Question 1174981: It is desired to estimate the mean number of days of continuous use until a certain kind of computer will first require repairs. If it can be assumed that σ = 48hours, how large a sample is needed so that one will be able to assert with 90% confidence that the sample mean is off by at most 10 hours?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the formula that is used here is:
z = (x - m) / s
z is the z-score.
x is the raw score.
m is the raw mean.
s is the standard error.
at 90% two tailed confidence level, the alpha on either end of the confidence interval is 5%.
the critical z-score is plus or minus 1.645, rounded to 3 decimal places.
if we want the range of the sample mean to be plus or minus 10 hours from the mean, then (x - m) must be equal to plus or minus 10.
if the z-score is minus, we can assume that (x - m) is negative which will make it equal to -10.
if the z-score is plus, we can assume that (x - m) is positive which will make it equal to 10.
x and m, in this case, can be anything, just as long as (x - m) is equal to plus or minus 10.
the formula for the standard error is:
s = sqrt(standard deviation) / square root of the sample size.
if we let n equal the sample size, and we are given that the standard deviation is 48, then the formula for standard error becomes s = 48 / sqrt(n)
the formula of z = (x - m) / s becomes z = (x m) / (48 / sqrt(n)) which becomes z = (x - m) * sqrt(n) / 48 because division by an expression is the same as multiplication by the reciprocal of that expression and the reciprocal of sqrt(n) / 48 is 48 / sqrt(n).
when z = -1.645 and (x - m) = -10 and standard deviation = 48, the formula of z = (x - m) * sqrt(n) / 48 becomes -1.645 = -10 * sqrt(n) / 48.
we solve for sqrt(n) to get:
sqrt(n) = 48 * -1.645 / 10.
the result is sqrt(n) = 7.896.
we square that to get n = 62.346816.
we don't need to test positive z-square because the results should be the same, but i'll do it anyway, just to show you.
when z = 1.645, the formula becomes 1.645 = 10 * sqrt(n) / 48.
solve for sqrt(n) to get:
sqrt(n) = 48 * 1.645 / 10 = 7.896.
square that to get n = 62.346816.
to confirm that we have the right sample size, we use this value of n to solve for s in the original equation for s.
we get s = 48 / sqrt(62.346816) = 48 / 7.896 = 6.079027356.
when z = -1.645, we get -1.645 = (x - m) / 6.079027356.
multiply both sides of this equation by 6.079027356 to get:
(x - m) = -1.645 * 6.079027356 = -10
similarly, when z = 1.645, we get:
(x - m) = 1.645 * 6.079027356 = 10.
as long as s = 6.079027356, this should work with any mean.
for example, if mean = 59670, we get:
when z = -1.645, the z-score formula becomes -1.645 = (x - 59670) / 6.079027356.
solve for x to get:
x = -1.645 * 6.079027356 + 59670 = 59660.
that's exactly 10 below 59670.
to make this happen, the standard deviation needs to be 48 and the sample size needs to be to 62.346816.
since the sample size needs to be an integer, then we just go up to the nearest integer to get:
sample size needs to be 63.
this will ensure that the margin of error is less than 10.
the standard error then becomes 48 / sqrt(63) = 6.047431568.
in the same problem we did above, when z = -1.645, the z-score formula becomes -1.645 = (x - 59670) / 6.047431568.
solve for x to get:
x = -1.645 * 6.047431568 + 59670 = 59660.05198
that's equal to 9.94802493 below 59670.
that's just a little bit less than 10 below.
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