Examples of When to Use Confidence, Prediction, and Tolerance Intervals

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Examples of When to Use Confidence, Prediction,
and Tolerance Intervals

• Engineering Experimental Design
• Valerie L. Young

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When Do I Do What?
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When Do I Do What?
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Example 1

• You measure the zinc concentration in the livers
  of 56 fish and find that the mean of these 56
  values is 9.15 ?g Zn / g liver and the standard
  deviation is 1.27 ?g Zn / g liver. What is the
  concentration of zinc in fish liver?
• n 56, so this is a large sample and we can use
  z
• For 95 confidence, we want 5 of the area in
  the 2 tails, or 2.5 of the area in each
• 1 0.025 0.975
• z(area 0.975) 1.96
• CI 1.96 ? 1.27 / sqrt(56) 0.3326
• Zinc concentration is 9.2 0.3 ?g Zn / g liver
  (95 confidence interval)
• Note if you use t-critical(?55) then you get
  t-critical2.000 and CI 0.3394. No important
  difference.

Based on a problem in Devore Farnum, Applied
Statistics for Engineers Scientists, 1999
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Example 2

• You measure the zinc concentration in the livers
  of 56 fish and find that the mean of these 56
  values is 9.15 ?g Zn / g liver and the standard
  deviation is 1.27 ?g Zn / g liver. What
  concentration of zinc do you expect to find in
  the next fish liver you eat?
• n 56, so we need t-critical for ?60 (closest
  we can get to 55) and ? 0.025 (half of 0.05).
• t-critical 2.000 (Table B2 in text)
• PI 2.000 ? 1.27 ? sqrt(1(1/56)) 2.562
• Concentration of zinc in next fish will be 9.2
  2.5 ?g Zn / g liver (95 prediction interval)

Based on a problem in Devore Farnum, Applied
Statistics for Engineers Scientists, 1999
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Example 3

• You measure the zinc concentration in the livers
  of 56 fish and find that the mean of these 56
  values is 9.15 ?g Zn / g liver and the standard
  deviation is 1.27 ?g Zn / g liver. What range
  of zinc concentrations will describe the livers
  of 90 of this type of fish?
• n 56 and p 0.90, so tol critical 1.6585
  (Table B.12 in text use the value for n60)
• TI 1.6585 ? 1.27 2.106
• 90 of these fish are likely to have 9.5 2.1
  ?g Zn / g liver in their livers (95 confidence
  level)

Based on a problem in Devore Farnum, Applied
Statistics for Engineers Scientists, 1999
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Example 4

• Measurements of stabilized viscosity were made on
  five asphalt specimens, resulting in values of
  2781, 2900, 3013, 2856, and 2888 cP. What is the
  viscosity of the asphalt?
• Mean 2887.60 cP, sample std dev 84.03 cP
• n5, so use t-critical for ?4 and ?0.025
• t-critical 2.776 (Table B2 in text)
• CI 2.776 ? 84.03 / sqrt(5) 104
• The viscosity of the asphalt is 2890 100 cP
  (95 confidence interval)

Based on a problem in Devore Farnum, Applied
Statistics for Engineers Scientists, 1999
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Example 5

• Measurements of stabilized viscosity were made on
  five asphalt specimens, resulting in values of
  2781, 2900, 3013, 2856, and 2888 cP. If you
  measured a sixth specimen, what would you expect
  its viscosity to be?
• Mean 2887.60 cP, sample std dev 84.03 cP
• n5, so use t-critical for ?4 and ?0.025
• T-critical 2.776 (Table B2 in text)
• PI 2.776 ? 84.03 ? sqrt(1 1/5) 256 cP
• The viscosity of the next asphalt will be 2890
  260 cP (95 prediction interval)

Based on a problem in Devore Farnum, Applied
Statistics for Engineers Scientists, 1999
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Example 6

• Measurements of stabilized viscosity were made on
  five asphalt specimens, resulting in values of
  2781, 2900, 3013, 2856, and 2888 cP. Give the
  range of viscosities within which you would
  expect 75 of specimens from this manufacturer
  to lie.
• Mean 2887.60 cP, sample std dev 84.03 cP
• n5. This sample is not large enough to derive a
  tolerance interval. At least one more specimen
  must be tested.

Based on a problem in Devore Farnum, Applied
Statistics for Engineers Scientists, 1999
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Example 7

• You find that the 50 Duralife batteries last an
  average of 4.15 hours (standard deviation 0.92
  hours). You find that the 45 Rayolife batteries
  last an average of 0.84 hours (standard deviation
  1.64 hours). How much longer do Rayolife
  batteries last?
• The difference between the means is 0.38.
• 95 CI for the difference is 1.96 ?
  sqrt((0.922/50)(0.842/45)) 0.3539
• Rayolife batteries last 0.38 0.35 hours longer
  than duralife batteries (95 confidence level).

Based on a problem in Devore Farnum, Applied
Statistics for Engineers Scientists, 1999
11
Example 8

• Youre on a budget, so you purchase 10 Duralife
  batteries and 10 Rayolife batteries. You find
  that the Duralife batteries last an average of
  4.15 hours (standard deviation 0.92 hours). You
  find that the Rayolife batteries last an average
  of 4.53 hours (standard deviation 0.84 hours).
  How much longer do Rayolife batteries last?
• CI 2.262 ? sqrt((0.922/10)(0.842/10)) 0.891
  hours
• Based on samples of 10, the difference in
  lifetime between the battery brands is not
  significant at the 95 confidence level.
• Note that if the sample sizes are not the same,
  there is a complex formula that must be used to
  calculate the degrees of freedom.

Based on a problem in Devore Farnum, Applied
Statistics for Engineers Scientists, 1999