Answer
a) Null: Mites do not induce resistance to wilt
b) Null in symbols: P1=P2
c) 1.3 + 1.62 + 1.928 + 2.382= 7.21
d)P-Value is between .0005 and .005 (Found using Table 9 or a calculator) so we have enough evidence to reject the null.
e) We can conclude that mites do induce resistance to wilt.
Work Step by Step
a) Think of this as the probability of wilt with mites is equal to probability to wilt without mites. This means that mites are not responsible for wilting.
b)If both probabilities are =, then P1 = P2
c)Use Chi Squared Formula X= (Observed value - estimated value)^2 / Estimated Value, where estimated value E = (Row total x column total)/ Grand total
-Use this formula for all 4 values in the chart. This should yield 1.3 + 1.62 + 1.928 + 2.382= 7.21
d) Chi Squared test yields 7.21, so the P-Value is between .0005 and .005 (Found using Table 9 or a calculator). Alpha is .05, so no matter what the P value is less than .05. Reject the null!
e) Now put your answer into context of the problem
