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Statistics / Probability and Statistical Inference 9 / Chapter 7.4 / Problem 7.4-3

Probability and Statistical Inference | 9th Edition | ISBN: 9780321923271 | Authors: Robert V. Hogg, Elliot Tanis, Dale Zimmerman

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A company packages powdered soap in 6-pound boxes. The sample mean and standard

Chapter 7, Problem 7.4-3

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QUESTION:

A company packages powdered soap in 6-pound boxes. The sample mean and standard deviation of the soap in these boxes are currently 6.09 pounds and 0.02 pound, respectively. If the mean fill can be lowered by 0.01 pound, $14,000 would be saved per year. Adjustments were made in the filling equipment, but it can be assumed that the standard deviation remains unchanged. (a) How large a sample is needed so that the maximum error of the estimate of the new is = 0.001 with 90% confidence? (b) A random sample of size n = 1219 yielded x = 6.048 and s = 0.022. Calculate a 90% confidence interval for . (c) Estimate the savings per year with these new adjustments. (d) Estimate the proportion of boxes that will now weigh less than 6 pounds.

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QUESTION:

A company packages powdered soap in 6-pound boxes. The sample mean and standard deviation of the soap in these boxes are currently 6.09 pounds and 0.02 pound, respectively. If the mean fill can be lowered by 0.01 pound, $14,000 would be saved per year. Adjustments were made in the filling equipment, but it can be assumed that the standard deviation remains unchanged. (a) How large a sample is needed so that the maximum error of the estimate of the new is = 0.001 with 90% confidence? (b) A random sample of size n = 1219 yielded x = 6.048 and s = 0.022. Calculate a 90% confidence interval for . (c) Estimate the savings per year with these new adjustments. (d) Estimate the proportion of boxes that will now weigh less than 6 pounds.

ANSWER:

Step 1 of 4

Given:

(a) We assume that the population standard deviation is equal to the sample standard deviation s, which is appropriate if the sample is large.

Formula sample size:

For confidence level , determine using table , which is the -value corresponding with a probability of .

Note: We take the average of and , because lies exactly in the middle between and

The sample size is then (round up to the nearest integer!):

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