In: Statistics and Probability
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 | 33 | 23 | ||
21 | 34 | 28 | ||
33 | 37 | 28 | ||
36 | 26 | 30 | ||
29 | 30 | 31 |
a. Construct an analysis of variance table (to 2 decimals but p-value to 4 decimals, if necessary).
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 p -value is?
What is your conclusion?
50°C | 60°C | 70°C | Total | |
Sum | 150 | 160 | 140 | 450 |
Count | 5 | 5 | 5 | 15 |
Mean, Sum/n | 30 | 32 | 28 | |
Sum of square, Ʃ(xᵢ-x̅)² | 128 | 70 | 38 |
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 = 40
SS(within) = SS1 + SS2 + SS3 = 236
SS(total) = SS(between) + SS(within) = 276
MS(between) = SS(between)/df(between) = 20
MS(within) = SS(within)/df(within) = 19.66667
F = MS(between)/MS(within) = 1.0169
p-value = F.DIST.RT(1.0169, 2, 12) = 0.3909
a)
ANOVA | |||||
Source of Variation | SS | df | MS | F | P-value |
Between Groups | 40.00 | 2 | 20.00 | 1.02 | 0.3909 |
Within Groups | 236.00 | 12 | 19.67 | ||
Total | 276.00 | 14 |
b)
Test statistic:
F = 1.02
p-value = F.DIST.RT(1.0169, 2, 12) = 0.3909
Conclusion:
P-value > α, Do not reject the null hypothesis.
There is not enough evidence to conclude that the temperature level has an effect on the mean yield of the process.