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Example 9.6 Analyzing Variability in Diameters of Machine Parts

Example 9.6 Analyzing Variability in Diameters of Machine Parts. Confidence Interval for a Standard Deviation. Objective.

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Example 9.6 Analyzing Variability in Diameters of Machine Parts

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  1. Example 9.6Analyzing Variability in Diameters of Machine Parts Confidence Interval for a Standard Deviation

  2. Objective To use StatPro’s one-sample procedure to find a confidence interval for the standard deviation of part diameters, and to see how variability affects the proportion of unusable parts produced.

  3. Background Information • A machine produces parts that are supposed to have diameter 10 centimeters. • However, due to inherent variability, some diameters are greater than 10 cm and some are less. • The production supervisor is concerned with two things: • First, he is concerned that the mean diameter might not be 10 centimeters. • Second, he is worried about the extent of variability in the diameters.

  4. Background Information -- continued • Even if the mean is on target, excessive variability implies that many of the parts will fail to meet specifications. • To analyze the process, he randomly samples 50 parts during the course of a day and measures the diameter of each part to the nearest millimeter.

  5. PARTS.XLS • This file shows the sample results in columns A and B. • Should the production supervisor be concerned about the results from this sample?

  6. Results

  7. Calculations • Because he is concerned about the mean and the standard deviation of diameters, we obtain 95% confidence intervals for both. • This is easy with StatPro’s One-Sample procedure. • The procedure is the same as before except this time we check the boxes for both confidence interval options - mean and standard deviation. • The confidence interval for the mean extends from 9.986 cm to 10.005 cm. Thus, the supervisor can be fairly confident that the mean diameter all parts is close to 10 cm.

  8. Calculations -- continued • The confidence interval for the standard deviation extends from 0.029 cm to 0.043 cm. Is this good or bad? • Lets assume the diameter is right on target at 10 cm, the standard deviation is at the upper limit, 0.043, and the population distribution is normal. The formula in cell F26 is=NORMDIST(10-F28,F29,F30,1) +(1-NORMDIST(10+F28,F29,F30,1)) • This shows that 13.1% of the parts are unusable.

  9. Calculations -- continued • To pursue this analysis further we form a two-way data table. • Each value in the data table is the resulting proportion of unusable parts. • The best-case scenario (where the mean is close to the target and the standard deviation is small) still results in 2.6% unusable parts. However, a mean off target and a large standard deviation can lead to as many as 15.1% unusable parts. • The message for the supervisor should be clear - he must work to reduce the underlying variability in the process.

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