A bottled water distributor wants to estimate the amount of water contained in 1

Question

A bottled water distributor wants to estimate the amount of water contained in 1 -gallon bottles purchased from a nationally known water bottling company. The water bottling company's specifications state that the standard deviation of the amount of water is equal to 0.01 gallon. A random sample of 50 bottles is selected, and the sample mean amount of water per

1-gallon bottle is 0.987 gallon.

(b) On the basis of these results, do you think that the distributor has a right to complain to the water bottling company? Why?

(d) Construct a 90% confidence interval estimate. How does this change your answer to part (b)?

Explanation / Answer

Z = (x-mean)/[sigma/sqrt(n)] = (0.987-1) / [0.01/sqrt(50)] = -9.19

For, z = 9.19, p value is <0.0005 and is significant, hence distributor has every right to complaint.

b - z for 90 % confidence level is 1.645

Hence, the interval is [1 - 1.645* 0.01/sqrt(50) , 1 + 1.645* 0.01/sqrt(50) ] = [0.9976, 1.0023]

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