In spite of the potential safety hazards, some people would like to have an internet connection in their car. A preliminary survey of adult Americans has estimated this proportion to be somewhere around. 30 (USA today, May 1, 2009). a. Use the given preliminary estimate to determine the sample size required to estimate the proportion of adult Americans who would like an internet connection in their car to within .02 with 95% confidence.

b. The formula for determine sample size given in this section corresponds to a confidence level of 95%. How would you modify this formula if a 99% confidence was desired?

c. Use the given preliminary estimate to determine the sample size required to estimate the proportion of adult Americans who would like an internet connection in their car to within .02 with 99% confidence.

step-by-step explanation:

points a and b are equidistant from the desired line, so the line you find by using them will go through point r.

using points other than a and b will give the desired line only if they are equidistant from r. if they are randomly located, the perpendicular line you create will not necessarily go through point r, hence will not be a solution to the problem.

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.

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