Can you please check my work for a-c

When only two treatments are involved, ANOVA and the Student's f-test (Chapter 11) result in the same conclusions. Also, for computed test statistics, I = F. To demonstrate this relationship, use the following example. Fourteen randomly selected students enrolled In a history course were divided into two groups, one consisting of six students who took the course In the normal lecture format. The other group of eight students took the course in a distance format. At the end of the course, each group was examined with a 50-Item test. The following is a list of the number correct for each of the two groups. Traditional Lecture Distance 36 45 31 35 45 34 35 33 45 36 Expicture Click here for the Excel Data File. Required a-1. Complete the ANOVA table. (Round your SS. MS. and Fvalues to 2 decimal places and p-value. Fcrit to 4 decimal places.) Source of Variation SS MS p-value F crit Treatment 96.01 96.00 5.67 0.0347 4.7472 Erro 203.21 16.93 Total 290.21 I decimal places required. a-2 Use a a = 0.01 level of significance, compute the critical value. (Round your answer to 2 decimal places.) The critical value of F is 0.52 b. Using the t-test from Chapter 11, compute z (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.) (2.250) 3 decimal places required. C. Is there any difference In the mean test scores? Fail to reject Ho. There is no statistically significant difference in the mean scores between lecture and internet-based formats