At the beginning of the year, Ms. Carter had 45 post it notes on her desk. Ms. Carter now has 60% more than the post its she sta
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Answer:
The 98% confidence interval for the mean purchases of all customers is ($37.40, $61.74).
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 is the standard deviation of the population and n is the size of the sample.
The lower end of the interval is the mean subtracted by M. So it is 49.57 - 12.17 = $37.40.
The upper end of the interval is the mean added to M. So it is 49.57 + 12.17 = $61.74.
The 98% confidence interval for the mean purchases of all customers is ($37.40, $61.74).