Respuesta :
Answer:
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.
Step-by-step 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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
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 [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
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:
[tex]p_P = \frac{23}{390} = 0.059[/tex]
[tex]s_P = \sqrt{\frac{0.059*0.941}{390}} = 0.0119[/tex]
Action:
46 out of 420, so:
[tex]p_A = \frac{46}{420} = 0.1095[/tex]
[tex]s_A = \sqrt{\frac{0.1095*0.8905}{420}} = 0.0152[/tex]
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
[tex]H_0: p_A - p_P = 0[/tex]
At the alternative hypothesis, we test if there is a difference, that is, the subtraction of the proportions is different of 0. So
[tex]H_1: p_A - p_P \neq 0[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{s}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, and s is the standard error.
0 is tested at the null hypothesis:
This means that [tex]\mu = 0[/tex]
From the samples:
[tex]X = p_A - p_P = 0.1095 - 0.059 = 0.0505[/tex]
[tex]s = \sqrt{s_A^2+s_P^2} = \sqrt{0.0119^2+0.0152^2} = 0.0193[/tex]
Value of the test statistic:
[tex]z = \frac{X - \mu}{s}[/tex]
[tex]z = \frac{0.0505 - 0}{0.0193}[/tex]
[tex]z = 2.62[/tex]
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.
