8.81) A quality of characteristic of interest for atea-bag-filling process is th

Question

8.81) A quality of characteristic of interest for atea-bag-filling process is the weight of the tea in the individualbags. In this example, the label weight on the package indicatesthat the mean amount is 5.5 grams of tea in a bag. If the bagsare under filled, two problems arise. First customers, may not beable to brew the tea to be as strong as they wish. Second, thecompany may be in violation of the truth-in-labeling laws/ On theother hand, if the mean amount of tea in a bag exceeds the labelweight, the company is giving away product. Getting an exact amountof tea in a bag is problematic because of variation in thetemperature and humidity inside the factory, difference in thedensity of the tea, and the extremely fast filling operation of themachine.

5.65,5.57,5.47,5.77,5.61,5.44,5.40,5.40,5.57,5.45,5.42,5.53,5.47,5.42,5.44,5.40,5.54,5.61,5.58,5.25,5.53,5.55,5.53,5.58,5.56,5.63,5.50,5.32,5.62,5.34,5.54,5.56,5.67,5.32,5.50,5.57,5.50,5.29,5.46,5.45,5.52,5.44,5.49,5.53,5.53,5.67,5.41,5.51,5.51,5.55,5.58,5.36

A) construct a 99% confidence interval estimate of thepopulation mean weight of the tea bags.
B) is the company meeting the requirement set forth on the labelthat the mean amount of tea in a bag is 5.5 grams?

Explanation / Answer

mean of the above data , = 5.502115 standard deviation of the above data, = 0.103816 A) construct a 99% confidence interval estimate of thepopulation mean weight of the tea bags.
at probability = 0.99, z value from the table = 2.33 confidence interval = ± z = 5.502115± (2.33x0.103816) = 5.502115 ± 0.24189128 = (5.26022372, 5.74400628) = (5.26,5.74)
B) is the company meeting the requirement set forth on the labelthat the mean amount of tea in a bag is 5.5 grams? P(X=5.5) P(Z=(X-)/)= P(Z=(5.5-5.502115)/0.103816) = P(Z=-0.02) = 0.4920 = (5.26,5.74)
B) is the company meeting the requirement set forth on the labelthat the mean amount of tea in a bag is 5.5 grams? P(X=5.5) P(Z=(X-)/)= P(Z=(5.5-5.502115)/0.103816) = P(Z=-0.02) = 0.4920

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