The Damon family owns a large grape vineyard in western New York along Lake Erie

Question

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, 400 of the vines treated with Pernod 5 were checked for infestation. Likewise, a sample of 400 vines sprayed with Action were checked. The results are: Insecticide Number of Vines Checked (sample size) Number of Infested Vines Pernod 5 400 29 Action 400 38 At the .02 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. 1. 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 > . Compute the pooled proportion. (Do not round the intermediate values. Round your answer to 3 decimal places.) Pooled proportion 2. 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 3. What is your decision regarding the null hypothesis? Decision

Explanation / Answer

STEP 1:
Null Hypothesis H0: P1 = P2
Alternate Hypothesis Ha: P1 P2

STEP 2:
Analysis Plan for significance level: 0.05

STEP 3:
Analyze Sample Data
n1: Group 1 sample size = 400
n2: Group 2 sample size = 400

Proportion p1: = 0.0675
Proportion p2: = 0.1

Pooled Sample Proportion
p = (p1 * n1 + p2 * n2) / (n1 + n2)
= [(0.0675 * 400) + (0.1 * 400)] / (400 + 400)
= 67 / 800
= 0.0838

Where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1 is the size of sample 2, and n2 is the size of sample 2.




Standard Error:
SE = sqrt{p * (1 - p ) * [(1/n1) + (1/n2)]}
= sqrt{0.0838 * (1 - 0.0838 ) * [(1/400) + (1/400)]}
= sqrt{0.000384}
= 0.0196


STEP 4:
Test Statistic z-score: z = (p1 - p2) / SE
= (0.0675 - 0.1)/0.0196
= -1.659


For two-tailed test, the p-value is the probability that the z-score is less than -1.659 and more than 1.659


Use the Normal Distribution Table to find P(z < -1.659) = 0.049, and P(z >1.659) = 0.049


The Table for Standard Normal Distribution is organized as a cummulative 'area' from the LEFT corresponding to the STANDARDIZED VARIABLE z. The Standard Normal Distribution is also symmetric (called a 'Bell Curve') which means its an interpretive procedure to Look-Up the 'area' from the Table. For STANDARDIZED VARIABLE z = -1.659 the corresponding LEFT 'area' = 0.049 And due to Table's cummulative nature, the corresponding RIGHT 'area' = 0.049

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