).


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 

