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
	37	35	30
	27	36	35
	39	39	35
	42	28	37
	35	32	38

A. Construct an analysis of variance table (to 2 decimals, if necessary). Round p-value to four decimal places.

Source of variation	sum of squares	degrees of freedom	mean square	F	P- Value
Treatments
Error
Total

Use a .05 level of significance to test whether the temperature level has an effect on the mean yield of the process.

The p-value is=

(less than .01; between .01 and .025; between .025 and .05; between .05 and .10; greater than .10)

What is your conclusion?

Answer 1

Here the given factor of variation is temperature (50C, 60C and 70C). The hypothesis to construct the analysis of variation can be assumed as:

Null Hypothesis;   H₀ : All the temperatures have an equal effect on the mean yield of the process.

Alternative Hypothesis;   H₁ : Atleast two temperatures do not have an equal effect on the mean yield of the process.

Level of Significance ;

Further other parameters to construct the ANOVA Table are:

n₁= n₂ = n₃ = 5 (Size of individual treatment)

k = 3 (Number of treatments)

N = n₁+ n₂ + n₃ = 15

The various sum of squares (SS) can be calculated as:

Temperature	1st	2nd	3rd	4th	5th	Total	Ti.2
50 C	37	27	39	42	35	T1.= 180	32400
60 C	35	36	39	28	32	T2.= 170	28900
70 C	30	35	35	37	28	T3.= 160	25600

Now after calculating all the required sum of sqaures , The ANOVA table can be obtained as:

Source of Variation	Sum Of Squares	degree of freedom	MSS	Variance Ratio
(1)	(2)		(4)= (2)/(3)
Treatments	40	k-1 = 3-1=2
Error	1241	N-k = 12
Total	1281	N-1 =14

P- value at (2,12, 0.1934) = 0.8267 ehich is greater than 0.1

Conclusion : Since p-value is greater than 0.05 so Do not reject the assumption that the mean yields for the three temperatures are equal.