Graphing is available with Two-Way ANOVA only. V-Window settings are performed automatically, regardless of Setup screen settings.

Using the Trace function automatically stores the number of conditions to variable A and the mean value to variable M, respectively.

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)

 

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.

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