A stationery store wants to estimate the mean retail value of greeting cards tha

Question

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 $ 3.82 and a standard deviation of $ 0.66 . a. Assuming a normal distribution, construct a 99 % confidence interval estimate of the mean value of all greeting cards in the store's inventory. b. Suppose there were 1,500 greeting cards in the store's inventory. How are the results in part (a) useful in assisting the store owner to estimate the total value of her inventory?

Explanation / Answer

standard error = 0.66/sqrt(121)

margin of error = z-score * standard error

= 2.576 * 0.66/sqrt(121)

= 0.15456

confidence interval is [3.82 - 0.15456 , 3.82 + 0.15456 ] which is [ 3.67 , 3.98 ]

b)

money in the inventory lies in the range of [ 3.67 * 1500 , 3.98 * 1500 ]

which is [ 5505 , 5970 ]

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