Prime factorisation of 500
1 answer:
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Complete question is;
A researcher wants to see if a kelp extract helps prevent frost damage on tomato plants. One hundred tomato plants in individual containers are randomly assigned to two different groups. Plants in both groups are treated identically, except that the plants in group 1 are sprayed weekly with a kelp extract, while the plants in group 2 are not. After the first frost, 12 of the 50 plants in group 1 exhibited damage and 18 of the 50 plants in group 2 showed damage. Let p1 be the actual proportion of all tomato plants of this variety that would experience damage under the kelp treatment, and let p2 be the actual proportion of all tomato plants of this variety that would experience damage under the no-kelp treatment. Is there evidence of a decrease in the proportion of
tomatoes suffering frost damage for tomatoes sprayed with kelp extract? To determine
this, yo u test the hypotheses H0: p1 = p2, Ha: p1 < p2. The p - value of your test is
A) greater than 0.10.
B) between 0.05 and 0.10.
C) between 0.01 and 0.05.
D)between 0.001 and 0.01.
E) below 0.001.
Answer:
B) between 0.05 and 0.10.
Step-by-step explanation:
We are given;
Number of plants in group that exhibited damage = 12
Number of group 1 plants; n₁ = 50
Number of plants in group 2 that exhibited damage = 18
Number of group 2 plants; n₂ = 50
Proportion of plants in group 1 that exhibited damage; p₁^ = 12/50 = 0.24
Proportion of plants in group 2 that exhibited damage; p₂^ = 18/50 = 0.36
Pooled proportion; p¯ = (12 + 18)/(50 + 50) = 0.3
q¯ = 1 - p¯
q¯ = 1 - 0.3 = 0.7
Z-score will be;
z = ((p₁^ - p₂^) - 0)/√((p¯•q¯)×((1/n₁) + (1/n₂))
Plugging in the relevant values;
z = (0.24 - 0.36)/√((0.3 × 0.7) × ((1/50) + (1/50))
z = -0.12/0.092
z = -1.3
From z-distribution table attached, we have;
p-value = 0.0968
Correct answer is option B.
