Tests (TEST)
kANOVA (Two-Way)
uDescription
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.
B (Heat Treatment Temperature) |
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A (Time) |
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A1 | 113 | , | 116 | 139 | , | 132 |
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A2 | 133 | , | 131 | 126 | , | 122 |
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Perform analysis of variance on the following null hypothesis, using a significance level of 5%.
Ho : No change in strength due to time
Ho : No change in strength due to heat treatment temperature
Ho : No change in strength due to interaction of time and heat treatment temperature
uSolution
Use
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 =
The level of significance (p =
The above test indicates that the time differential is not significant, the temperature differential is significant, and interaction is highly significant.
20011101