In: Statistics and Probability
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 400 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 400 39 At the .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 as the first sample. 1. State the decision rule. (Negative amounts should be indicated by a minus sign. Do not round the intermediate values. Round your answers to 2 decimal places.) H0 is rejected if z < or z > . Compute the pooled proportion. (Do not round the intermediate values. Round your answer to 3 decimal places.) Pooled proportion 2. Compute the value of the test statistic. (Negative amount should be indicated by a minus sign. Do not round the intermediate values. Round your answer to 2 decimal places.) Value of the test statistic 3. What is your decision regarding the null hypothesis? Decision
let p1 be the proportion of vines infested using Pernod 5 and p2 be the proportion of vines infested using Action.
we want to check whether there is a difference in the proportion of vines infested using Pernod 5 as opposed to Action.
1. so the null hypothesis is H0: p1=p2
and the alternative hypothesis is H1: p1p2
now from the data at hand, we have
a sample of n1=400 vines sprayed with Pernod 5 and a sample of n2=400 vines sprayed with Action.
let f1 be the number of vines in the first sample infested and f2 be the same from sample 2
so f1~Bin(n1,p1) and f2~Bin(n2,p2) independently
then E[f1]=n1*p1 and E[f2]=n2*p2 and V[f1]=n1*p1*(1-p1) and V[f2]=n2*p2*(1-p2)
then E[f1/n1]=p1 and E[f2/n2]=p2 V[f1/n1]=p1*(1-p1)/n1 V[f2/n2]=p2*(1-p2)/n2
then f1/n1 and f2/n2 are the respective sample proportions of infested vines using Pernod 5 and Action.
then E[f1/n1-f2/n2]=p1-p2 V[f1/n1-f2/n2]=p1*(1-p1)/n1+p2*(1-p2)/n2
since n1=n2=400 is very large , hence by central limit theorem the distribution of f1/n1-f2/n2 can be approximated by normal distribution.
2. so f1/n1-f2/n2~N(p1-p2,p1*(1-p1)/n1+p2*(1-p2)/n2)
now under H0, p1=p2=p(say)
then under H0, f1/n1-f2/n2~N(0,p*(1-p)*[1/n1+1/n2])
but since p is unknown, it is estimated by the pooled proportion
phat=(f1+f2)/(n1+n2)
hence the test statistic is given as Z=(f1/n1-f2/n2)/sqrt[phat*(1-phat)*[1/n1+1/n2]]
which under H0 follows N(0,1)
now since the alternative hypothesis is not equal to type, H0 will be rejected iff |z|>taoalpha/2
where z is the observed value of Z and taoalpha/2 is the upper alpha/2 point of N(0,1) and alpha is the level of significance
now f1=26 f2=39 n1=n2=400 phat=(26+39)/(400+400)=0.08125
so value of test statistic=z=(26/400-39/400)/sqrt[0.08125*(1-0.08125)*(1/400+1/400)]=-1.68
3. we have level of significance as alpha=0.10
so tao0.1/2=tao0.05= 1.644854
so |z|=1.68>tao0.05
hence H0 is rejected. and the conclusion is yes, there is a difference in the proportion of vines infested using Pernod 5 as opposed to Action