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 | 26 | 24 | ||
20 | 27 | 29 | ||
32 | 30 | 29 | ||
35 | 19 | 31 | ||
28 | 23 | 32 |
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 .05 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 is - Select your answer -less than .01between .01 and .025between .025 and .05between .05 and .10greater than .10Item 12
What is your conclusion?
Solution:
We can use the excel ANOVA: Single Factor data analysis
tool to find the answer to the given questions.
The excel steps are:
Enter the data in excel.
Click on Data > Data Analysis > ANOVA: Single Factor > OK
Input Range: Select the data range for all the data including labels
Mark Labels in the first row
Alpha = 0.05
Choose the output range and click OK.
The output is given below:
Anova: Single Factor | ||||||
SUMMARY | ||||||
Groups | Count | Sum | Average | Variance | ||
50°C | 5 | 145 | 29 | 32 | ||
60°C | 5 | 125 | 25 | 17.5 | ||
70°C | 5 | 145 | 29 | 9.5 | ||
ANOVA | ||||||
Source of Variation | SS | df | MS | F | P-value | F crit |
Between Groups | 53.33333333 | 2 | 26.66666667 | 1.355932203 | 0.294497217 | 3.885293835 |
Within Groups | 236 | 12 | 19.66666667 | |||
Total | 289.3333333 | 14 |
a. Construct an analysis of variance table (to 2 decimals but p-value to 4 decimals, if necessary).
Answer:
Source of Variation | Sum of Squares | Degrees of Freedom | Mean Square | F | p-value |
Treatments | 53.33 | 2 | 26.67 | 1.36 | 0.2945 |
Error | 236 | 12 | 19.67 | ||
Total | 289.33 | 14 |
b. Use a .05 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: The value of the test statistic is:
The p-value is greater than .10
What is your conclusion?
Since the p-value is greater than the significance level, we, therefore, fail to reject the null hypothesis and conclude that the temperature level has no significant effect on the mean yield of the process.