B. Use the table below (degrees of freedom for each test = 1) Accept the null hypothesis Reject the null hypothesis Probability Degrees of Freedom 0.90 0.50 0.25 0. 10 0.05 0.01 1 0.016 0.46 1.32 2.71 3.84 6.64 2 0.21 1.39 2.77 4.61 5.99 9.21 3 0.58 2.37 4.11 6.25 7.82 11.35 4 1.06 3.36 5.39 7.78 9.49 13.28 5 1.61 4.35 6.63 9.24 11.07 15.09 Note that the chi-square value increases as the probability that there is no difference in the control and the variable decreases. The column we are concerned with is the "0.05." Here we can say that if the probability of getting the observed deviation is greater than 5%, then we can accept the null hypothesis. In other words, there is no significant difference in the treatments of lectin (etc.) and water on the Allium root cells. If the probability of getting the observed deviation is less than 5%, we should reject the null hypothesis. In other words, there is a significant difference and some factor is at work. -4. Based on the data above, should you accept or reject the null hypothesis? Why