/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 The accompanying data resulted f... [FREE SOLUTION] | 91Ó°ÊÓ

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The accompanying data resulted from an experiment to compare the effects of vitamin \(\mathrm{C}\) in orange juice and in synthetic ascorbic acid on the length of odontoblasts in guinea pigs over a 6-week period ("The Growth of the Odontoblasts of the Incisor Tooth as a Criterion of the Vitamin C Intake of the Guinea Pig," J. Nutrit., 1947: 491-504). Use the Wilcoxon rank-sum test at level \(.01\) to decide whether true average length differs for the two types of vitamin \(C\) intake. Compute also an approximate \(P\)-value. [Hint: See Exercise 14.] \(\begin{array}{lrrrrrr}\text { Orange Juice } & 8.2 & 9.4 & 9.6 & 9.7 & 10.0 & 14.5 \\ & 15.2 & 16.1 & 17.6 & 21.5 & & \\ \text { Ascorbic Acid } & 4.2 & 5.2 & 5.8 & 6.4 & 7.0 & 7.3 \\ & 10.1 & 11.2 & 11.3 & 11.5 & & \end{array}\)

Short Answer

Expert verified
Reject H0. There is a significant difference in odontoblast length between the two groups.

Step by step solution

01

State the Hypotheses

We will perform a Wilcoxon rank-sum test. The null hypothesis (H0) states that there is no difference in the true average length of odontoblasts between the two groups, orange juice and ascorbic acid. The alternative hypothesis (H1) states that there is a difference.
02

Combine and Rank the Data

Combine the data from both groups: Orange Juice = [8.2, 9.4, 9.6, 9.7, 10.0, 14.5, 15.2, 16.1, 17.6, 21.5] and Ascorbic Acid = [4.2, 5.2, 5.8, 6.4, 7.0, 7.3, 10.1, 11.2, 11.3, 11.5]. Rank all observations from smallest to largest, assigning average ranks where there are ties.
03

Calculate the Rank Sums

Calculate the sum of the ranks for each group. Let's denote the sum of the ranks for the Orange Juice group as R1 and for the Ascorbic Acid group as R2.
04

Determine Critical Value and Decision Rule

Using the Wilcoxon rank sum test table for a significance level of 0.01, find the critical value for n1 = 10 and n2 = 10. If the value of R1 or R2 is less than or exceeds the critical value, we reject the null hypothesis.
05

Calculate the P-value

Approximate the P-value using the ranks and standard normal distribution properties. This helps in determining how likely it is to observe such data if the null hypothesis were true.
06

Make a Conclusion

Compare the test statistic to the critical value and the P-value to the significance level. If the test statistic is in the rejection region or if the P-value is less than 0.01, reject the null hypothesis; otherwise, do not reject it.

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Key Concepts

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

Nonparametric Tests
Nonparametric tests are a set of statistical methods that do not rely on data belonging to any certain distribution, such as the normal distribution. This makes them particularly useful in situations where the data does not meet the assumptions necessary for parametric tests, like the t-test. The Wilcoxon rank-sum test is a type of nonparametric test. It's often used to compare differences between two groups when the sample size is small or the data isn't normally distributed.

Nonparametric tests are quite flexible because they rank the data rather than relying on the raw data itself. This approach lessens the impact of outliers and skewed data. By ranking all observations, nonparametric tests focus on the medians rather than the means. This is useful in many practical situations, as it allows for a more robust comparison of the central tendencies of two groups.
Hypothesis Testing
Hypothesis testing is a statistical method that allows us to make decisions and inferences about a population based on a sample. In the context of the Wilcoxon rank-sum test, we start by stating our null hypothesis ( H0), which usually suggests no effect or no difference between groups. The alternative hypothesis ( H1) states that there is an effect or a difference.

For the guinea pig odontoblast length data, the null hypothesis would be that there's no difference in true average length between those receiving vitamin C from orange juice versus ascorbic acid. The alternative hypothesis would be that there is a difference.

The goal of hypothesis testing is to evaluate the null hypothesis against the alternative by assessing the likelihood of an observable outcome assuming the null hypothesis is true. This is where the concepts of test statistics and P-values come into play.
P-value Calculation
Calculating the P-value is a critical step in hypothesis testing since it indicates the probability of obtaining the observed test results under the assumption that the null hypothesis is correct. A P-value provides a measure of the strength of evidence against the null hypothesis.

In a Wilcoxon rank-sum test, the P-value is approximated using the ranks of the data combined from both groups. If the P-value is lower than the pre-determined significance level, often expressed as alpha (\( \alpha \)), we reject the null hypothesis. In this exercise, an alpha of 0.01 is used. This means that there's only a 1% chance that the observed difference, or something more extreme, would occur if the null hypothesis were true.

Finding a low P-value signifies strong evidence against the null hypothesis; hence, the smaller the P-value, the higher the evidence against the null hypothesis.
Rank-based Analysis
Rank-based analysis is an important aspect of nonparametric tests, like the Wilcoxon rank-sum test. It involves ranking all data points without considering the group they belong to. For each observation in the combined data, we assign ranks, from smallest to largest. If there are ties, we assign average ranks to the tied observations.

This ranked dataset is then split back into their original groups, and the sum of the ranks ( R1) for the first group and ( R2) for the second group is computed.

The advantage of using ranks is that it eliminates the impact of skewness or outliers on statistical testing. By focusing on the position of each data point in the overall dataset, rather than its absolute value, rank-based methods provide a more robust comparison between groups. In this exercise, calculating ranks and comparing the rank sums helps assess whether one group's distribution differs from the other's, which forms the basis for the test's conclusion.

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Most popular questions from this chapter

The accompanying data is a subset of the data reported in the article "Synovial Fluid \(\mathrm{pH}\), Lactate, Oxygen and Carbon Dioxide Partial Pressure in Various Joint Diseases" (Arthritis Rheum., 1971: 476-477). The observations are \(\mathrm{pH}\) values of synovial fluid (which lubricates joints and tendons) taken from the knees of individuals suffering from arthritis. Assuming that true average \(\mathrm{pH}\) for non-arthritic individuals is \(7.39\), test at level \(.05\) to see whether the data indicates a difference between average \(\mathrm{pH}\) values for arthritic and nonarthritic individuals. $$ \begin{array}{lllllll} 7.02 & 7.35 & 7.34 & 7.17 & 7.28 & 7.77 & 7.09 \\ 7.22 & 7.45 & 6.95 & 7.40 & 7.10 & 7.32 & 7.14 \end{array} $$

Consider a random sample \(X_{1}, X_{2}, \ldots, X_{n}\) from the normal distribution with mean 0 and precision \(\tau\) (use \(\tau\) as a parameter instead of \(\sigma^{2}=1 / \tau\) ). Assume a gamma-distributed prior for \(\tau\) and show that the posterior distribution of \(\tau\) is also gamma. What are its parameters?

The accompanying 25 observations on fracture toughness of base plate of \(18 \%\) nickel maraging steel were reported in the article "Fracture Testing of Weldments" (ASTM Special Publ. No. 381, 1965: 328-356). Suppose a company will agree to purchase this steel for a particular application only if it can be strongly demonstrated from experimental evidence that true average toughness exceeds 75. Assuming that the fracture toughness distribution is symmetric, state and test the appropriate hypotheses at level \(.05\), and compute a \(P\)-value. $$ \begin{array}{lllllllll} 69.5 & 71.9 & 72.6 & 73.1 & 73.3 & 73.5 & 74.1 & 74.2 & 75.3 \\ 75.5 & 75.7 & 75.8 & 76.1 & 76.2 & 76.2 & 76.9 & 77.0 & 77.9 \\ 78.1 & 79.6 & 79.7 & 80.1 & 82.2 & 83.7 & 93.7 & & \end{array} $$

Suppose we wish to test \(H_{0}\) : the \(X\) and \(Y\) distributions are identical versus \(H_{\mathrm{a}}\) : the \(X\) distribution is less spread out than the \(Y\) distribution The accompanying figure pictures \(X\) and \(Y\) distributions for which \(H_{\mathrm{a}}\) is true. The Wilcoxon rank-sum test is not appropriate in this situation because when \(H_{\mathrm{a}}\) is true as pictured, the \(Y\) 's will tend to be at the extreme ends of the combined sample (resulting in small and large \(Y\) ranks), so the sum of \(X\) ranks will result in a \(W\) value that is neither large nor small. $$ \text { "Ranks" : } \begin{array}{cccccccc} \cline { 7 - 8 } & 1 & 1 & 1 & & 1 & 1 & 1 \\ \hline & 3 & 5 & \ldots & 6 & 4 & 2 \end{array} $$ Consider modifying the procedure for assigning ranks as follows: After the combined sample of \(m+n\) observations is ordered, the smallest observation is given rank 1 , the largest observation is given rank 2, the second smallest is given rank 3, the second largest is given rank 4 , and so on. Then if \(H_{\mathrm{a}}\) is true as pictured, the \(X\) values will tend to be in the middle of the sample and thus receive large ranks. Let \(W^{\prime}\) denote the sum of the \(X\) ranks and consider rejecting \(H_{0}\) in favor of \(H_{\mathrm{a}}\) when \(w^{\prime} \geq c\). When \(H_{0}\) is true, every possible set of \(X\) ranks has the same probability, so \(W^{\prime}\) has the same distribution as does \(W\) when \(H_{0}\) is true. Thus \(c\) can be chosen from Appendix Table A.13 to yield a level \(\alpha\) test. The accompanying data refers to medial muscle thickness for arterioles from the lungs of children who died from sudden infant death syndrome \((x\) 's \()\) and a control group of children ( \(y\) 's). Carry out the test of \(H_{0}\) versus \(H_{\mathrm{a}}\) at level .05. $$ \begin{array}{llllll} \hline \text { SIDS } & 4.0 & 4.4 & 4.8 & 4.9 & \\ \text { Control } & 3.7 & 4.1 & 4.3 & 5.1 & 5.6 \\ \hline \end{array} $$ Consult the Lehmann book (in the chapter bibliography) for more information on this test, called the Siegel-Tukey test.

The article "Effects of a Rice-Rich Versus PotatoRich Diet on Glucose, Lipoprotein, and Cholesterol Metabolism in Noninsulin-Dependent Diabetics" (Amer. J. Clin. Nutrit., 1984: 598-606) gives the accompanying data on cholesterol-synthesis rate for eight diabetic subjects. Subjects were fed a standardized diet with potato or rice as the major carbohydrate source. Participants received both diets for specified periods of time, with cholesterolsynthesis rate ( \(\mathrm{mmol}\) /day) measured at the end of each dietary period. The analysis presented in this article used a distribution- free test. Use such a test with significance level \(.05\) to determine whether the true mean cholesterol-synthesis rate differs significantly for the two sources of carbohydrates. $$ \begin{array}{lcccccccc} \hline \text { Subject } & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \mathbf{6} & \mathbf{7} & \mathbf{8} \\ \hline \text { Potato } & 1.88 & 2.60 & 1.38 & 4.41 & 1.87 & 2.89 & 3.96 & 2.31 \\ \text { Rice } & 1.70 & 3.84 & 1.13 & 4.97 & .86 & 1.93 & 3.36 & 2.15 \\ \hline \end{array} $$

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