6-5-24 Tests

k ANOVA (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)

 

B1

 

 

B2

 

A (Time)

 

 

 

 

 

 

 

 

 

 

 

 

 

A1

113

,

116

139

,

132

 

 

 

 

 

 

 

A2

133

,

131

126

,

122

 

 

 

 

 

 

 

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 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.

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