Respuesta :
Answer:
a. H₀ is rejected if z ≤ -2.58 or z ≥ 2.58.
b. Pooled proportion : 0.079
c. Value of the test statistic : Z= 1.66
d. The decision is to not reject the null hypothesis.
p-value: 0.0969
Step-by-step explanation:
Hello!
The objective of this experiment is to test if there is a difference in the effectiveness of two new insecticides Pernod 5 and Action. To do so, two random samples of vines, one is treated with Pernod 5 and the other with Action. At the end of the season, it was documented the number of infested vines in each sample.
Sample 1 (Pernod 5)
X₁: Number of infested vines, after being treated with Pernod 5, in a sample of 425.
n₁= 425
x₁= 40
Sample 2 (Action)
X₂: Number of infested vines, after being treated with Action, in a sample of 425.
n₂= 410
x₂= 26
The hypothesis to test is if there is any difference between the proportions of infested vines in both populations, symbolically:
H₀: ρ₁ = ρ₂
H₁: ρ₁ ≠ ρ₂
α: 0.01
The critical region for this type of hypothesis is two-tailed, the critical number is:
[tex]Z_{\alpha /2} = Z_{0.005} = -2.586[/tex]
[tex]Z_{1-\alpha /2} = Z_{0.995} = 2.586[/tex]
The decision rule is: H₀ is rejected if z ≤ -2.58 or z ≥ 2.58.
The statistic is:
Z= [tex]\frac{(p_1-p_2) - (P_1-P_2)}{\sqrt[]{p(1-p)[1/n_1+1/n_2]} }[/tex] ≈ N(0;1)
P₁ and P₂ represent the population proportions
p₁ and p₂ represent the sample proportions
p is the pooled proportion
p= (x₁ + x₂)/(n₁ + n₂) = (40+26)/(425+410)= 0.0790
p₁= x₁/n₁= 40/425= 0.094
p₂=x₂/n₂= 26/410= 0.063
Z= [tex]\frac{(0.094-0.063) - (0)}{\sqrt[]{0.079(1-0.079)[1/425+1/410]} }[/tex]
Z= 1.66
Since the calculated Z value is between the two critical values, the decision is to not reject the null hypothesis. This means, at a significance level of 1% that there is no difference between the population proportions of infected vines treated with Pernod 5 and Action.
The p-value is, as the critical region, two-tailed.
To calculate it: P(Z<-1.66) + P(Z>1.66) = P(Z<-1.66) + (1 - P(Z<1.66)) = 0.0485 + ( 1 - 0.9515)= 0.0969
I hope it helps!
-rest of the question-
Insecticides; Number of vines checked sample size; Number of infected vines
Pernod 5 425 40
Action 410 26
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At the 0.01 significance level, can we conclude that there is a difference in the proportion of vines infested using Pernod 5 as opposed to Action? Hint: For the calculations, assume the Pernod as the first sample.
a. State the decision rule. (Negative amounts should be indicated by a minus sign. Do not round the intermediate values. Round your answers to 2 decimal places.)
H0 is rejected if z < or z > .
b. Compute the pooled proportion. (Do not round the intermediate values. Round your answer to 3 decimal places.)
Pooled proportion
c. Compute the value of the test statistic. (Negative amount should be indicated by a minus sign. Do not round the intermediate values. Round your answer to 2 decimal places.)
Value of the test statistic
d. What is your decision regarding the null hypothesis? (Round your p-value answer to 4 decimal places.)
p value=