Question

A chemical manufacturer has been researching new formulas to provide quicker relief of minor pains. His laboratories have produced three different formulas, which he wanted to test. Fifteen people who complained of minor pains were recruited for an experiment. Five were given formula 1, five were given formula 2, and the last five were given formula 3. Each was asked to take the medicine and report the length of time until some relief was felt. The results below shows the time until relief is Felt.

Formula -1 : 4 8 6 9 8

Formula - 2 : 2 5 3 7 1

Formula - 3 : 6 7 7 8 6

SST = 78.4, SSE=42
(a) Write down the model and the ANOVA table to test whether there exits any differences in the time of relief exist among the three formulas? Use α =0.05.
(b) Is the Formula-1 different from Formula-3 at 5% level

Answer 1

(a)
H0: Null Hypothesis: ( There do not exist any differences in the time of relief exist among the three formulas)

HA: Alternative Hypothesis: ( There exist any differences in the time of relief exist among the three formulas)

From the given data, the following Table is calculated:

	Formula - 1	Formula - 2	Formula - 3	Total
N	5	5	5	15
	35	18	34	87
Mean	35/5=7	18/5=3.6	34/5=6.8	87/15=5.8
	261	88	234	583
Std. Dev.	2	2.4083	0.8367	2.3664

From the above Table, ANOVA Table is got as follows:

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Sum of Squares	F
Between treatments	36.4	2	36.4/2=18.2	18.2/3.5=5.20
With treatments	42	12	42/12=3.5
Total	78.4	14

F - Ratio = 18.2/3.5=5.20

Degrees of Freedom for numerator = 2

Degrees of Freedom for denominator = 12

By Technology, p - value = 0.0236\Since p - value = 0.0236 is less than = 0.05, the difference is significant. Reject null hypothesis.

Conclusion:

The data support the claim that there exist any differences in the time of relief exist among the three formulas.

(b)

Formula 1 & Formula - 3 give:

H0: Null Hypothesis: ( The Formula-1 is not different from Formula-3)

HA: Alternative Hypothesis: ( The Formula-1 is different from Formula-3)

n1 = 5

1 = 7

s1 = 2

n2 = 5

2 = 5.8

s2 = 0.8367

Test Statistic is given by:

t = (7 - 6.8)/0.9695

= 0.2063

= 0.05

ndf = n1 + n2 - 2 = 5 + 5 - 2 = 8

From Table, critical values of t = 2.3060

Since calculated value of t = 0.2063 is less than critical value of t = 2.3060, the difference is not significant. Fail to reject null hypothesis.

Conclusion:

The data do not support the claim that the Formula-1 is different from Formula-3.