A recent campaign was designed to convince car owners that they should fill their tires with nitrogen instead of air. At a cost of about $5 per tire, nitrogen supposedly has the advantage of leaking at a much slower rate than air, so that the ideal tire pressure can be maintained more consistently. Before spending huge sums to advertise the nitrogen, it would be wise to conduct a survey to determine the percentage of car owners who would pay for the nitrogen. How many randomly selected car owners should be surveyed? Assume that we want to be 98% confident that the sample percentage is within three percentage points of the true percentage of all car owners who would be willing to pay for the nitrogen.
(2.29)^2 x (.25)/(.03^2)=1457
but the correct answer is 1509.
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The only mistake you made is the Z value for the 98% confidence level.
For 98% confidence, z value = 2.33
We will also use P= 0.5 as the population is not known.
The confidence interval (c) is known as 3% = 0.03
SS= (z^2)*P (1-P)/ c^2
SS= [(2.33)^2 (.5)(1-0.5)]/(0.03)^2
= [(2.33)^2 (0.25)]/0.0009
= 1508. 027
==> SS = 1509 is the minimum sample size.
From the values you have taken it could be observed that you take the proportional value P = 0.5, Q = 1-P = 0.5. Therefore the estimayed sample standard deviation is sqrt (PQ/n) for the sample size n is sqrt((0.5)(0.5)/n) = sqrt(0.25/n)
Let the observed value of the sample proportion be p
|p-P| = 3% of P= 0.03*(1/2) = 0.015.
So, (Observed value, p - population Prportion P)/sqrt(0.5^2/n is a normal variate p with mean P and variance PQ/n = 0.25/n
Therefore Pr( 0.5-0.015 < p < 0.5+0.015) = 0.98 Or
Pr( Z < 0.515) < 0.99.
Therefore (0.515-0.5)/sqrt(0.5^2/n) = 2.3266.
(0.015)(sqrtn)/0.25) = 2.3266
n = 2.3266^2*0.25/0.015^2 = 6015 .
So I differ with your book also. The reason is that instead of the sample proportion to vary 0.03 % of population proportion on either side, the work out shows p-P = 0.03 which amounts 0.03/(1/2)*100 % = 6% on either side of the population proportion P =1/2.
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