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 26
Action 380 40

Required:
At the 0.05 significance level, can we conclude that there is a difference in the proportion of vines infested using Pernod 5 as opposed to Action?

Answer 1

We conclude that there is a difference in the proportion of vines infested using Pernod 5 as opposed to Action.

Explanation:

We are given that Two new insecticides have just been marketed: Pernod 5 and Action. The results are:

Insecticide Number of Vines Checked (sample size) Number of Infested Vines

Pernod 5 400 26

Action 380 40

Let
`p_1` = proportion of vines infested using Pernod 5.

`p_2` = proportion of vines infested using Action.

So, Null Hypothesis,
`H_0` :
`p_1-p_2` = 0 {means that there is no difference in the proportion of vines infested using Pernod 5 as opposed to Action}

Alternate Hypothesis,
`H_A` :
`p_1-p_2\\eq` 0 {means that there is a difference in the proportion of vines infested using Pernod 5 as opposed to Action}

The test statistics that would be used here Two-sample z proportion statistics;

T.S. =
`\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+(\hat p_2(1-\hat p_2))/(n_2) } }` ~ N(0,1)

where,
`\hat p_1` = sample proportion of vines infested using Pernod 5 =
`(26)/(400)` = 0.065

`\hat p_1` = sample proportion of vines infested using Action =
`(40)/(380)` = 0.105

`n_1` = sample of Pernod 5 vines = 400

`n_2` = sample of Action vines = 380

So, the test statistics =
`\frac{(0.065-0.105)-(0)}{\sqrt{(0.065(1-0.065))/(400)+(0.105(1-0.105))/(380) } }`

= -2.002

The value of z test statistics is -2.002.

Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistics does not lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that there is a difference in the proportion of vines infested using Pernod 5 as opposed to Action.