Answer to Question #281076 in Statistics and Probability for kcrodrigo

Question

A campaign was designed to convince car owners that they should fill their tires with nitrogen instead of air. At a cost of about ₱60 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 99% confident that the sample percentage is within four percentage points of the true percentage of all car owners who would be willing to pay for the nitrogen.

Answer 1

Solution:

For a population proportion P , confidence interval is

(P-ME ,P+ME)

here ME is margin of error and equation to find margin of error is

$M E=z \times \sqrt{\dfrac{P(1-P)}{n}}$

here z is z score for confidence interval and n is total sample size

On rearranging this , equation to find sample size is

$n=\left(\dfrac{z}{M E}\right)^{2} \times P(1-P)$

Here proportion is unknown. So take P=0.5 (when P is unknown, Take P as 0.5)

Here sample percentage should be within four percentage points of the true percentage. So margin of error is 4%. So ME=0.04

z score for 99% confidence interval is z=2.576

Using these values on equation for sample size we get

$n=\left(\dfrac{2.576}{0.04}\right)^{2} \times 0.5 \times(1-0.5)=1037$

So we need to survey at least 1037 Car owners.