In a random sample of
sixsix
mobile devices, the mean repair cost was
$70.0070.00
and the standard deviation was
$14.0014.00.
Assume the population is normally distributed and use a t-distribution to find the margin of error and construct a
9090%
confidence interval for the population mean. Interpret the results.
Question content area bottom
Part 1
The
9090%
confidence interval for the population mean
muμ
is
(58.4858.48,81.5281.52).
(Round to two decimal places as needed.)
Part 2
The margin of error is
$enter your response here.
(Round to two decimal places as needed.)