excel 15.29.1 for mac error bars problem

Should we adjust for ties as there are ties presented in some columns.

We will use the terminology from Kruskal-Wallis Test and Two Factor ANOVA without Replication.

If the null hypothesis that the sum of the ranks of the groups are the same, then.

They invited 12 people to taste one red, one white and one rose wine with the order of tasting chosen at random and a suitable interval between tastings.

Each person was asked to evaluate each wine with the scores tabulated in the table on the left side of Figure 1.

Since p-value CHISQ.TEST(1.79, 2) 0.408.05, we conclude there is no significant difference between the three types of wines.

Note that Kendalls W can be calculated from Friedmans Q as follows.

On the examples workbook, I give two examples: one with p-value.408 and another with p-value.0408.

The first of these shows a non-significant result, while the second shows a significant result.

Charles.

So I performed an ANOVA for repeated measures; it resulted highly significant as expected, and then two orthogonal contrasts, the only ones interesting for this experiment, that is: A vs (BC) B vs C Even after applying Bonferroni correction with k2, both resulted significant with p.

I tried to rebuild the sheet with the wines, but i am unable to do the second ranking for the adding.

What ERxcel command is associated to build the ranking you display In your set which i downloaded, this Friedman with the wines is unfortunately not there.

If possible maybe you can mail mec the sheet as presented in this document.

You can download this workbook at Real Statistics Examples Workbooks Charles.

The data is scheduled to be ranked on the scale of 1-5 (1less important, 5highly important).

In the survey, you ranked the factors on the scale of 1-5 (1 to mean very strong and 5 to mean very weak).

Suppose you have 5 factors and each of 10 subjects rates each of the 5 factors on a scale of 1 to 5.

The factor that has the lowest total rating from 10 to 50 (or mean score from 1 to 5) would be strongest.

You could now use a statistical test to determine whether there are significant differences between the factors for the entire population of potential subjects.

Suppose factors A, B, C, D, E had mean scores of 1.3, 1.4, 4.1, 4.2, 4.2 respectively and A and B are shown to be significantly different from C, D and E, but not significantly different from each other, then you would conclude that A and B are the strongest.

Note that I didnt actually perform a Friedman post-hoc analysis to reach this conclusion, but this the general idea.) Charles.

Is there a non-parametric test for 2 factors repeated measures (more than 2 treatment groups and individual subject within each group have multiple assessments over time) Since the Friedmans statistic is distributed as a chi-square, can we extend this test the difference in treatment effect Each treatment group is independent from each other, so the test statistic can be added with modified df (ie, sum).

I dont know whether the approach that you are suggesting will work.

I found the following reference on the Internet, but I dont know whether or not it is useful to you.

If the means are significantly different between the two groups tested, the t test statistic value is greater than the critical value.

I have now make your suggested correction on the webpage.

Excel 15.29.1 For Mac Error Bars Problem