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? Hint: For the calculations, assume the Pernod 5 as the first sample.

a) state the decision rule

b) compute the pooled proportion

c) Compute the value of the test statistic

d) What is your decision regarding the null hypothesis?

Answer 1

p1: Proportion of  vines infested using Pernod 5

p2 : Proportion of  vines infested using Action

Null Hypothesis : Ho : p1=p2 or p1-p2 =0

Alternative hypothesis : H1: p1p2 or p1-p2 0

Two tailed test.

a) state the decision rule

Given, Significance level : =0.10

For two tailed test:

Critical values : -Z/2, Z/2

=0.10 ; /2 =0.10/2=0.05

Critical values : -Z/2, Z/2 : (-Z0.05,Z0.05) (-2.5758,2.5758);

Decision rule : Reject the null hypothesis if Test statistic < -2.5758 or Test statistic >2.5758;

P-value Approach :

If p-value < Level of significance : :0.10 ; Rejet the null hypothesis.

b) compute the pooled proportion

Insecticide	Checked (sample size)	Infested Vines	Sample proportion vines infested using
Pernod 5	n1=440	x2 =42	=42/440=0.0955
Action	n2=360	x2=22	=22/360=0.0611

Pooled proportion = 0.08

c) Compute the value of the test statistic

value of the Test Statistic = 1.7842.

d) What is your decision regarding the null hypothesis?

Critical value approach

As Value of the test statistic is with in the Critical Values i.e.( -2.5758 < 1.7842 < 2.5758 )

Fail To Reject Null Hypothesis

p-value approach

As P-Value i.e. is greater than Level of significance i.e (P-value:0.0744 > 0.01:Level of significance);

Fail to Reject Null Hypothesis
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At the 0.10 significance level. There is not sufficient evidence to conclude that there is a difference in the proportion of vines infested using Pernod 5 as opposed to Action