Question

n 2000​, researchers investigated the effect of​ weed-killing herbicides on house pets. They examined 832 cats from homes where herbicides were used​ regularly, diagnosing malignant lymphoma in 420 of them. Of the 145 cats from homes where no herbicides were​ used, only 17 were found to have lymphoma. Find the standard error of the difference in the two proportions.

Answer 1

`SE=\sqrt{(0.505 (1-0.505))/(832)+(0.0311(1-0.0311))/(145)}=0.0226`

Explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

1) Data given and notation

`X_(1)=420` represent the number of cats diagnosing malignant lymphoma from homes where herbicides were used​ regularly

`X_(2)=17` represent the number of cats diagnosing malignant lymphoma from homes where NO herbicides were used​ regularly

`n_(1)=832` sample 1 selected

`n_(2)=145` sample 2 selected

`\hat p_(1)=(420)/(832)=0.505` represent the proportion of of cats diagnosing malignant lymphoma from homes where herbicides were used​ regularly

`\hat p_(2)=(17)/(145)=0.0311` represent the proportion of cats diagnosing malignant lymphoma from homes where NO herbicides were used​ regularly

z would represent the statistic (variable of interest)

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

`p_1 -p_2` parameter of interest

2) Solution to the problem

We are interested on the standard error for the difference of proportions and is given by this formula:

`SE=\sqrt{(\hat p_1 (1-\hat p_1))/(n_(1))+(\hat p_2 (1-\hat p_2))/(n_(2))`

And if we replace the values given we got:

`SE=\sqrt{(0.505 (1-0.505))/(832)+(0.0311(1-0.0311))/(145)}=0.0226`