Question

A researcher was interested in the effects of different study methods on learning, among a class of children. Following one of three methods, 6 students were divided into groups.

Method 1 (book alone)	Method 2 (borrowing notes)	Method 3 (taking notes)
10	12	18
8	6	40
15	30	35
26	24	29
28	18	30
12	13	25

1. Using manual computation, perform the appropriate hypothesis test at α = 0.01

Answer 1

The one way ANOVA analysis is performed to test the hypothesis whether there is difference in means among there methods.

The null and alternative hypotheses are defined as,

Null Hypothesis: All the group means are equal,  

Alternative Hypothesis: At least one mean differ significantly.

The test is performed in following steps,

From the data values,

	Method 1 (book alone)	Method 2 (borrowing notes)	Method 3 (taking notes)
	10	12	18
	8	6	40
	15	30	35
	26	24	29
	28	18	30
	12	13	25
	99	103	177
	16.5	17.167	29.5
	1993	2149	5515
SD	8.479	8.727	7.662
	359.5	380.83	293.5
n	6	6	6

Where, 'i' represent rows and 'j' represents column

The degree of freedom values are,

The sum of square values are,

Where,

The means squares are computed using the formula,

The F-statistic values is obtained using the formula,

The F-critical value

The F-critical value is obtained from F-distribution table for significance level = 0.01, numerator degree of freedom = 2 and denominator degree of freedom = 15

Conclusion,

Since,

The null hypothesis is not rejected, hence we can conclude that there is no significant difference in mean among three groups.

The ANOVA table is shown below,

Source of Variation	SS	df	MS	F	F crit
Between Groups	643.111	2	321.556	4.665	6.359
Within Groups	1033.833	15	68.922
Total	1676.944	17