Homework Soursea 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 60 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)
Note that
p^ = point estimate of the population proportion = x / n = 0.266666667
Also, we get the standard error of p, sp:
sp = sqrt[p^ (1 - p^) / n] = 0.057089923
Now, for the critical z,
alpha/2 = 0.025
Thus, z(alpha/2) = 1.959963985
Thus,
Margin of error = z(alpha/2)*sp = 0.111894192
lower bound = p^ - z(alpha/2) * sp = 0.154772475
upper bound = p^ + z(alpha/2) * sp = 0.378560859
Thus, the confidence interval is
( 0.154772475 , 0.378560859 ) [ANSWER]
****************
b)
Note that
n = z(alpha/2)^2 p (1 - p) / E^2
where
alpha/2 = 0.025
Using a table/technology,
z(alpha/2) = 1.959963985
Also,
E = 0.02
p = 0.266666667
Thus,
n = 1878.046535
Rounding up,
n = 1879 [ANSWER]
******************
c)
Note that
n = z(alpha/2)^2 p (1 - p) / E^2
where
alpha/2 = 0.025
As there is no previous estimate for p, we set p = 0.5.
Using a table/technology,
z(alpha/2) = 1.959963985
Also,
E = 0.02
p = 0.5
Thus,
n = 2400.911763
Rounding up,
n = 2401 [ANSWER]
