The Shapiro-Wilk test For each Environment, using data from Table 1.3, was used to perform the Shapiro Wilk test to assess whether data are normally distributed or not. Results for Environment at 15 C: W(14) = .8, p = .005, p-value = 0.004969, W = 0.7998, KS - D = large (0.298). The Shapiro-Wilk test showed a significant departure from normality. Since p-value<, we reject the null hypothesis, assuming that the data is not normally distributed.The difference between the data sample and the normal distribution is big enough to be statistically significant. The p-value means that the chance of type I error is small . The smaller the p-value the more it supports the working hypothesis. The test statistic is not in the 95% region of acceptance. The observed effect size is large and indicates that the magnitude of the difference between the sample distribution and the normal distributions is large. The Shapiro-Wilk test result indicates that the data from the environment at 15 C does not follow a normal distribution, supported by a low p-value and additional evidence from the W statistic and KS-D value. Results for Environment at 25 C: W(14) = .76, p = .002, p-value = 0.001628, W = 0.7591, KS - D = large (0.2717) The Shapiro-Wilk test showed a significant departure from normality. Since p-value<, we reject the null hypothesis, assuming that the data is not normally distributed.The difference between the data sample and the normal distribution is big enough to be sta