1. I don’t have any further advice, except to try other values for lambda.

2. 10 point Likert has better chances than 5 point Likert of satisfying normality. I haven’t come across many ways of calculating the minimum sample size for nonparametric tests. I know that when normality is satisfied then power of the Mann-Whitney test is about 94% of the power of the t test, everything else being equal. This means that the sample size required is only slightly higher than for the t test (when normality is satisfied).

Charles ]]>

1. I have applied Box Cox transformations as suggested by you but it didn’t worked. I am using 5 point likert scale in my research and majority of responses lies between 3 to 5. I used the value of λ as 1 as no clue was given in your literature. I was able to calculate the value of X & Y but the value of Z and r didn’t come when I have applied formula for Z as =NORM.S.INV((H$4-0.5)/H$203). My data started from cell H4 to H203.

2. What should the optimum sample size for using non-parametric tests. Do the 5 point likert test is less efficient than a likert test of say 10 or 11 points for normality check of the data.

Best regards.

Gulab Kaliramna

The ranking is done in the same way< the fact that there are negative observations doesn't change anything. (2,0,-1,5,6,1) has ranks (4,2,1,5,6,3) if lowest value is ranked 1 or (3,5,6,2,1,4) if highest value is ranked 1. Charles ]]>

Something like this (2,0,-1,5,6,1) . ]]>

Yes, there are ways to perform regression which takes autocorrelation into account. Typical approaches use GLS instead of OLS and Newey-West method.

I am planning to add this topic to the website and software shortly.

Charles ]]>

Regarding the definition of L, please see Maximum Log-Likelihood

Yes, y_i is n_i/n.

Charles ]]>

This is a strange error message since I believe that it is generated by Excel’s data analysis tool and not Real Statistics. Very weird.

I suggest that you close Excel and start again. Perhaps it is some sort of memory overload problem.

Charles ]]>

Yes, that is a common recommendation and I believe that is the threshold that SPSS uses.

Charles ]]>

If you enter the total scores you will only have two measurements: one for each rater. ICC won’t be able to do much with this.

For this situation, it sounds like Cohen’s kappa will be a good fit: two raters, categorical data.

Some of these interrater measurements give low rating even when there is almost perfect agreement. This what I meant by”in extreme cases”. As long as you don’t have close to perfect agreement, you should be ok.

Charles ]]>

This is because SHAPLEY is an array function. To see how to get all the output, see the following webpage:

Array Formulas and Functions

Charles ]]>