1/6 as a percent to the nearest tenth
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Answer:
The 99% confidence interval for the true mean number of hours a union member is absent per month is (0 hours, 16.5125 hours).
Step-by-step explanation:
We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of .
So it is z with a pvalue of , so
Now, find M as such
In which s is the standard deviation of the sample.
So
The lower end of the interval is the mean subtracted by M. So it is 7.5 - 9.0125 = -...
There is not a negative number of hours. So the lower end of the interval is 0 hours.
The upper end of the interval is the mean added to M. So it is 7.5 + 9.0125 = 16.5125 hours.
The 99% confidence interval for the true mean number of hours a union member is absent per month is (0 hours, 16.5125 hours).