SOLUTION: True or False: Jennifer wants to find a 95% confidence interval for the time it takes her to get to work. She kept records for 30 days and found her average time to commute to work
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-> SOLUTION: True or False: Jennifer wants to find a 95% confidence interval for the time it takes her to get to work. She kept records for 30 days and found her average time to commute to work
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Question 1204928: True or False: Jennifer wants to find a 95% confidence interval for the time it takes her to get to work. She kept records for 30 days and found her average time to commute to work was 20.5 minutes with a standard deviation for the population of 3.9 minutes. Jennifer's margin of error would be 1.4 minutes. Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! sample mean is 20.5 minutes.
sample size is 30
population standard deviation is 3.9
sample margin of error is supposed to be 1.4 minutes.
standard error = standard deviation / sqrt(sample size) = 3.9/sqrt(30) = .71204.
critical z-score at 95% confidence interval is plus or minus z = 1.96.
z-score formula is z = (x-m)/s)
z is the critical z-score
x is the sample mean
m is the assumed mean
(x-m) is the margin of error)
s is the standard error.
to find the margin of error, we solve for (x-m).
from the z-score formula, we solve for (x-m) to get:
(x-m) = z * s.
we get:
(x-m) = 1.96 * .71204 = 1.3956
it becomes 1.4 when we round to 1 decimal place.
that was on the high side of the confidence interval.
on the low side of the confidence interval, the results should be the same, only negative.
(x-m) = z * s becomes (x-m) = -1.96 * .71204 = -1.3956.
that becomes -1.4 when we round to 1 decimal place.
it looks like the statement is true.
use of the z-score was indicated because the sample size was 30 and we were using the population standard deviation.
note that we know what the margin of error is, but we don't know what the population mean is assumed to be.
it could be anything.
the margin of error would always be 1.4 as long as the standard error is the same at .71204.
here's what it looks like on the normal distribution graph.
i assumed a population mean of 50 just because the calculator requires a mean when it's dealing with raw scores.
the margin of error is shown as plus or minus 1.396 which rounds to 1.4.
