C8Pack*?8 @EACTDEACT10flPȁ$DH T́Ё ======060524 =======vANOVA (Two-Way)  NDescription 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. ThPerform 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 |Solution Use two-way ANOVA to test the above hypothesis. eInput the above data as shown below. @STAT[EXE]8Pack*?( @GUIDEKEYLOG(NOTE00 &(%2$2$2$22$000000000)xz |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=0.04222398836caThe level of significance (p=0.04222398836) is less than the significance level (0.05), so the hypothesis is rejected. Interaction (AB) level of significance P=2.78169946e3The level of significance (p=2.78169946e3) 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.