Another example is the Friedmann Test, we will use the data of a wine test at https://www.real-statistics.com/anova-repeated-measures/friedman-test/
Ho: No wine preference vs Ha: some wine preference (white, red or rose) . Ranking the wine taste results as shown below we obtain a Test Statistic of 1.79 which results in a p-value of 0.408267 indicating no significant wine preference.

Here is the same Friedmann Test performed using the TiNspire’s Statistics Made Easy app at https://www.TiNspireapps.com/?a=SPME . First we enter the wine taste data as 3 rows, the ranks are shown to the right accounting for ties as well.

Next , we compute the sum of ranks and their averages.

Now, squaring the ranks and adding them yields 1749.5. The Test Statistics uses that, the number of tasters and the number wines and comes out to be 1.79167

Using the Chi Squared Distribution yields the p-value of 0.40827 which is not below 5% and we thus fail to reject our above stated Null-Hypotheses Ho :

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