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Johnson transformation example analysis, many novices are not very clear about this, in order to help you solve this problem, the following editor will explain in detail for you, people with this need can come to learn, I hope you can gain something.

Johnson transform

Johnson transform (Johnson transformation) can choose an optimal function from three distribution families of variables (SB, SL and SU) to easily transform the data into standard normal distribution (standard normal distribution), where SB, SL and SU represent bounded (bounded) transformations (

), lognormal (lognormal) transformation (), and unbounded (unbounded) transformation (

). Minitab can output the normal probability diagram and its P value of the original data and the transformed data in order to compare the distribution before and after the transformation. Although Box-Cox transform is easy to understand, its function is very limited, and it is usually difficult to find a suitable transformation. When the Box-Cox transformation is not successful, the Johnson transformation can be achieved. The Johnson transform function is complex, but the function of finding an appropriate transformation is very powerful. When the Johnson transform can not fully transform the data, the effect of the Box-Cox transform may be better.

[example 17-2] try to perform Johnson transform on the data of case 17-1. (calcium content .MTW)

First, open the worksheet: "calcium content. MTW".

Select [Statistics (Stat)] → [quality tools (QualityTools)] → [Johnson transform (Johnson transformation)] menu, open the Johnson transform (Johnson transformation) main dialog box, see figure 17-10.

Select [single column (Single column)] in [data arrangement as (Data are arranged as)], and select C1 (calcium content).

Third, options (Options) dialog box (see figure 17-11), [selected best fit selected P value is (P-valueto select best fit)] is the default value of 0.10.

IV. Main results and analysis

As can be seen from the Johnson transform diagram, the graph is divided into four parts (see figure 17-12):

1. Normal probability map of original data: individual data points of calcium content fall outside the confidence band of fitting distribution line. Anderson-Darling normality test, AD=0.754,P=0.046 < 0.05.According to the level of α = 0.05, it can not be considered that the calcium content of the vitamin capsule obeys normal distribution.

2. Select the transformation graph, the vertical coordinate is the P value of the AD test, the horizontal axis is the fitting Z value horizontal reference line is located at 0.10, when the Z value is 0.66 (vertical reference line), the P value (0.99) of the AD test reaches the maximum.

3. The transformed data points all fall within the confidence band of the fitting distribution line, and Anderson-Darling normality test shows that AD=0.125,P=0.99 > 0.05.According to the α = 0.05level, it can be considered that the transformed data obeys the normal distribution.

4. The best transformation is the SB distribution family, and the best transformation function is

The best fitting Z value is 0.66 and 0.99 > 0.05.

[example 17-3] A researcher collected and measured some physiological indexes of 82 30-year-old men, and tried to classify the data distribution. (male physiological index. MTW)

First, open the worksheet: "male physiological indicators. MTW".

Second, Johnson transform (Johnson transformation) main dialog box (see figure 17-10), [data arrangement as (Data are arranged as)] is [single column (Single column)], and select "C1 (indicator)". Other options are the default.

III. Main results and analysis

Since all the fitting Z values in the selected transformation map correspond to the P values of the AD test, the Johnson transformation is not performed, as shown in figure 17-13.

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