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 30 29 28 20 30 33 32 33 33 35 22 35 28 26 36 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). The p-value is What is your conclusion?
a)
Null hypothesis H0: All means are equal
i.e. temperature level has not an effect on the mean yield of the process
Alternative hypothesis H1: Not all means are equal
i.e. temperature level has an effect on the mean yield of the process
Significance level α = 0.05
Equal variances were assumed for the analysis.
From the given data
Total | Ti^2/ni | ||||||
50°C | 30 | 20 | 32 | 35 | 28 | 145 | 4205 |
60°C | 29 | 30 | 33 | 22 | 26 | 140 | 3920 |
70°C | 28 | 33 | 33 | 35 | 36 | 165 | 5445 |
Total G | 450 | 13570 |
P-value = 0.2105
b. Use a level of significance to test whether the temperature level has an effect on the mean yield of the process.
H0:
H1: temperature level has an effect on the mean yield of the process
Test Statistic = 1.78
The p-value is 0.2105 > alpha 0.05 so we accept H0
thus we conclude that All means are equal i.e. temperature level has not an effect on the mean yield of the process