Non-parametric tests do not rely on data belonging to any particular distribution. These distribution free methods do not rely on assumptions that the data come from a given probability distribution.
Purpose
To compare the medians of two or more populations on a continuous CT characteristic. For data that comes from a non-normal distribution, this test is comparable to the One way ANOVA. Since we don't know the population medians, an analysis of samples of data is required.
This test is usually used to determine if there is a statistically significant change in the median of a CT characteristic under two or more different conditions introduced by one factor (see concept Factor and Levels).
Anatomy
Reference: Business Statistics by D. Downing and J. Clark
Terminology
A. Null (H0) and alternative (Ha) hypotheses, where the medians (M1, M2, …, Mg) of the g levels of the factor are compared. There is only one alternative hypothesis which is that at least the medians of two levels are significantly different.
B. Minitab Session Window Output.
C. Sample medians of the ordered data.
D. The z-value indicates how the mean rank for the group i differs from the mean rank for all N observations.
E. Kruskal-Wallis test statistic.
F. P-Value – This value has to be compared with the alpha level and the following decision rule is used: if P < alpha, reject H0 and accept Ha with (1-P)100% confidence; if P > alpha, don't reject H0.
Major Considerations
The Kruskal-Wallis test assumes that the data come from g independent random samples from continuous distributions, all having the same shape.
Application Cookbook
1. Define problem and state the objective of the study.
2. Establish hypotheses. State the Null hypothesis and the Alternative hypothesis.
3. Establish alpha level, usually alpha is 0.05.
4. Select the random samples.
5. Measure the CT characteristic.
6. Analyze the data with Minitab:
7. Make statistical decision from the Session Window output of Minitab. Either accept or reject H0. If H0 is rejected, we can conclude that there is a significant difference between the medians of the levels.
8. Translate statistical conclusion to practical decision about the CT characteristic.
Purpose
To compare the medians of two populations on a continuous CT characteristic. For data that comes from a non-normal distribution, this test is comparable to the Two Samples T-Test. Since we don't know the population medians, an analysis of two samples of data is required.
This test is usually used to determine if there is a statistically significant change in the median of a CT characteristic under two different conditions.
Anatomy
Reference: Business Statistics by D. Downing and J. Clark
Terminology
A. Null (H0) and alternative (Ha) hypotheses, where the two population medians (M1, M2) are compared. For the alternative hypothesis, one of the three hypotheses has to be chosen before collecting the data to avoid being biased or influenced by the observations of the sample.
B. Minitab Session Window output.
C. Sample medians of the ordered data.
D. The 95.1% confidence interval for the difference in population medians.
E. Mann-Whitney test statistic.
F. P-Value – This value has to be compared with the alpha level and the following decision rule is used: if P < alpha, reject H0 and accept Ha with (1-P)100% confidence; if P > alpha, don't reject H0.
Major Considerations
The Mann-Whitney test assumes that the data are independent random samples from two populations that have the same shape (hence the same variance), and a scale that is at least ordinal in nature (see concept Measurement Scale – Ordinal).
For normal populations, this test is less powerful than the Two Samples T-Test when variances are assumed to be equal (see tool T-Test - Two Samples).
However, for other populations, the Mann-Whitney test is more powerful than the T-Test. If the two populations have different shapes or standard deviations, then the Two Samples T- Test with variances not assumed equal is a more appropriate test.
Application Cookbook
1. Define problem and state the objective of the study.
2. Establish hypotheses. State the Null hypothesis and the Alternative hypothesis.
3. Establish alpha level, usually alpha is 0.05.
4. Select the random samples.
5. Measure the CT characteristic.
6. Analyze the data with Minitab:
7. Make statistical decision from the Session Window output of Minitab. Either accept or reject H0.
8. Translate statistical conclusion to practical decision about the CT characteristic.
Purpose
To compare the medians of two or more populations on a continuous CT characteristic. For data that comes from a non-normal distribution, this test is comparable to the One way ANOVA.
Since we don't know the population medians, an analysis of data samples is required. This test is usually used to determine if there is a statistically significant change in the median of a CT characteristic under two or more different conditions introduced by one factor (see concept Factor and Levels).
Anatomy
Reference: Business Statistics by D. Downing and J. Clark
Terminology
A. Null (H0) and alternative (Ha) hypotheses, where the medians (M1, M2, …, Mg) of the g levels of the factor are compared. There is only one alternative hypothesis: at least the medians of two levels are significantly different.
B. Minitab session window output.
C. Number of measurements for each sample that falls above or below the overall median.
D. The interquartile range for each of each samples.
E. Individual sample 95% Confidence Intervals, shown as a graphical output.
F. The Chi-squared test value.
G. P-Value – This value has to be compared with the alpha level and the following decision rule is used: if P < alpha, reject H0 and accept Ha with (1-P)100% confidence; if P > alpha, don't reject H0.
Major Considerations
The Mood's Median test assumes that the data come from g independent random samples from continuous distributions, all having the same shape.
Compared to the Kruskal-Wallis test (see above), this test is more robust to the presence of outliers in data, and is particularly appropriate in the preliminary stages of analysis. However the Mood's Median test is less efficient for data coming from many distributions, including the normal.
Application Cookbook
1. Define problem and state the objective of the study.
2. Establish hypotheses. State the Null hypothesis and the Alternative hypothesis.
3. Establish alpha level, usually alpha is 0.05.
4. Select the random samples.
5. Measure the CT characteristic.
6. Analyze the data with Minitab:
7. Make a Statistical decision from the session window output of Minitab. Either accept or reject Ho. If H0 is rejected, we can conclude that there is a significant difference between the medians of the levels.
8. Translate statistical conclusion into practical decision about the CT characteristic.
Purpose
To compare the population median of a continuous CT characteristic with a value such as the target. For data that comes from a non-normal distribution, this test is comparable to the One Sample T-Test.
Since we don't know the population median, an analysis of a sample of data is required. This test is usually used to determine if the median of a CT characteristic is on target.
Anatomy
Reference: Business Statistics by D. Downing and J. Clark
Terminology
A. Null (H0) and alternative hypotheses (Ha), where the population median (M) is compared to a value (M0) such as a target.
For the alternative hypothesis, one of the three hypotheses has to be chosen before collecting the data to avoid being biased or influenced by the observations of the sample.
B. Minitab Session Window output.
C. Sample size, N for test (Sample size minus measurements equal to the hypothesized median).
D. Wilcoxon test statistic.
E. P-Value – This value has to be compared with the alpha level and the following decision rule is used: if P < alpha, reject H0 and accept Ha with (1-P)100% confidence; if P > alpha, don't reject H0.
Major Considerations
The Wilcoxon test assumes that the data comes from a random sample and the distribution is symmetrical. For a normal population, this test is less powerful than Student's t test. However, for other populations, the Wilcoxon test may be more powerful than Student's t test.
Application Cookbook
1. Define problem and state the objective of the study.
2. Establish hypotheses. State the Null hypothesis and the Alternative hypothesis.
3. Establish alpha level, usually alpha is 0.05.
4. Select a random sample.
5. Measure the CT characteristic.
6. Analyze the data with Minitab:
7. Make statistical decision from the Session Window output of Minitab. Either accept or reject H0.
8. Translate statistical conclusion to practical decision about the CT characteristic.
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