HW 47 A random sample of 50 suspension helmets used by motor

HW 4.7. A random sample of 50 suspension helmets used by motorcycle riders and automobile race-car drivers was marketed to an impact test, and some damage was observed on 18 of these helmets. (a) Find a 95% two-sided confidence interval on the true proportion of helmets, denoted by p, that would show damage from this test. (1)) Using the point estimate of p from the 50 helmets, how many helmets must be tasted to be 95% confident that the error in estimating 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)
Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=18
Sample Size(n)=50
Sample proportion = x/n =0.36
Confidence Interval = [ 0.36 ±Z a/2 ( Sqrt ( 0.36*0.64) /50)]
= [ 0.36 - 1.96* Sqrt(0.005) , 0.36 + 1.96* Sqrt(0.005) ]
= [ 0.227,0.493]

b)
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.05 is = 1.96
Samle Proportion = 0.2
ME = 0.02
n = ( 1.96 / 0.02 )^2 * 0.2*0.8
= 1536.64 ~ 1537

c)
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.05 is = 1.96
Samle Proportion = 0.5
ME = 0.02
n = ( 1.96 / 0.02 )^2 * 0.5*0.5
= 2401 ~ 2401