Question

When only two treatments are involved, ANOVA and the Student's t test (Chapter 11) result in the same conclusions. Also, for computed test statistics, t 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 8 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 group:s Traditional Lecture Distance 36 31 35 30 43 31 36 35 46 43 37 cel a-1. Complete the ANOVA table. (Round your SS, MS, and F values to 2 decimal places and p value to 4 decimal places.) rce df MS Factors Error Total a-2. Use a a 0.01 level of significance. Round your answer to 2 decimal places.) e critical value of F b. Using the t test from Chapter 11, compute t. (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.) C. There is any difference in the mean test scores HO. There is the mean scores between lecture and internet-based formats.

Answer 1

Here we have data:

Traditional Lecture	Distance
36	43
31	31
35	44
30	36
33	44
37	35
	46
	43

Here we calculate by EXCEL.

a) - 1

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Traditional Lecture	6	202	33.667	7.867
Distance	8	322	40.25	29.643
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	148.60	1	148.60	7.22	0.0198	9.33
Within Groups	246.83	12	20.57
Total	395.43	13

a)-2

The critical value of F = 9.33

b) Here we have

F = 7.22

t² = F

t = $\sqrt{}$7.22

t = 2.687

c) Fail to reject H_{o. There is not difference in the mean scores between lecture and internet-based formals.}