Question

Number 9?

Use the following information/data to answer questions 8 and 9: ou have invented a new biodegradable chemical spray that will protect developing almond nuts from seed eating weevil insects. In a "clinical" trial, the almonds on one side of an almond tree were sprayed with your chemical while th Treatment Type Chemical Spray Controls e other side almonds were left untreated to serve as the controls. The results are as follows: # of seeds with weevils % seeds weevil infected # of almond seeds 245 213 182 74.2857143% 191 89.6713615% 8. Calculate the 95% confidence interval for the percentage (proportion) of sprayed seeds that were weevil infected. Remember, z 1.960: a.) Standard Method b.) Wilson Method c.) Safe/Conservative Method 9. Calculate the 95% confidence interval for the difference in percentage of seeds that were infected for these two treatments using the standard method ANDİNTERPRETTHISCONFIDENCEINTERVAL. Remember, z = 1.960.

Answer 1

The 95% per cent confidence interval for p1-p2 is given by:
$(\hat{p}_1-\hat{p}_2)\pm z_{\alpha/2} \sqrt{\dfrac{\hat{p}_1(1-\hat{p}_1)}{n_1}+\dfrac{\hat{p}_2(1-\hat{p}_2)}{n_2}}$

are the proportions of two groups.

pı and fz

$z_{\alpha/2}$is the critical value
             

$\sqrt{\dfrac{\hat{p}_1(1-\hat{p}_1)}{n_1}+\dfrac{\hat{p}_2(1-\hat{p}_2)}{n_2}}$ is the SE of the difference

So confidence interval is

$(0.89671-0.74286)\pm 1.96* \sqrt{\dfrac{0.74286(1-0.74286)}{245}+\dfrac{0.89671(1-0.89671)}{213}}$

or

So 95% confidence interval for the difference in the percentage of seeds that were infected for these two treatments is between 8.5to 22.2 %