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 380 vines sprayed with Action were checked. The results are:

Insecticide	Number of Vines Checked (sample size)	Number of Infested Vines
Pernod 5	400	25
Action	380	42

At the 0.10 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 amounts should be indicated by a minus sign. Do not round the intermediate values. Round your answers to 2 decimal places.)

2. Compute the pooled proportion. (Do not round the intermediate values. Round your answer to 2 decimal places.)

3. 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.)

4. What is your decision regarding the null hypothesis?

• Reject

• Fail to reject

Answer 1

Let be the true proportion of vines infested using Pernod 5 and be the true proportion of vines infested while using Action

We want to test if there is a difference in the proportion of vines infested using Pernod 5 as opposed to Action, that is we want to test if

The following are the hypotheses

We have the following information from the sample

Since are all greater than 5, we will use normal distribution as the sampling distribution of p. That is we will use z-test to test the hypotheses

This is a 2 tailed test (the alternative hypothesis has "not equal to")

The right tail critical value is

Using the standard normal table we get that for z=1.645, P(Z<1.645) = 0.95.

Hence the right tail critical value is

Since this is a 2 tailed test, the acceptance region is -1.64 to +1.64.

a. State the decision rule

ans: We will reject the null hypothesis if the test statistics lies outside the acceptance region -1.64 to +1.64.

b. The pooled proportion is

ans: The pooled proportion is 0.09

c. The hypothesized value of the difference in proportions is (from the null hypothesis)

The standard error of the difference in proportions is

The test statistics is

ans:  the value of the test statistic is -2.39

d) As per the decision rule in a) We will reject the null hypothesis if the test statistics lies outside the acceptance region -1.64 to +1.64.

Here, the test statistics is -2.39 and it lies outside the acceptance region -1.64 to +1.64. Hence we reject the null hypothesis.

Ans: Reject

e) We conclude that there is sufficient evidence to support the claim that there is a difference in the proportion of vines infested using Pernod 5 as opposed to Action.