Question 1141192: A travel agent wants to estimate the mean price of a flight from Boston to Denver. A different airline study show that the standard deviation of flight tickets is $56.92. The agent wants her estimate to be within $25 of the true mean flight cost. How many flights must she include in her sample in order to be 90% in her estimate?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the critical z-score at 90% confidence level is plus or minus 1.645.
the z-score formula is z = (x - m) / s
z is the z-score.
x is the raw score
m is the mean
s is the standard error.
the formula for s is:
s = standard deviation / square root of sample size = sd / sqrt(n).
for the score to be within 25 of the mean, then (x - m) must be equal to plus or minus 25.
we'll work the plus side.
the minus side should give you the same answer.
z = (x - m) / s becomes 1.645 = 25 / s which becomes 25 / (56.92/sqrt(n))
simplify to get 1.645 = 25 * sqrt(n) / 56.92
solve for sqrt(n) to get sqrt(n) = 1.645 * 56.92 / 25 = 3.745336.
solve for n to get n = 3.745336 ^ 2 = 14.02754175.
when the sample size is 14.02754175, s = sd / sqrt(n) becomes s = 56.92 / sqrt(14.02754175) = 15.19756839.
when z = 1.645, the z-score formula of z = (x - m) / s becomes 1.645 = (x - m) / 15.19756839.
solve for (x - m) to get (x - m) = 1.645 * 15.19756839) = 25.
if the z-score is -1.645, then z = (x - m) / s becomes -1.645 = (x - m) / 15.19756839 and you get (x - m) = -1.645 * 15.19756839 = -25.
when x - m = 25, then x = m + 25.
when x - m = -25, then x = m - 25.
your score will be within 25 dollars of the mean.
the mean could be anything and the formula will get you within 25 dollars of the mean.
for example, let the mean be 100 dollars.
at the 90% confidence level, the z-score is plus or minus 1.645.
the positive z-score formula becomes 1.645 = (x - 100) / 15.19756839.
solve for x to get x = 1.645 * 15.19756839 + 100 = 125.
the negative z-score formula becomes -1.645 = (x - 100) / 15.19756839.
solve for x to get x = -1.645 * 15.19756839 + 100 = 75.
125 is 25 dollars from the mean of 100.
75 is 25 dollars from the mean of 100.
you can substitute any other mean and you will get the same distance from the mean, i.e. 25 dollars plus or minus from the mean.
to be exact, the sample size needed to be 14.02754175.
however, the number of flights is not a fraction, therefore you would go to the next higher integer.
she must include at least 15 flights to be 25 dollars or less from the true mean.
the following excel spreadsheet shows some of the figures from 10 to 20 flights in the sample size, including the one where we calculated 14.02754175.
|
|
|
