Question

In 1990, researchers investigated the effect of woeding herbicides on house pets. They caminad 10 dogs from homes where herbicides were used regularly diagnosing malignant lymphoma in 446 of them. Of the 140 dogs from homes where no herbicides were used, only 18 were found to have lymphoma aj What is the standard error of the difference in the two proportions? by Create a 99% confidence interval for this difference c) S e an appropriate conclusion aj let, be the sample proportion of dogs with lymphoma from homes where herbicides were used regularly and be the sample proportion of dogs with lymphoma from homes where no herbicides were used. The standard error of the difference in the two proportions SEPP (Round to four decimal places as needed.) b) Create a 90% confidence interval for the d ance between the two proportions.p. - Round to three decimal places as needed) c) Choose an appropriate conclusion below O A. There is 99% confidence that the lymphoma rate of dogs in homes where herbicides we used is greater by an amount in the interval OB. There is 90% confidence that the lymphoma te of pets in homes where herbicides are used is greater by an amount in the interval OC. There is 90% confidence that the lymphoma mensages of dogs from homes where herties are used will be greater by an amount in the interval Click to select your answers MacBook Pro vió å löô IwER T oe soFHK)

Answer 1

a)

x1                =	446	x2                =	18
p̂1=x1/n1 =	0.5506	p̂2=x2/n2 =	0.1286
n1                       =	810	n2                       =	140
estimated difference in proportion   =p̂1-p̂2   =			0.4220
std error Se =√(p̂1*(1-p̂1)/n1+p̂2*(1-p̂2)/n2) =			0.0333

b)

for 99 % CI value of z=		2.576
margin of error E=z*std error =		0.0857
lower bound=(p̂1-p̂2)-E=		0.3364
Upper bound=(p̂1-p̂2)+E=		0.5077
from above 99% confidence interval for difference in population proportion =(0.336,0.508)

c)

option A is correct