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 26 Action 400 39 At the .10 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 5 as the first sample. (1) State the decision rule. (Negative values 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 value 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? (Round the value of the test statistic to 2 decimal places. Round your p-value answer to 4 decimal places.) p-value Decision

Answer 1

Given that,

For sample 1 : n1 = 400, x1 = 26 and

For sample 2 : n2 = 400, x2 = 39 and

The null and alternative hypotheses are,

H0 : p1 = p2

Ha : p1 ≠ p2

1) critical values at significance level of 0.10 are, z^* = ± 1.645

Decision Rule : H0 is rejected if z < -1.645 or z > 1.645

Pooled proportion is,

=> Pooled proportion = 0.08125

2) Test statistic is,

=> Test statistic = -1.68

3) Since, test statistic = -1.68 < -1.645, we reject the null hypothesis.

p-value = 2 * P(Z < -1.68) = 2 * 0.0465 = 0.0930

=> p-value = 0.0930