Question

The following scores are from an independent-measures study comparing two treatment conditions. (Be sure to include your critical values)

a. Use an independent -measures t-test with a = 0.05 to determine whether there is a significant mean difference between the two treatments.

b. Use an ANOVA with a = 0.05 to determine whether there is a significant mean difference between the two treatments. You should find that F = t².

Treatment I	Treatment II
10	7	N = 16
8	4	G =120
7	9	Σx2 = 1036
9	3
13	7
7	6
6	10
12	2

Answer 1

Treatment I	Treatment II
10	7	N = 16
8	4	G =120
7	9	Σx2 = 1036
9	3
13	7
7	6
6	10
12	2

Treatment 1

N₁: 8
df₁ = N - 1 = 8 - 1 = 7
(Mean) M₁ : 9
(Sum of squared value )SS₁: 44
s²₁ = SS₁/(N - 1) = 44/(8-1) = 6.29


Treatment 2

N₂: 8
df₂ = N - 1 = 8 - 1 = 7
M₂: 6
SS₂: 56
s²₂ = SS₂/(N - 1) = 56/(8-1) = 8


T-value Calculation

s²_p

= ((df₁/(df₁ + df₂)) * s²₁) + ((df₂/(df₂ + df₂)) * s²₂)

= ((7/14) * 6.29) + ((7/14) * 8)

= 7.14

s²_M1

= s²_p/N₁

= 7.14/8

= 0.89

s²_M2

= s²_p/N₂

= 7.14/8

= 0.89

t = (M₁ - M₂)/√(s²_M1 + s²_M2)

= 3/√1.79

= 2.24

The p-value for t value of 2.24 at 14 degrees of freedom is .041443. The result is significant at p < .05

On the other side, The critical value of t at 14 df and 0.05 significance level is 2.14.

Calculated t value is greater than critical t value. Hence, the result is significant at 0.05 significance level.

	Treatment 1	Treatment 2	Total
N	8	8	16
Sum of X	72	48	120
Mean	9	6	7.5
sum of X^2	692	344	1036
std. dev	2.5	2.8	3.0

Source	SS	df	MS
Between treatments	36	1	36
Within treatments	100	14	7.14
Total	136	15

The F ratio is 36 / 7.14

= 5.04

The critical F value for numerator df =1, denominator df = (8+8-2) and significance level 0.05 is 4.6

the calculated f value is greater than critical f value. The result is significant at 0.05 significance level.

Also F = t^2

Since, 5.04 = 2.24^2