SOLUTION: In a random sample of 75 shoppers at a grocery store ,20 were found to spend in excess of 30 minutes shopping. Assuming the times are approximately normally distributed, determine

Algebra ->  Probability-and-statistics -> SOLUTION: In a random sample of 75 shoppers at a grocery store ,20 were found to spend in excess of 30 minutes shopping. Assuming the times are approximately normally distributed, determine       Log On


   

Question 868957: In a random sample of 75 shoppers at a grocery store ,20 were found to spend in excess of 30 minutes shopping. Assuming the times are approximately normally distributed, determine the maximum error of estimate for the proportion of the population with 80% confidence.
Found 2 solutions by ewatrrr, stanbon:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!



Hi

p = 20/75 = 4/15

ME = 1.282sqrt%28%28%284%2F15%29%2811%2F15%29%29%2F75%29

CI: 20-ME < u <  20 + ME

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
In a random sample of 75 shoppers at a grocery store ,20 were found to spend in excess of 30 minutes shopping. Assuming the times are approximately normally distributed, determine the maximum error of estimate for the proportion of the population with 80% confidence.
----
ME = z*sqrt[pq/n]
-----
ME = 1.2816*sqrt[(20/75)((55/75)/75] = 0.0654
--------------
Cheers,
Stan H.
=============