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Chapter 5: Q34E (page 316)

Plastic fill process. University of Louisville operators examined the process of filling plastic pouches of dry blended biscuit mix (Quality Engineering, Vol. 91, 1996). The current fill mean of the process is set at μ= 406 grams, and the process fills standard deviation is σ= 10.1 grams. (According to the operators, “The high level of variation is since the product has poor flow properties and is, therefore, difficult to fill consistently from pouch to pouch.”) Operators monitor the process by randomly sampling 36 pouches each day and measuring the amount of biscuit mix in each. Considerx the mean fill amount of the sample of 36 products. Suppose that on one particular day, the operators observe x= 400.8. One of the operators believes that this indicates that the true process fill mean for that day is less than 406 grams. Another operator argues thatμ = 406, and the small observed value is due to random variation in the fill process. Which operator do you agree with? Why?

Short Answer

Expert verified

The operator, who got the mean value of 406 grams, can be considered.

Step by step solution

01

Given information

There is the process of filling plastic pouches with dry-blended biscuit mix. The mean of the filling is 406 grama, and the standard deviation is 10.1 grams.

Two operators took 36 pouches as the sample, and one of them observed that the mean of the filling is 400.8 grams, and another one observed that the mean is 406 grams.

Null hypothesis: Ho

μ=406

Alternate hypothesis: Ha

μ<406

Significance level:α=0.05

Statistic test:

Z=−μσ/nZ=400.8−40610.1/36Z=−3.08910891

P value:

Z = - 3.089 and Z test p-value will be 0.001

02

Conclusion

Simply by the outputs of those two operators, there can be easily concluded that the second operators get the absolute mean compared to the original population of pouches. He gets the mean value of 406 grams, the same as the population. Other while the first operator gets the mean value of 400.8 grams.So, there can be agreed with the operator who got the mean value of 407 grams.

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Most popular questions from this chapter

Question:Who prepares your tax return? As part of a study on income tax compliance (Behavioral Research and Accounting, January 2015), researchers found that 37% of adult workers prepare their own tax return. Assume that this percentage applies to all U.S. adult workers. Now consider a random sample of 270 adult workers.

a. Find the probability that more than 112 of the workers prepare their own tax return.

b. Find the probability that between 100 and 150 of the workers prepare their own tax return

Question: Refer to Exercise 5.5, in which we found the sampling distribution of the sample median. Is the median an unbiased estimator of the population mean m?

Will the sampling distribution ofχ¯ always be approximately normally distributed? Explain

Soft-drink bottles. A soft-drink bottler purchases glass bottles from a vendor. The bottles are required to have an internal pressure of at least 150 pounds per square inch (psi). A prospective bottle vendor claims that its production process yields bottles with a mean internal pressure of 157 psi and a standard deviation of 3 psi. The bottler strikes an agreement with the vendor that permits the bottler to sample from the vendor’s production process to verify the vendor’s claim. The bottler randomly selects 40 bottles from the last 10,000 produced, measures the internal pressure of each, and finds the mean pressure for the sample to be 1.3 psi below the process mean cited by the vendor.

a. Assuming the vendor’s claim to be true, what is the probability of obtaining a sample mean this far or farther below the process mean? What does your answer suggest about the validity of the vendor’s claim?

b. If the process standard deviation were 3 psi as claimed by the vendor, but the mean were 156 psi, would the observed sample result be more or less likely than in part a? What if the mean were 158 psi?

c. If the process mean were 157 psi as claimed, but the process standard deviation were 2 psi, would the sample result be more or less likely than in part a? What if instead the standard deviation were 6 psi?

Surface roughness of pipe. Refer to the Anti-CorrosionMethods and Materials(Vol. 50, 2003) study of the surface roughness of oil field pipes, Exercise 2.46 (p. 96). Recall that a scanning probe instrument was used to measure thesurface roughness x(in micrometers) of 20 sampled sectionsof coated interior pipe. Consider the sample mean,X¯.

  1. Assume that the surface roughness distribution has a mean of = 1.8 micrometers and a standard deviation of = .5 micrometer. Use this information to find theprobability thatexceeds 1.85 micrometers.
  2. The sample data are reproduced in the following table.Compute.
  3. Based on the result, part b, comment on the validity ofthe assumptions made in part a.

1.72

2.50

2.16

2.13

1.06

2.24

2.31

2.03

1.09

1.40

2.57

2.64

1.26

2.05

1.19

2.13

1.27

1.51

2.41

1.95
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