User Manual

6-35
k ANOVA (Two-Way)
u Description
The nearby table shows measurement results for a metal product produced by a heat
treatment process based on two treatment levels: time (A) and temperature (B). The
experiments were repeated twice each under identical conditions.
Perform analysis of variance on the following null hypothesis, using a significance level of 5%.
H
o
: No change in strength due to time
H
o
: No change in strength due to heat treatment temperature
H
o
: No change in strength due to interaction of time and heat treatment temperature
u Solution
Use Two-Way ANOVA to test the above hypothesis.
Input the above data as shown below.
List1={1,1,1,1,2,2,2,2}
List2={1,1,2,2,1,1,2,2}
List3={113,116,139,132,133,131,126,122}
Define List 3 (the data for each group) as Dependent. Define List 1 and List 2 (the factor
numbers for each data item in List 3) as Factor A and Factor B respectively.
Executing the test produces the following results.
• Time differential (A) level of significance P = 0.2458019517
The level of significance (
p = 0.2458019517) is greater than the significance level (0.05), so
the hypothesis is not rejected.
• Temperature differential (B) level of significance P = 0.04222398836
The level of significance (
p = 0.04222398836) is less than the significance level (0.05), so
the hypothesis is rejected.
• Interaction (A × B) level of significance P = 2.78169946e-3
The level of significance (
p = 2.78169946e-3) is less than the significance level (0.05), so the
hypothesis is rejected.
The above test indicates that the time differential is not significant, the temperature differential
is significant, and interaction is highly significant.
B (Heat Treatment Temperature) B1 B2
A1 113 , 116
133 , 131
139 , 132
126 , 122
A2
A (Time)