Question

A stationery store wants to estimate the mean retail value of greeting cards that it has in its inventory. A random sample of 100 greeting cards indicates a mean value of $2.93 and a standard deviation of $0.58.

a.	Assuming a normal​ distribution, construct a 90% confidence interval estimate of the mean value of all greeting cards in the​ store's inventory. $___ ≤ μ ≤ ​$___
b.	Suppose there were 2,000 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?

There are many reasons adults use credit cards. A recent survey found that 47​%

of adults used credit cards for convenience.

a. To conduct a​ follow-up study that would provide 95​% confidence that the point estimate is correct to within

±0.02 of the population​ proportion, how many people need to be​ sampled?

b. To conduct a​ follow-up study that would provide 95​% confidence that the point estimate is correct to within

±0.06 of the population​ proportion, how many people need to be​ sampled?

c. Compare the results of​ (a) and​ (b).

Answer 1

(a)

Answer: ($2.834, $3.036)

(B)

We can be 90% confident that with 2000 cards total value of her​ inventory is in the interval ($5668, $6072).