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
	31	32	20
	21	33	25
	33	36	25
	36	25	27
	29	29	28

1. 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

   
   
2. 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 Selectless than .01between .01 and .025between .025 and .05between .05 and .10greater than .10Item 11
   
   What is your conclusion?
   SelectConclude that the mean yields for the three temperatures are not all equalConclude that the mean yields for the three temperatures are all equal

Answer 1

Ans:

See calculations below:

a)

Source of Variation	SS	df	MS	F	P-value
Between Groups	103.33	2	51.67	2.63	0.1132
Within Groups	236.00	12	19.67
Total	339.33	14

b)p-value is greater than .10

Conclude that the mean yields for the three temperatures are all equal.

Analysis of variance:

	Group 1	Group 2	Group 3	Total
Sum	150	155	125	430
Count	5	5	5	15
Mean, Sum/n	30	31	25
Sum of square, Ʃ(xᵢ-x̅)²	128	70	38
Standard deviation	5.6569	4.1833	3.0822

Number of treatment, k =	3
Total sample Size, N =	15
df(between) = k-1 =	2
df(within) = N-k =	12
df(total) = N-1 =	14
SS(between) = (Sum1)²/n1 + (Sum2)²/n2 + (Sum3)²/n3 - (Grand Sum)²/ N =	103.3333
SS(within) = SS1 + SS2 + SS3 =	236
SS(total) = SS(between) + SS(within) =	339.3333
MS(between) = SS(between)/df(between) =	51.6667
MS(within) = SS(within)/df(within) =	19.6667
F = MS(between)/MS(within) =	2.6271
p-value = F.DIST.RT(2.6271, 2, 12) =	0.1132

Null and Alternative Hypothesis:
Ho: µ1 = µ2 = µ3
H1: At least one mean is different.
Test statistic:
F =	2.63
Critical value:
Critical value Fc = F.INV.RT(0.05, 2, 12) =	3.89
p-value:
p-value = F.DIST.RT(2.6271, 2, 12) =	0.1132
Decision:
P-value > α, Do not reject the null hypothesis.