Question

To study the effect of temperature on yield in a chemical process, five batches were produced at each of three temperature levels. The results follow.

Temperature
	50°C	60°C	70°C
	32	34	29
	22	35	34
	34	38	34
	37	27	36
	30	31	37

a. Construct an analysis of variance table (to 2 decimals but p-value to 4 decimals, if necessary).

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F	p-value
Treatments
Error
Total

b. Use a  level of significance to test whether the temperature level has an effect on the mean yield of the process.

Calculate the value of the test statistic (to 2 decimals).

Answer 1

Here, there are 3 levels of temperature namely 50°C, 60°C and 70°C. There are five batches which can be considered as the replicates of each treatment.

Yij: jth batch (response) receiving ith temperature (treatment); i=1,2,3 and j=1,2,3,4,5.
k=levels of temperature=3, n=5

H0: Temperature (treatment) effect is Insignificant VS H1: at least effect of one of the temperature level is different from others

R-Software commands and outputs:
T1=c(32,22,34,37,30)
T2=c(34,35,38,27,31)
T3=c(29,34,34,36,37)
y=c(T1,T2,T3) ##Response
yibar=c(mean(T1),mean(T2),mean(T3))
yibar
[1] 31 33 34

mean(y) ##Grand mean
[1] 32.66667
temp=rep(1:3,each=5)
temperature=factor(temp)

ANOVA=aov(y~temperature)
ANOVA
Call:
aov(formula = y ~ temperature)
Terms:
temperature Residuals
Sum of Squares 23.33333 236.00000
Deg. of Freedom 2 12
Residual standard error: 4.434712
Estimated effects may be unbalanced

summary(ANOVA)
Df Sum Sq Mean Sq F value Pr(>F)
temperature 2 23.33 11.67 0.593 0.568
Residuals 12 236.00 19.67   

qf(1-0.05,2,12) ###Tabulated (or critical) value
[1] 3.885294
## Since, calculated F-value=0.593 is less than tabulated F=3.885294, we Fail to Reject H0.
1-pf(0.593,2,12) ### p-value=P(F>0.593)
[1] 0.5680794

Final answer:

a. Construct an analysis of variance table (to 2 decimals but p-value to 4 decimals, if necessary).

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F	p-value
Treatments	23.33	2	11.67	0.593	0.568079
Error	236	12	19.67
Total	259.33	14

b. Use a  level of significance to test whether the temperature level has an effect on the mean yield of the process. Calculate the value of the test statistic (to 2 decimals).

The value of test statistic (F-statistic)=0.593.

Since, p-value is greater than alpha (0.05) i.e. level of significance, we fail to reject the null hypothesis H₀. This indicates that the temperature level has an NO significant effect on the mean yield of the process.