Mathematics, 13.12.2019 21:31, samueldfhung

The damon family owns a large grape vineyard in western new york along lake erie. the grapevines must be sprayed at the beginning of the growing season to protect against various insects and diseases. two new insecticides have just been marketed: pernod 5 and action. to test their effectiveness, three long rows were selected and sprayed with pernod 5, and three others were sprayed with action. when the grapes ripened, 425 of the vines treated with pernod 5 were checked for infestation. likewise, a sample of 410 vines sprayed with action were checked. the results are:

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answered: markel19

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:

$Z_{\alpha /2} = Z_{0.005} = -2.586$

$Z_{1-\alpha /2} = Z_{0.995} = 2.586$

The decision rule is: H₀ is rejected if z ≤ -2.58 or z ≥ 2.58.

The statistic is:

Z= $\frac{(p_1-p_2) - (P_1-P_2)}{\sqrt[]{p(1-p)[1/n_1+1/n_2]} }$ ≈ 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= $\frac{(0.094-0.063) - (0)}{\sqrt[]{0.079(1-0.079)[1/425+1/410]} }$

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

--------------------------------------------------------------------------------

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=

answered: Guest

-362

step-by-step explanation:

(12)(4)+2(12)-34(12)+2(12)-3

\/ \/ \/ \/

48 + 24 - 408 + 24 - 3

\ / \ /

72 - 432 - 3 = -363

answered: Guest