Question 481101: Please help me solve:
A stationery store wants to estimate the mean retail value of greeting cards that it has in its inventory. A random sample of 121 greeting cards indicates a mean value of $3.92 and a standard deviation of $0.45.
a. Assuming a normal distribution, construct a 90% confidence interval estimate of the mean value of all greeting cards in the store's inventory.
b. Suppose there were 3,000 greeting cards in the inventory. How are the results useful in assisting the owner to estimate the total value of her inventory.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A stationery store wants to estimate the mean retail value of greeting cards that it has in its inventory. A random sample of 121 greeting cards indicates a mean value of $3.92 and a standard deviation of $0.45.
a. Assuming a normal distribution, construct a 90% confidence interval estimate of the mean value of all greeting cards in the store's inventory.
x-bar = 3.92
ME = 1.645*0.45/sqrt(121) = 0.07
--
90% CI: 3.92-0.07 < u < 3.92+0.07
==========================================
b. Suppose there were 3,000 greeting cards in the inventory. How are the results useful in assisting the owner to estimate the total value of her inventory.
With 90% confidence she can conclude the total value is between
3000*3.85 and 3000*3.99
===========================
Cheers,
Stan H.
==========
|
|
|
