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, 390 of the vines treated with Pernod 5 were checked for infestation. Likewise, a sample of 420 vines sprayed with Action were checked. The results are:

Insecticide Number of Vines Checked (sample size) Number of Infested Vines
Pernod 5 390 23
Action 420 46

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

The p-value of the test is 0.0088 < 0.05, which means that at the 0.05 significance level, we can conclude that there is a difference in the proportion of vines infested using Pernod 5 as opposed to Action.

Explanation:

Before testing the hypothesis, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean
`\mu = p` and standard deviation
`s = \sqrt{(p(1-p))/(n)}`

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Pernod 5:

23 out of 390, so:

`p_P = (23)/(390) = 0.059`

`s_P = \sqrt{(0.059*0.941)/(390)} = 0.0119`

Action:

46 out of 420, so:

`p_A = (46)/(420) = 0.1095`

`s_A = \sqrt{(0.1095*0.8905)/(420)} = 0.0152`

Test if there is a difference in proportions:

At the null hypothesis, we test if there is not a difference, that is, the subtraction of the proportions is 0. So

`H_0: p_A - p_P = 0`

At the alternative hypothesis, we test if there is a difference, that is, the subtraction of the proportions is different of 0. So

`H_1: p_A - p_P \\eq 0`

The test statistic is:

`z = (X - \mu)/(s)`

In which X is the sample mean,
`\mu` is the value tested at the null hypothesis, and s is the standard error.

0 is tested at the null hypothesis:

This means that
`\mu = 0`

From the samples:

`X = p_A - p_P = 0.1095 - 0.059 = 0.0505`

`s = √(s_A^2+s_P^2) = √(0.0119^2+0.0152^2) = 0.0193`

Value of the test statistic:

`z = (X - \mu)/(s)`

`z = (0.0505 - 0)/(0.0193)`

`z = 2.62`

P-value of the test and decision:

The p-value of the test is the probability of a difference in proportions of at least 0.0505 to either side, which is P(|z| > 2.62), that is, 2 multiplied by the p-value of z = -2.62.

Looking at the z-table, z = -2.62 has a p-value of 0.0044.

2*0.0044 = 0.0088

The p-value of the test is 0.0088 < 0.05, which means that at the 0.05 significance level, we can conclude that there is a difference in the proportion of vines infested using Pernod 5 as opposed to Action.