A nurse is ambulating a patient who has had a stroke. the patient has paresis on the right side of the upper body. where would t
2 answers:
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Answer: Χ²=8.18
Explanation: In this population, the frequency of homozygote poisonous toads is:
p=0.82
For the homozygote non-poisonous toads:
q = 0.18
The frequency for heterozygote poisonous toad, we can use the Hardy-Weinberg equation: , in which, heterozygote frequency is given by 2pq
2pq = 2*0.82*0.18 = 0.3
Now, to compute the chi-square test, follow the instructions:
1) Find the observed values: in this case, they are the found frequency:
0.82 0.3 0.18
2) Find the expected values: As the question mentioned, the expected proportion is:
16 8 1
3) Subtract the observed value from the expected value:
0.82 - 16 = - 0.184 0.3 - 8 = -7.7 0.18 - 1 = - 0.818
4) Square each value from above:
(-0.184)² = 0.034 (-7.7)² = 59.3 (-0.818)² = 0.77
5) Divide each value by expected value:
= 0.0021
= 7.41
= 0.77
6) Add all the values and we will have the chi-square test:
Χ² = 0.0021 + 7.41 + 0.77 = 8.18
The chi-square test is Χ² = 8.18