A researcher wants to see if a kelp extract helps prevent frost damage on tomato plants. One hundred tomato plants in individual
1 answer:
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.
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Answer:
6.18% probability that more than 150 first-year students attend this college.
Step-by-step explanation:
We use the binomial approximation to the normal to solve this question.
For each item selected, there are only two possible outcomes. Either it is defective, or it is not. This means that we use concepts of the binomial probability distribution to solve this problem.
However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
The standard deviation of the binomial distribution is:
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that ,
.
In this problem, we have that:
So
Approximate the probability that more than 150 first-year students attend this college.
This is 1 subtracted by the pvalue of Z when X = 150. So
has a pvalue of 0.9382
1 - 0.9382 = 0.0618
6.18% probability that more than 150 first-year students attend this college.