Need help 1 and 3 The Damon family owns a large grape vineyard in western New Yo

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Need help 1 and 3

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 thei 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 Number of Vines Checked (sample size) 400 400 Number of Insecticide Pernod 5 Action Infested Vines 27 40 At the.02 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 (1) State the decision rule. (Negative values should be indicated by a minus sign. Do not round the intermediate values. Round your answers to 2 decimal places.) Ho is rejected if z

Explanation / Answer

For sample 1, we have that the sample size is N1=400, the number of favorable cases is X1=27, so then the sample proportion is p^1= N1/X1=27/400=0.0675

For sample 2, we have that the sample size is N2 =400, the number of favorable cases is X2=40, so then the sample proportion p^2=N2/X2=40/400=0.1

The value of the pooled proportion is compute as P =27+40/400+400=0.0838

Also, the given significance level is =0.02.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho:p1=p2H

Ha: p1p2

This corresponds to a two-tailed test, for which a z-test for two population proportions needs to be conducted.

(2) Rejection Region

Based on the information provided, the significance level is =0.02, and the critical value for a two-tailed test is zc=2.33.

The rejection region for this two-tailed test is R={z:z>2.33}

Z<-2.33 or Z> 2.33

(3) Test Statistics

The z-statistic is computed as follows:

z=p^1p^2/sqrt(P(1-P)*(1/N1+1/N2))=(0.06750.1)/sqrt((0.0838)*(1-0.0838)(1/400+1/400)) =1.659= -1.66

(4) Decision about the null hypothesis

Since it is observed that z=1.659zc=2.33, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: p=0.0971, and since p = 0.0971 p=0.09710.02, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population proportion p1 is different than p2, at the 0.02 significance level.

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