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, 440 of the vines treated with Pernod 5 were checked for infestation. Likewise, a sample of 360 vines sprayed with Action were checked. The results are: Insecticide Number of Vines Checked (sample size) Number of Infested Vines Pernod 5 440 42 Action 360 22 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

Answer 1

Answer:

$z=\frac{0.0955-0.0611}{\sqrt{0.08(1-0.08)(\frac{1}{440}+\frac{1}{360})}}=1.784$    

$p_v =2*P(Z>1.784) =0.074$  

Comparing the p value with the significance level given $\alpha=0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the two proportions differes at 10% of significance.

Step-by-step explanation:

Data given and notation  

$X_{1}=42$ represent the number of infested vines for Pernod5

$X_{2}=22$ represent the number of infested vines for Action

$n_{1}=440$ sample of Pernod5

$n_{2}=360$ sample of Action

$\hat p_{1}=\frac{42}{440}=0.0955$ represent the proportion of residents of  infested vines for Pernod5

$\hat p_{2}=\frac{22}{360}=0.0611$ represent the proportion of infested vines for Action

z would represent the statistic (variable of interest)  

$p_v$ represent the value for the test (variable of interest)  

$\alpha=0.1$ significance level given

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if is there is a difference between thw two population proportions, the system of hypothesis would be:  

Null hypothesis:$p_{1} - p_{2}=0$  

Alternative hypothesis:$p_{1} - p_{2} \neq 0$  

We need to apply a z test to compare proportions, and the statistic is given by:  

$z=\frac{\hat p_{1}-\hat p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{1}})}}$   (1)  

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{42+22}{440+360}=0.08$  

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

Calculate the statistic  

Replacing in formula (1) the values obtained we got this:  

$z=\frac{0.0955-0.0611}{\sqrt{0.08(1-0.08)(\frac{1}{440}+\frac{1}{360})}}=1.784$    

Statistical decision

Since is a two sided test the p value would be:  

$p_v =2*P(Z>1.784) =0.074$  

Comparing the p value with the significance level given $\alpha=0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the two proportions differes at 10% of significance.