a Find a 95 twosided confidence interval on the true proport

(a) Find a 95% two-sided confidence interval on the true proportion of helmets of this type that would show damage from this test. Round the answers to 3 decimal places.

(b) Using the point estimate of p obtained from the preliminary sample of 70 helmets, how many helmets must be tested to be 95% confident that the error in estimating the true value of p is less than 0.02?

(c) How large must the sample be if we wish to be at least 95% confident that the error in estimating p is less than 0.02, regardless of the true value of p?

Solution

a)
Proportion p = 18/70 = 0.2571
z for 95% confidence interval =1.96
Confidence interval :
p +- z*sqrt(p(1-p)/n)
=(0.2571 - 1.96*sqrt(0.2571*(1-0.2571)/70) ,0.2571 + 1.96*sqrt(0.2571*(1-0.2571)/70) )
=(0.155 , 0.359) Answer

b)
Let n be the number of helmets that must be tested to be 95% confident that the error bound is less than 0.02
alpha=1-0.95 = 0.05
Using ^p=02571 to estimate p,then
n = p(1-p)*(z(alpha/2)/B)^2
= 0.2571*(1-0.2571) *(1.96 /0.02)^2
= 1834.36
~ 1835 Helmets Answer

c)
Using the estimate of p:
n =0.2571*(1-0.2571) *(1.96 /0.02)^2
= 1834.36
~ 1835 Helmets Answer

Without using estimate of p:
n = (1.96 /0.02)^2 *(0.5)*(0.5)
= 2401 Answer