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, t2 = 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 groups.

Traditional Lecture (first column) Distance (second column)

38 42

35 33

37 46

35 35

32 49

37 33

      44

      41

a-1.) Complete the ANOVA table. (Round your SS, MS, and F values to 2 decimal places and p value to 4 decimal places.)

a-2.) Use a α = 0.05 level of significance. (Round your answer to 2 decimal places.)

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.

Answer 1

a)

Using Excel

data -> data analysis -> Anova single factor

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
traditional	6	214	35.66666667	4.6666667
distance	8	323	40.375	37.125
ANOVA
Source of Variation	SS	df	MS	F	P-value
Between Groups	76.01	1	76.01	3.22	0.0979
Within Groups	283.21	12	23.60
Total	359.21	13

ANova table is above

F = 3.22

b)

using Excel

data -> data analysis ->

t-Test: Two-Sample Assuming Equal Variances
	traditional	distance
Mean	35.667	40.375
Variance	4.667	37.125
Observations	6	8
Pooled Variance	23.601
Hypothesized Mean Difference	0
df	12
t Stat	-1.795
P(T<=t) one-tail	0.049
t Critical one-tail	1.782
P(T<=t) two-tail	0.098
t Critical two-tail	2.179

TS = -1.795

c)

p-value = 0.098 > 0.05

hence we fail to reject the null hypothesis

there is not significant difference

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