Hi

p = 20/75 = 4/15

ME = 1.282

CI: 20-ME < u <  20 + ME 

Answer by stanbon(75887)   (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.

=============

 

RELATED QUESTIONS

Here are the shopping times (in minutes) for a sample of 6 shoppers at a particular... (answered by Mathtut)

A large department store has on average 448 shoppers in an hour, with a standard... (answered by Boreal)

People end up tossing 12% of what they buy at the grocery store (Reader's Digest, March,... (answered by Theo)

A random sample of 450 shoppers at Quincy Mall found that 125 favored longer shopping... (answered by stanbon)

Executives of a supermarket chain are interested in the amount of time that customers... (answered by textot)

at a shopping mall, 40% of the shoppers are willing to take a survey. estimate the... (answered by Boreal)

A random sample of 200 male shoppers in pavilion shopping mall were interviewed to... (answered by Boreal)

SUPPOSE THAT DURING ANY HOUR IN A LARGE DEPARTMENT STORE, THE AVERAGE NUMBER OF SHOPPERS... (answered by stanbon)

 Executives of a supermarket chain are interested in the amount of time that customers... (answered by stanbon)