# Randomly selected 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. My work: (2.29)^2 x (.25)/(.03^2)=1457 but the correct answer is 1509.

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

Now substitute:

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.

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