**Exhibit 13-21**| Source of Variation | Sum of Squares | Degrees of Freedom | Mean Square $ F $ ||---------------------|----------------|--------------------|---------------------|| Between Groups (Treatment) | 6,000.0 | 5 | 1,200 || Within Groups (Error) | 24 | 288 | || **Total** | | **29** | |Refer to Exhibit 13-21. The test statistic to test the null hypothesis is what type of distribution ____?- ○ F-Distribution- ○ Chi Square Distribution- ○ T-Test- ○ Normal Distribution---**Analysis:**This exhibit presents an analysis of variance (ANOVA) table. The table summarizes the variance among different groupings within a dataset.1. **Between Groups (Treatment):** - Sum of Squares: 6,000.0 - Degrees of Freedom: 5 - Mean Square: 1,200 (calculated by dividing the Sum of Squares by Degrees of Freedom) 2. **Within Groups (Error):** - Sum of Squares: 24 - Degrees of Freedom: 2883. **Total:** - Degrees of Freedom: 29In ANOVA, the F-Distribution is used to test the null hypothesis that group means are equal. If the calculated F-value is greater than the critical value from the F-distribution (depending on the significance level), the null hypothesis is rejected, indicating that there are significant differences between groups. The questioning piece prompts the reader to identify the type of distribution used in this context, emphasizing understanding of statistical testing and distribution application.

Hint: The source of variation given in the problem above tells the best test statistic to test the…

Answered: Sum of Squares Between Groups…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Sum of Squares Between Groups (Treatment) 6,000.0 Within Groups (Error) Total Source of Variation Degrees of Mean Freedom Square 5 24 29 1,200 288 Refer to Exhibit 13-21. The test statistic to test the null hypothesis is what type of distribution