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Dylan wants to determine a 80 percent confidence interval for the true proportion of high school students in the area who attend their home basketball games. How large of a sample must he have to get a margin of error less than 0. 02? [to find n, use the value p* = 1/2 f

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To find the required sample size, we can use the formula:

n = (z_alpha/2)^2 * p* (1-p*) / ME^2

where:

z_alpha/2 is the critical value of the standard normal distribution for the desired confidence level. For an 80% confidence interval, this value is approximately 1.28.
p* is an estimate of the true proportion of high school students who attend their home basketball games. The problem suggests using p* = 1/2, which corresponds to maximum variability and provides the most conservative estimate for the sample size.
ME is the desired margin of error, which is given as 0.02 in the problem.
Substituting the given values into the formula, we get:

n = (1.28)^2 * (1/2) * (1 - 1/2) / (0.02)^2
n = 615.04

Rounding up to the nearest integer, we get:

n = 616

Therefore, Dylan needs to survey at least 616 high school students to obtain an 80% confidence interval with a margin of error less than 0.02, assuming that the true proportion of students who attend their home basketball games is around 50%.
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