Question

In a certain country, pesticide T is used for controlling insects that are harmful to crops. According

to standards set by government, drinking water wells with level of pesticide T above 1 part per billion

are considered contaminated. A study has been conducted by the Environmental Protection

Department (EPD) which has selected a random sample of 250 wells and tested them, and it is found

that 50 wells are contaminated.

(a) Construct a 90% confidence interval for the true proportion of wells contaminated with

pesticide T in the country. State the condition(s) required for this confidence interval to be

valid?

(b) How large a sample is required if the EPD wants to be 95% confident that the estimation

error is within ±0.01?

Answer 1

Solution-A:

z crit for 90%=1.45

p^=sample proportion of wells contaminated=x/n=50/250=0.20

90% confidence interval for the true proportion of wells contaminated with

pesticide T in the country. is

p^-z*sqrt(p^*(1-p^)/n,p^+z*sqrt(p^*(1-p^)/n

0.20-1.645*sqrt(0.20*(1-0.20)/250),0.20+1.645*sqrt(0.20*(1-0.20)/250)

0.1584,0.2416

0.1584<p<0.2416

90% confidence interval lower limit=0.1584

90% confidence interval upper limit=0.2416

Assumptions needed re:

samples to be choosen randomly

sample follows normal distribution that is sample needs to be large ,n>30

10% rule :sample size no more than 10% of the population verify this condition by checking

np^>=10 and nq>=10

Solution-b:

z crit for 95%=1.96

E=0.01

no prior estimation available so p^=1/2=0.5

n=(Z/E)^2*p*(1-p^)

=(1.96/0.01)^2*0.5*(1-0.5)

=9604

required sample size,n=9604