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, 410 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 5 410 26

Action 400 39

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?

Step-by-step explanation:

This is a test of 2 population proportions. Let 1 and 2 be the subscript for the the vines infested using Pernod 5 and vines infested using Action. The population proportion of the vines infested using Pernod 5 and vines infested using Action would be p1 and p2 respectively.

p1 - p2 = difference in the proportion of the vines infested using Pernod 5 and vines infested using Action.

The null hypothesis is

H0 : p1 = p2

p1 - p2 = 0

The alternative hypothesis is

Ha : p1 ≠ p2

p1 - p2 ≠ 0

it is a two tailed test

Sample proportion = x/n

Where

x represents number of success (number of complaints)

n represents number of samples

For vines infested using Pernod 5,

x1 = 26

n1 = 410

p1 = 26/410 = 0.063

For vines infested using Action,

x2 = 39

n2 = 400

P2 = 39/400 = 0.098

The pooled proportion, pc is

pc = (x1 + x2) / (n1 + n2)

pc = (26 + 39) / (410 + 400) = 0.08

1 - pc = 1 - 0.08 = 0.92

z = (p1 - p2) / √pc (1 - pc) (1/n1 + 1/n2)

z = (0.063 - 0.098) / √ (0.08) (0.92) (1/410 + 1/400) = - 0.035/0.019066

z = - 1.84

Since it is a two tailed test, the curve is symmetrical. We will look at the area in both tails. Since it is showing in one tail only, we would double the area

From the normal distribution table, the area below the test z score in the left tail 0.033

We would double this area to include the area in the right tail of z = 1.84 Thus

p = 0.033 * 2 = 0.066

By using the p value,

Since alpha, 0.01 < than the p value, 0.066, then we would fail to reject the null hypothesis.

Therefore, At a 1% level of significance, we cannot conclude that there is a difference in the proportion of vines infested using Pernod 5 as opposed to Action