Question 481101
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
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90% CI: 3.92-0.07 < u < 3.92+0.07
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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
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Cheers,
Stan H.
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