Question

In 2001​,researchers investigated the effect of​ weed-killing herbicides on house pets. They examined 827 cats from homes where herbicides were used​ regularly, diagnosing malignant lymphoma in 439 of them. Of the 121 cats from homes where no herbicides were​ used, only 19 were found to have lymphoma.
​a) What is the standard error of the difference in the two​ proportions?
​b) Create a 99% confidence interval for this difference.
​c) State an appropriate conclusion.

Answer 1

a)
Here, , n1 = 827 , n2 = 439
p1cap = 0.1463 , p2cap = 0.0433

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.1463 * (1-0.1463)/827 + 0.0433*(1-0.0433)/439)
SE = 0.0157

b)

For 0.99 CI, z-value = 2.58
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.1463 - 0.0433 - 2.58*0.0157, 0.1463 - 0.0433 + 2.58*0.0157)
CI = (0.0625 , 0.1435)

c)

As zero is not included in the interval we reject the null hypothesis