Question 1153276: In a random sample of six mobile devices, the mean repair cost was $60.00 and the standard deviation was $11.00. Assume the population is normally distributed and use at-distribution to find the margin of error and construct a 99% confidence interval for the population mean. Interpret the results.
The 99% confidence interval for the population mean mu is ( ), ( )
(Round to two decimal places as needed.)
The margin of error is ( _ )(Round to two decimal places asneeded.)
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! half-interval is tdf=5, 0.005*s/sqrt(n)
=4.032 (value of t)*11.00*/sqrt(6)
=$18.11 and this is margin of error
add and subtract to mean
($41.89, $78.11)
We are not exactly sure what the true value is, but we are very highly confident it lies in the above interval.
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