Casio FX-CG10 Software User Guide - Page 237
ANOVA Two-Way, Description, Solution
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• 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. 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. 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 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. 6-44