Question

Chapter 12 Extra Practice (Optional 1 03 pon When only two treatments are involved. ANOVA and the Student's test (Chapter 19 result in the same conclusions. Also, for computed test statistics, 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 6 students who took the course in the normal lecture format. The other group of students took the course as a distance course 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 Distance eBook 45 16 45 44 40 42 References Click here for the Excel Data File --1. Complete the ANOVA table. (Round your SS, MS, and Fvalues to 2 decimal places and p value to 4 decimal places.) Source 55 MS Factor Error Total -2. Use a 0.01 level of significance. (Round your answer to 2 decimal places.) Theorical value of F

Answer 1

Excel output is attached below:

Anova: Single Factor Count traditional distance 45 42 35 38 45 42 36 44 43 40 38 46 42 44 SUMMARY Groups traditional distance 6 Sum Average Variance 242 40.3333 20.6667 338 42.25 6.21429 00 df ANOVA Source of Variation Between Groups Within Groups Total F 1.03 MS 12.60 12.24 P-value 0.3303 1 SS 12.60 146.83 159.43 12 13

### Answers:

a-1] ANOVA table is attached above.

a-2] the critical value of F = 9.33

b] t=sqrt(F) = 1.015

c] since test statistics F is less than 9.33 we fail to reject H0.

Fail to reject H0. there is not statistically significant difference in the mean scores.