University of Southeastern Philippines
eBA 201 Managerial Statistics
Exercise 5 - ANOVA
Name: Nathan Kit A. Berro
Six different machines are being considered for use in the manufacturing of rubber seals. The
machines are being compared with respect to tensile strength of the product. A random sample of
four seals from each machine is used to determine whether or not the mean tensile strength varies
from machine to machine. The following are the tensile strength measurements in kilograms per
square centimeters x 10.
Machine
1 2 3 4 5 6
17.5 16.4 20.3 14.6 17.5 18.3
16.9 19.2 15.7 16.7 19.2 16.2
15.8 17.7 17.8 20.8 16.5 17.5
18.6 15.4 18.9 18.9 20.5 20.1
Perform the analysis of variance at the 0.05 level of significance and indicate whether or not the
treatment means differ significantly. Present also the statistical output using SPSS or MS Excel.
Note: You can just refer to any statistics book for the tabular value of F for the ANOVA.
Answer:
Let µ1 = the mean tensile strength machine 1
µ2 = the mean tensile strength machine 2
µ3 = the mean tensile strength machine 3
µ4 = the mean tensile strength machine 4
µ5 = the mean tensile strength machine 5
µ6 = the mean tensile strength machine 6
Step One: State the null hypothesis and the alternate hypothesis.
H0: µ1 = µ2 = µ3 = µ4 = µ5 = µ6
H1: µ1 ≠ µ2 ≠ µ3 ≠ µ4 ≠ µ5 ≠ µ6
Step Two: Select the level of significance. This is given in the problem statement as
0.05.
Step Three: Determine the test statistic. The test statistic follows the F distribution.
Step Four: Formulate the decision rule.
The numerator degrees of freedom, k-1, equal 6-1 or 5. The denominator degrees of freedom, n-
k, equal 24-6 or 18. The value of F at 5 and 18 degrees of freedom is 2.77. Thus, H0 is rejected if
F>2.77 or p< α of 0.05.
eBA 201 Managerial Statistics
Exercise 5 - ANOVA
Name: Nathan Kit A. Berro
Six different machines are being considered for use in the manufacturing of rubber seals. The
machines are being compared with respect to tensile strength of the product. A random sample of
four seals from each machine is used to determine whether or not the mean tensile strength varies
from machine to machine. The following are the tensile strength measurements in kilograms per
square centimeters x 10.
Machine
1 2 3 4 5 6
17.5 16.4 20.3 14.6 17.5 18.3
16.9 19.2 15.7 16.7 19.2 16.2
15.8 17.7 17.8 20.8 16.5 17.5
18.6 15.4 18.9 18.9 20.5 20.1
Perform the analysis of variance at the 0.05 level of significance and indicate whether or not the
treatment means differ significantly. Present also the statistical output using SPSS or MS Excel.
Note: You can just refer to any statistics book for the tabular value of F for the ANOVA.
Answer:
Let µ1 = the mean tensile strength machine 1
µ2 = the mean tensile strength machine 2
µ3 = the mean tensile strength machine 3
µ4 = the mean tensile strength machine 4
µ5 = the mean tensile strength machine 5
µ6 = the mean tensile strength machine 6
Step One: State the null hypothesis and the alternate hypothesis.
H0: µ1 = µ2 = µ3 = µ4 = µ5 = µ6
H1: µ1 ≠ µ2 ≠ µ3 ≠ µ4 ≠ µ5 ≠ µ6
Step Two: Select the level of significance. This is given in the problem statement as
0.05.
Step Three: Determine the test statistic. The test statistic follows the F distribution.
Step Four: Formulate the decision rule.
The numerator degrees of freedom, k-1, equal 6-1 or 5. The denominator degrees of freedom, n-
k, equal 24-6 or 18. The value of F at 5 and 18 degrees of freedom is 2.77. Thus, H0 is rejected if
F>2.77 or p< α of 0.05.
