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 64 greeting cards indicates a mean value of $3.53 and a standard deviation of $0.23. 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 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

Mean =3.53 Std Deviation =0.23 n=population=1500 z value at 90% confidence level=+/-1.645 Interval = Mean +/- Std dev*1.645/Root1500 = 3.53   +/- 0.23*1.645/Root1500 =3.53+/-0.0098 So the range is between 3.520 & 3.540 As the estimate gives a range of mean value for the Greetings card , the store owner knows the value of the inventory with 90% confidence level. This helps   him to understand the average asset   value he has and the aprroximate fund requirement of the store .

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