26 27 28 Normal-plott for Eksempel 3 (MINITAB 14) Probability - - PDF document

26 27 28 normal plott for eksempel 3 minitab 14
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26 27 28 Normal-plott for Eksempel 3 (MINITAB 14) Probability - - PDF document

26 27 28 Normal-plott for Eksempel 3 (MINITAB 14) Probability Plot of C1 Normal - 95% CI 99 Mean 40,55 StDev 1,986 N 10 95 AD 0,607 90 P-Value 0,082 80 70 Percent 60 50 40 30 20 10 5 1 35,0 37,5 40,0 42,5 45,0 47,5


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SLIDE 1

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SLIDE 2

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SLIDE 3

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SLIDE 4

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C1 Percent

47,5 45,0 42,5 40,0 37,5 35,0

99 95 90 80 70 60 50 40 30 20 10 5 1 Mean 0,082 40,55 StDev 1,986 N 10 AD 0,607 P-Value

Probability Plot of C1

Normal - 95% CI

Normal-plott for Eksempel 3 (MINITAB 14)

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SLIDE 5

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Example 1: The industrial experment

An experiment was performed on a manufacturing plant by making in sequence 10 batches using the standard production method (A), followed by 10 batches of a modified method (B). The results from these trials are given in the table on the next slide. What evidence do the data provide that method B is better than method A?

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SLIDE 6

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SLIDE 7

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SLIDE 8

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(Assuming equal variances:)

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SLIDE 9

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(Not assuming equal variances:)

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SLIDE 10

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SLIDE 11

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SLIDE 12

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Metode 95% Bonferroni Confidence I ntervals for StDevs

2 1 8 7 6 5 4 3 2

Metode X

2 1 92 90 88 86 84 82 80 F-Test 0,390 Test Statistic 1,58 P-Value 0,505 Levene's Test Test Statistic 0,78 P-Value

Test for Equal Variances for X

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SLIDE 13

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SLIDE 14

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Reference distribution based on historical data:

On the next slide are shown 210 earlier observations of the process for method A (upper curve) And the 191 differences between 10 and 10 succeeding averages. A table of the differences as well as a histogram are also given,

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SLIDE 15

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SLIDE 16

41 The histogram expresses the reference distribution based on historical data, and indicates which differences that naturally would be found if there was no difference between the methods A and B. By a direct count we find that 9 differences exceed the value 1.3 which we observed in our

  • experiment. The p-value is hence 9/191 = 0.047 (and there is thus reason to reject the

hypothesis that the methods are equally good)