XBb 76570829%5blackboardcdn.com/5c1802599953e/76570829?X-Blackboard-Expiration=1671084000000&X-Blackboard-Sign... 1☆TG3/3aabbCX Bb 76419361-These data are evaluations of 4 job candidates (can.1 through can.4) by 3 raters (a,b & c).You are testing for differences in the candidates.rater can.1 can.2 can.3 can.47665857459758167496361467471858966848079Questions 11-20 refer to the above:6100% + HY11. The design is unbalanced?12. The design is balanced?13. The null hypothesis of equal candidate rating means is rejected via a two way ANOVA?14. Compute the SST15. Compute the SSTr for candidates.16. Compute the SSTr for raters17. Compute the observed f statistic18. The critical f statistic at a = .05 for these numerator and denominator degrees offreedom is less than the observed f statistics from these data?19. Fisher's LSD identifies which rater pairs as unequal? (not just a T/F question)20. The within candidate variances are unequal?HMacBook Pro7UJ81X +K69900LP{~~[+ 11

Solution: From the given information,11. The design is said to be balanced if the number of rows per sample should be same for all three caters. Otherwise, it is unbalanced. From the given data it can be observed that the number of rows per sample for cater a is 2, number of rows per sample for cater b is 2 and number of rows per sample for cater c is 1. Here, the number of rows per sample is not same for all caters. Thus, the design is unbalanced.12. Since the number of rows per sample is not same for all caters the design is not balanced.13. State the hypotheses. Null hypothesis: H0: The candidate rating means are equal. Alternative hypothesis: H1: At least one mean rating is different. Number of raters a=3, number of job candidates b= 4 and the number of treatment groups is a*b=12. Total number of observations N=20. The correction factor can be calculated as follows: The total sum of square can be calculated as follows: The sum of squares for raters can be calculated as follows: The sum of squares for candidates can be calculated as follows: The sum of squares for interaction is calculated as follows: The sum of squares for error is calculated as follows: The mean sum of squares for raters is The mean sum of squares for Error is The F statistic for raters can be calculated as follows: The mean sum of squares for Candidates is The F statistic for candidates can be calculated as follows: The critical value is Rejection rule: If the F-statistic is greater than F critical value, then reject the null hypothesis. Conclusion: Here, the F statistic (0.648) is less than F critical value (4.459). By the rejection rule, do not reject the null hypothesis. Thus, the null hypothesis of equal candidate rating means is not rejected via two way ANOVA. 14. From the part (13), the value of SST is 2678.95. 15. From the part (13), the value of SSTr for candidates is 463.75. 16. From the part (13), the value of SSTr for raters is 289.2. 17. From the part (13), the observed f statistic is 0.648. 18. From the part (13), the critical f statistic is 4.459 and the observed f statistic values are 0.648 and 0.693. These f statistics are less than the critical f statistic. Thus, critical f statistic at α=0.05 for theses numerator and denominator degrees of freedom is not less than the observed f statistics from these data.