Question

A psychologist would like to examine the effects of sugar consumption on the activity level of preschool children. Three samples of children are selected with n = 5 in each sample. One group gets no sugar, one group gets a small dose, and the third group gets a large dose. The psychologist records the activity level for each child. The data from this experiment are presented below. a. Clearly state the findings of your analysis. Report p-value. b. Compute η2 , the percentage of variance accounted for by the treatment. c. Test for the homogeneity of variance assumption. No sugar Small dose Large dose 5 3 6 3 3 8 1 7 7 G = 71 1 4 9 ΣX2 = 419 5 3 6 T = 15 T = 20 T = SS = 16 SS = 12 SS =

Answer 1

Result:

A psychologist would like to examine the effects of sugar consumption on the activity level of preschool children. Three samples of children are selected with n = 5 in each sample. One group gets no sugar, one group gets a small dose, and the third group gets a large dose. The psychologist records the activity level for each child. The data from this experiment are presented below. a. Clearly state the findings of your analysis. Report p-value. b. Compute η2 , the percentage of variance accounted for by the treatment. c. Test for the homogeneity of variance assumption. No sugar Small dose Large dose 5 3 6 3 3 8 1 7 7 G = 71 1 4 9 ΣX2 = 419 5 3 6 T = 15 T = 20 T = SS = 16 SS = 12 SS =

Ho: µ₁= µ₂= µ₃

H₁: At least one of the mean is different from the others

	No sugar	small dose	large dose
	5	3	6
	3	3	8
	1	7	7
	1	4	9
	5	3	6	Total
sum	15	20	36	71
sum of squares	61	92	266	419
SS	16	12	6.80	34.8

Between group sum of square = 82.9333-34.8 = 48.1333

ANOVA
Source of Variation	SS	df	MS	F
Between Groups	48.1333	2	24.0667	8.2989
Within Groups	34.8000	12	2.9000
Total	82.9333	14

P value = 0.0055

Table value of F with (DF1=2, DF2=12 ) at 0.05 level=3.89

Calculated F=8.2989 > 3.89 the table value

The null hypothesis is rejected.

The data indicate there is a significant difference among the three groups

b).

η2 = 48.1333/82.9333

=0.5804

percentage of variance accounted for by the treatment =58.04%

c).

MINITAB used.

Test for Equal Variances: No sugar, small dose, large dose

Method

Null hypothesis	All variances are equal
Alternative hypothesis	At least one variance is different
Significance level	α = 0.05

95% Bonferroni Confidence Intervals for Standard Deviations

Sample	N	StDev	CI
No sugar	5	2.00000	(0.961059, 7.9855)
small dose	5	1.73205	(0.388082, 14.8317)
large dose	5	1.30384	(0.505920, 6.4470)

Individual confidence level = 98.3333%

Tests

Method	Test Statistic	P-Value
Multiple comparisons	—	0.488
Levene	0.42	0.667

Calculated Levene test 0.42, P=0.667 which is > 0.05 level of significance. Ho is not rejected.

We conclude that homogeneity of variance assumption. Is not violated.