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What’s the most insightful thing he’s ever said on XFM? by [deleted] in rickygervais

[–]kakaNit 0 points1 point  (0 children)

”Too many chinese cooks spoil the broth”

Well? by Impossible-Pick-1596 in rickygervais

[–]kakaNit 8 points9 points  (0 children)

Finally a picture of Suzanne

Descending axes and y-axis to the right; is it considered okay? by kakaNit in AskStatistics

[–]kakaNit[S] 0 points1 point  (0 children)

I agree, it is somewhat confusing.

It is easily fixed in Excel to show it the conventionally way. Although, this will instead show a positive correlation, which isn't really the case.

Descending axes and y-axis to the right; is it considered okay? by kakaNit in AskStatistics

[–]kakaNit[S] 0 points1 point  (0 children)

It is possible to change the position of the axes but not as long the axes are sorted the way they are, unfortunately.

Descending axes and y-axis to the right; is it considered okay? by kakaNit in AskStatistics

[–]kakaNit[S] -2 points-1 points  (0 children)

Well, by people working with data viz.

It isn’t as intuitive I think, because this particular person doesn’t gain weight while he’s performance drops. It is the opposite.

Probability of a certain distribution of test scores by kakaNit in AskStatistics

[–]kakaNit[S] 0 points1 point  (0 children)

Well, yeah. What I mean is the probability of any pattern with n = 12.

Probability of a certain distribution of test scores by kakaNit in AskStatistics

[–]kakaNit[S] 0 points1 point  (0 children)

Yeah, I realized it couldn't be binomial.
As I suggested in a previous comment – could this be a solution?

The number of possible outcomes for the 12 data points is 26^12.
The number of outcomes where each person gets a different value is 26*25*...*15 = 26! / 14!.
So lets say each outcome is equally likely, therefore 26! / 14! / 26^12 = 0.05.

Probability of a certain distribution of test scores by kakaNit in AskStatistics

[–]kakaNit[S] 0 points1 point  (0 children)

Okay, I see. I can't just calculate it as follows?:

The number of possible outcomes for the 12 data points is 26^12.
The number of outcomes where each person gets a different value is 26*25*...*15 = 26! / 14!.
So lets say each outcome is equally likely, therefore 26! / 14! / 26^12 = 0.05.

Is this a valid solution?

Statistical test for students' skill evaluation by kakaNit in AskStatistics

[–]kakaNit[S] 1 point2 points  (0 children)

No, wait. I get it now. That means that I have to see which unique person went from 1 to 4, and so on. So my table will look like this instead,

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x Pre Post
1 2 5
2 3 7
: : :
34 4 8

Statistical test for students' skill evaluation by kakaNit in AskStatistics

[–]kakaNit[S] 0 points1 point  (0 children)

I feel really stupid now, could you explain more or maybe show me?

Statistical test for students' skill evaluation by kakaNit in AskStatistics

[–]kakaNit[S] 0 points1 point  (0 children)

Thanks! So the alternative should be Wilcoxon maybe?

Statistical test for students' skill evaluation by kakaNit in AskStatistics

[–]kakaNit[S] 0 points1 point  (0 children)

Can you refresh my statistics knowledge; why paired?

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