The Damon family owns a largo grape vineyard in western Now York along new York
ID: 3228749 • Letter: T
Question
The Damon family owns a largo grape vineyard in western Now York along 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 sprayed. 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: At the .05 significance level, can we conclude that there is a difference in the proportion of vines infested using Pernod 5 as opposed to Action? State the decision rule Compute the pooled proportion Compute the value of the test statistic: What is your decision regarding the null hypothesis?Explanation / Answer
Solution:-
1) Decision rule is
+ 1.96 < z < 1.96
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P1 = P2
Alternative hypothesis: P1 P2
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the proportion from population 1 is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a two-proportion z-test.
Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).
p = (p1 * n1 + p2 * n2) / (n1 + n2)
p = 0.08625
SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
SE = 0.01985
z = (p1 - p2) / SE
z = - 1.39
where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.
Since we have a two-tailed test, the P-value is the probability that the z-score is less than - 1.39 or greater than 1.39.
Thus, the P-value = 0.165
Interpret results. Since the P-value (0.165) is greater than the significance level (0.05), we have to accept the null hypothesis.
From the above test we do not have sufficient evidence in the favor of the claim that there is difference between the proportions.
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