91Ó°ÊÓ

The single-factor ANOVA model considered in Chapter 11 assumed the observations in the \(i\) th sample were selected from a normal distribution with mean \(\mu_{i}\) and variance \(\sigma^{2}\), that is, \(X_{i j}=\mu_{i}+\varepsilon_{i j}\) where the \(\varepsilon\) 's are normal with mean 0 and variance \(\sigma^{2}\). The normality assumption implies that the \(F\) test is not distribution-free. We now assume that the \(\varepsilon\) 's all come from the same continuous, but not necessarily normal, distribution, and develop a distribution-free test of the null hypothesis that all \(I \mu_{i}\) 's are identical. Let \(N=\sum J_{i}\), the total number of observations in the data set (there are \(J_{i}\) observations in the \(i\) th sample). Rank these \(N\) observations from 1 (the smallest) to \(N\), and let \(\bar{R}_{i}\) be the average of the ranks for the observations in the ith sample. When \(H_{0}\) is true, we expect the rank of any particular observation and therefore also \(\bar{R}_{i}\) to be \((N+1) / 2\). The data argues against \(H_{0}\) when some of the \(\bar{R}_{i}\) 's differ considerably from \((N+1) / 2\). The Kruskal-Wallis test statistic is $$ K=\frac{12}{N(N+1)} \sum J_{i}\left(\bar{R}_{i}-\frac{N+1}{2}\right)^{2} $$ When \(H_{0}\) is true and either (1) \(I=3\), all \(J_{i} \geq 6\) or (2) \(I>3\), all \(J_{i} \geq 5\), the test statistic has approximately a chi-squared distribution with \(I-1\) df. The accompanying observations on axial stiffness index resulted from a study of metal-plate connected trusses in which five different plate lengths-4 in., 6 in., 8 in., 10 in., and 12 in. were used ("Modeling Joints Made with LightGauge Metal Connector Plates," Forest Products \(J ., 1979: 39-44)\). \(\begin{array}{lllll}i=1(4 \text { in. }): & 309.2 & 309.7 & 311.0 & 316.8 \\\ & 326.5 & 349.8 & 409.5 & \\ i=2(6 \text { in. }): & 331.0 & 347.2 & 348.9 & 361.0 \\ & 381.7 & 402.1 & 404.5 & \\ i=3(8 \text { in. }): & 351.0 & 357.1 & 366.2 & 367.3 \\ & 382.0 & 392.4 & 409.9 & \\ i=4(10 \text { in. }): & 346.7 & 362.6 & 384.2 & 410.6 \\ & 433.1 & 452.9 & 461.4 & \\ i=5(12 \text { in. }): & 407.4 & 410.7 & 419.9 & 441.2 \\ & 441.8 & 465.8 & 473.4 & \end{array}\) Use the \(K-W\) test to decide at significance level \(.01\) whether the true average axial stiffness index depends somehow on plate length.

Step by step solution

01

Collect and List All Observations

First, collect all observations from the different groups. We have five groups corresponding to plate lengths 4, 6, 8, 10, and 12 inches. These values will be ranked collectively, regardless of group.

02

Rank All Observations

Rank all the observations from smallest to largest. If two observations are equal, assign each of them the average rank for those positions.

03

Calculate Average Ranks for Each Group

For each group, calculate the average of the ranks obtained in Step 2. This average rank is denoted as \( \bar{R}_i \) for the ith group.

04

Calculate the Kruskal-Wallis Test Statistic

Use the formula for the Kruskal-Wallis test statistic: \[K = \frac{12}{N(N+1)} \sum J_i \left( \bar{R}_i - \frac{N+1}{2} \right)^2\]where \(N\) is the total number of observations and \(J_i\) is the number of observations in the ith group. Compute this sum to get the test statistic \(K\).

05

Determine Degrees of Freedom

The degrees of freedom for the Kruskal-Wallis test statistic is given by \(I - 1\), where \(I\) is the number of groups. In this problem, \(I = 5\), so the degrees of freedom is 4.

06

Compare Test Statistic to Critical Value

For a significance level of 0.01 and 4 degrees of freedom, find the critical value from chi-squared tables. Compare the calculated \(K\) to this critical value to determine if we can reject the null hypothesis \(H_0\).

07

Draw a Conclusion

If \(K\) is greater than the critical chi-squared value, reject the null hypothesis; otherwise, do not reject the null hypothesis. This tells us whether the true average axial stiffness index is influenced by plate length.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Non-Parametric Statistics

Unlike parametric statistics, which rely on assumptions about the distribution of data (usually normal), non-parametric statistics do not make such assumptions. The Kruskal-Wallis test is a prime example of non-parametric statistics.
It is particularly useful when you suspect that data does not follow a normal distribution or your data is ordinal rather than interval or ratio.
This makes non-parametric tests more flexible and widely applicable when data does not meet the strict criteria of parametric tests. Non-parametric methods are often used in situations where:

• The sample size is small.
• The data includes outliers that cannot be removed.
• The data is ordinal, ranked, or not normally distributed.

By using ranks rather than raw data, non-parametric methods can handle a wider variety of data types and situations.

Chi-Squared Distribution

The chi-squared distribution is critical for the Kruskal-Wallis test as it's used to approximate the distribution of the test statistic under the null hypothesis. Essentially, when we compute the test statistic from our data, it follows an approximate chi-squared distribution when the null hypothesis is true. This distribution is specifically characterized by degrees of freedom, which depend on the number of groups being compared, in this case, the plate lengths. For the Kruskal-Wallis test, the degrees of freedom are calculated as the number of groups minus one, ( I - 1 ). Hence, it is highly important for hypothesis testing, especially for non-parametric techniques.
When you compare your test statistic against this distribution, you determine the likelihood that the observed data could have occurred under the null hypothesis. The critical value from this distribution helps in deciding whether to reject the null hypothesis.

Analysis of Variance

Analysis of Variance, or ANOVA, typically involves comparing three or more group means to see if at least one of the means is different from the others, under the assumption that the data follows a normal distribution. The Kruskal-Wallis test serves a similar purpose to ANOVA but without relying on normality assumptions. Using rank-transformed data instead of raw data, the Kruskal-Wallis test checks for differences in distributions across the groups. This makes it non-parametric ANOVA. While ANOVA uses the F-distribution, the Kruskal-Wallis uses the chi-squared distribution to infer statistical significance.
ANOVA is highly effective with larger sample sizes and normally distributed data, whereas Kruskal-Wallis shines under less rigid conditions, handling ordinal data and non-uniform distributions gracefully.

Rank-Based Testing

Rank-based testing is a fundamental concept utilized by the Kruskal-Wallis test. This method involves replacing raw data values with their rank when sorted from the smallest to the largest. By focusing on these ranks instead of the actual data points, this testing approach diminishes the influence of extreme values or outliers. Through rank-based testing:

• It becomes easier to detect differences between groups, particularly non-normal distributions.
• The impact of outliers is minimized.
• Data only needs to be ordinal; exact magnitude isn't essential.

Rank-based methods ensure a certain level of robustness, making them highly applicable in flexible, non-parametric test designs such as the Kruskal-Wallis test, where distinct but non-numerical patterns are more crucial than precise data values.