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Chapter 15: Problem 23

Two plastics, each produced by a different process, were tested for ultimate strength. The measurements in the accompanying table represent breaking loads in units of 1000 pounds per square inch. Do the data present evidence of a difference between the locations of the distributions of ultimate strengths for the two plastics? Test by using the Mann-Whitney \(U\) test with a level of significance as near as possible to \(\alpha=.10\).$$\begin{array}{cc} \hline \text { Plastic 1 } & \text { Plastic 2 } \\\\\hline 15.3 & 21.2 \\\18.7 & 22.4 \\\22.3 & 18.3 \\\17.6 & 19.3 \\\19.1 & 17.1 \\\14.8 & 27.7 \\\\\hline\end{array}$$

Short Answer

Expert verified
There is not enough evidence to suggest a difference between the two plastics' distributions at \( \alpha = 0.10 \).

Step by step solution

01

Combine and Rank Data

First, we need to combine the data from Plastic 1 and Plastic 2 and then rank all these values from smallest to largest. Assign ranks across both samples. If there are ties (same values), assign the average rank for those values.
02

Calculate Sum of Ranks

Separate the ranks back into their respective plastic groups and calculate the sum of the ranks for each group: Let \( R_1 \) and \( R_2 \) be the sums of ranks for Plastic 1 and Plastic 2, respectively.
03

Compute Mann-Whitney U Statistics

Compute the Mann-Whitney U statistics for both samples using the formulas: \[ U_1 = n_1n_2 + \frac{n_1(n_1+1)}{2} - R_1 \]\[ U_2 = n_1n_2 + \frac{n_2(n_2+1)}{2} - R_2 \] where \( n_1 \) and \( n_2 \) are the sample sizes for Plastic 1 and Plastic 2, respectively.
04

Identify the Test Statistic

The test statistic \( U \) is the smaller of \( U_1 \) and \( U_2 \). This statistic will be compared against critical values of the Mann-Whitney U distribution or used to calculate a p-value.
05

Determine Critical Value or P-value

For a significance level of \( \alpha = 0.10 \), refer to Mann-Whitney U tables or use statistical software to find the critical value or compute the p-value for a two-tailed test, based on the sample sizes.
06

Make a Decision

Compare the calculated \( U \) statistic to the critical value or p-value. If \( U \) is less than the critical value or the p-value is less than 0.10, reject the null hypothesis, indicating a significant difference in the distributions of ultimate strengths.

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Key Concepts

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

Non-Parametric Test
A non-parametric test is a type of statistical test that does not assume a specific distribution for the data. This makes non-parametric tests ideal for analyzing data that may not follow a normal distribution.
- They're useful when the data has outliers or is skewed. - They handle ordinal data (data that can be ranked). - They are versatile, applicable to various data types. The Mann-Whitney U test is an example of a non-parametric test. It is used to compare the difference between two independent groups to determine if they come from the same distribution. Unlike parametric tests, which rely on means and variances, the Mann-Whitney U test focuses on the medians of the samples. This test is particularly useful when sample sizes are small or when normality assumptions might not be satisfied. It鈥檚 used in scenarios where calculating a mean could be misleading due to the data structure.
Difference in Distributions
Determining the difference in distributions is crucial in statistical analysis, especially when comparing two samples like the ultimate strength of plastics. The Mann-Whitney U test tackles this by evaluating if the two samples differ significantly in their central location - essentially their median.
To conduct this test, ranks of the combined sample data from both groups are calculated. This step involves: - Assigning rank values across both samples combined. - Addressing ties by assigning tied ranks their average. Once ranked, the sum of ranks for each group is calculated, highlighting differences between the groups. If the sum of ranks significantly differs, it suggests a significant difference in the distribution location (median) of the two groups. This approach focuses on whether one group tends to have higher or lower values than the other, thus highlighting a "shift" in distribution.
Ultimate Strength Comparison
The comparison of ultimate strength between materials, like the two types of plastic in the exercise, is common in material science. Here, the ultimate strength refers to the maximum stress a material can withstand before breaking.
To compare the ultimate strength:
  • Data for ultimate strength is collected for each material type.
  • Statistical tests like the Mann-Whitney U test are used to evaluate differences.
  • The test checks if the distribution of strengths is the same across materials.
In the exercise, each plastic sample鈥檚 strength was compared to check if differences were statistically significant. By applying the Mann-Whitney U test, we determine if any observed differences are likely due to real variations in ultimate strength or just random chance. This information is vital for making informed decisions on material selection based on performance criteria.

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Most popular questions from this chapter

In some tests of healthy, elderly men, a new drug has restored their memories almost to the level of young adults. The medication will soon be tested on patients with Alzheimer's disease, the fatal brain disorder that eventually destroys the minds of those afflicted. According to Dr. Gary Lynch of the University of California, Irvine, the drug, called ampakine CX-516, accelerates signals between brain cells and appears to significantly sharpen memory. \(^{\star}\) In a preliminary test on students in their early 20 s and on men aged \(65-70\), the results were particularly striking. The accompanying data are the numbers of nonsense syllables recalled after 5 minutes for ten men in their 20 s and ten men aged \(65-70\) who had been given a mild dose of ampakine \(\mathrm{CX}-516\). Do the data provide sufficient evidence to conclude that there is a difference in the number of nonsense syllables recalled by men in the two age groups when older men have been given ampakine CX-516? Give the associated \(p\) -value.

If a matched-pairs experiment using \(n\) pair of observations is conducted, if \(T^{+}=\) the sum of the ranks of the absolute values of the positive differences, and \(T^{-}=\) the sum of the ranks of the absolute values of the negative differences, why is \(T^{+}+T^{-}=n(n+1) / 2 ?\)

A comparison of reaction (in seconds) to two different stimuli in a psychological word-association experiment produced the results in the accompanying table when applied to a random sample of 16 people. Do the data present sufficient evidence to indicate a difference in location for the distributions of reaction times for the two stimuli? Use the Mann-Whitney \(U\) statistic and test with \(\alpha=.05 .\) (Note: This test was conducted by using Student's \(t\) in Exercise 13.3 . Compare your results.)

Calculate the probability that the Wilcoxon \(T\) (Section 15.4 ) is less than or equal to 2 for \(n=3\) pairs. Assume that no ties occur and that \(H_{0}\) is true.

Assuming no ties, obtain the exact null distribution of the Kruskal-Wallis \(H\) statistic for the case \(k=3, n_{1}=n_{2}=n_{3}=2 .\) [Because the sample sizes are all equal, if ranks 1 and 2 are assigned to treatment 1, ranks 3 and 4 are assigned to treatment 2, and ranks 5 and 6 are assigned to treatment 3, the value of \(H\) is exactly the same as if ranks 3 and 4 are assigned to treatment 1 , ranks 5 and 6 are assigned to treatment \(2,\) and ranks 1 and 2 are assigned to treatment 3 . That is, for any particular set of ranks, we may interchange the roles of the \(k\) populations and obtain the same values of the \(H\) statistic. Thus, the number of cases that we must consider can be reduced by a factor of \(1 / k ! .\) Consequently, \(H\) must be evaluated only for \((6 ! /[2 ! \cdot 2 ! \cdot 2 !]) / 3 !=15\) distinct arrangements of ranks.]
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