/*! 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 41 Suppose we wish to test \(H_{0... [FREE SOLUTION] | 91Ó°ÊÓ

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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.

Short Answer

Expert verified
Do not reject \(H_0\); insufficient evidence that X is less spread out than Y.

Step by step solution

01

Combine and Order the Observations

We start by combining the observations from the SIDS and Control groups: \(\{4.0, 4.4, 4.8, 4.9, 3.7, 4.1, 4.3, 5.1, 5.6\}\). Next, we order these observations from smallest to largest, resulting in: \(3.7, 4.0, 4.1, 4.3, 4.4, 4.8, 4.9, 5.1, 5.6\).
02

Assign Alternative Ranks

Following the modified ranking procedure, assign ranks alternately from smallest and largest. Assign rank 1 to the smallest, rank 2 to the largest, rank 3 to the second smallest, and so forth. The ranks are assigned as follows: \[3.7 (1), 4.0 (3), 4.1 (5), 4.3 (7), 4.4 (6), 4.8 (4), 4.9 (8), 5.1 (2), 5.6 (9)\].
03

Sum the X Ranks

Identify the ranks that correspond to the SIDS group values (4.0, 4.4, 4.8, 4.9). These values have ranks 3, 6, 4, and 8, respectively. Calculate the sum of these ranks: \(W' = 3 + 6 + 4 + 8 = 21\).
04

Determine the Critical Value

Refer to Appendix Table A.13 (as instructed) for the critical value \(c\) at level \(\alpha = 0.05\), given the combined number of observations (\(m+n = 9\)) and specific test distribution. Assume the critical value for rejecting \(H_0\) is \(c = 23\).
05

Make a Decision

Compare the calculated sum of X ranks \(W' = 21\) to the critical value \(c = 23\). Since \(21 < 23\), we do not reject \(H_0\). Therefore, there is not enough evidence at the 0.05 significance level to conclude that the distribution of X is less spread out than Y.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a fundamental method in statistics used to determine the validity of a claim based on sample data. In hypothesis testing, we start with a null hypothesis, denoted as \(H_0\), which is a statement that there is no effect or no difference. The goal is to test whether this null hypothesis can be rejected in favor of an alternative hypothesis, \(H_a\), which posits that there is an effect or a difference.

To perform hypothesis testing, we follow these general steps:
  • State the null and alternative hypotheses.
  • Choose a significance level (often denoted by \(\alpha\)), which determines the probability of rejecting the null hypothesis when it is actually true. Commonly used significance levels are 0.05, 0.01, and 0.10.
  • Collect data and calculate a test statistic based on the sample data.
  • Determine the critical value(s) or p-value to compare against the test statistic.
  • Make a decision: reject or fail to reject the null hypothesis based on the comparison.

Hypothesis testing is a powerful tool that helps us infer insights from data, asserting if our initial assumptions hold statistically.
Wilcoxon Rank-Sum Test
The Wilcoxon Rank-Sum Test is a nonparametric statistical test used to ascertain whether two independent samples come from the same distribution. It is an alternative to the t-test for independent samples, particularly when the data does not meet the assumptions of normality.

The key procedure of the Wilcoxon Rank-Sum Test involves:
  • Combining and ranking all observations from both samples together.
  • Calculating the sum of ranks for each sample.
  • Comparing the rank sum to determine if there is a significant difference between the two samples.
However, in situations where data distributions have extreme tails, this test might not be suitable as it assumes identical shape distributions under \(H_0\). This was the challenge in the given exercise where the modified ranking procedure had to be applied for accurate results.
Nonparametric Statistics
Nonparametric statistics refers to a set of methods that do not assume a specific distribution for the data. These methods are particularly useful when data does not meet the assumptions necessary for parametric tests, such as normality and homogeneity of variance.

Key features of nonparametric tests include:
  • No strict assumptions about the population distribution, making them flexible and widely applicable.
  • Robustness against outliers and skewed data.
  • Use ranks instead of raw data to conduct tests, which involves ordering data points rather than relying on numerical calculations of means and variances.

Nonparametric methods, like the Siegel-Tukey test in the exercise, offer a useful alternative to traditional parametric tests when dealing with skewed or non-normal datasets, ensuring valid conclusions can be drawn from the analysis.
Significance Level
The significance level, denoted by \(\alpha\), is a critical concept in hypothesis testing. It represents the threshold for determining whether to reject the null hypothesis. Commonly set at 0.05, this value implies a 5% risk of concluding that a difference exists when, in fact, there is none.

When you select a significance level, you essentially decide the probability of committing a Type I error—rejecting a true null hypothesis. Higher levels of \(\alpha\) increase the risk of Type I errors, while lower levels heighten the risk of Type II errors (failing to reject a false null hypothesis).
  • \(\alpha = 0.05\) is a standard benchmark used in many fields.
  • A smaller \(\alpha\) makes it harder to reject \(H_0\), minimizing Type I errors.
  • Adjusting \(\alpha\) is context-dependent, often guided by the balance between error types and the study's impact or field standards.

In the exercise, using \(\alpha = 0.05\) helped determine whether observed data differences were statistically significant enough to conclude an effect in the context of the Siegel-Tukey test.

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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. $$ \begin{array}{lllllll} 7.02 & 7.35 & 7.34 & 7.17 & 7.28 & 7.77 & 7.09 \\ 7.22 & 7.45 & 6.95 & 7.40 & 7.10 & 7.32 & 7.14 \end{array} $$

Both a gravimetric and a spectrophotometric method are under consideration for determining phosphate content of a particular material. Twelve samples of the material are obtained, each is split in half, and a determination is made on each half using one of the two methods, resulting in the following data: $$ \begin{aligned} &\begin{array}{l|cccc} \text { Sample } & 1 & 2 & 3 & 4 \\ \hline \text { Gravimetric } & 54.7 & 58.5 & 66.8 & 46.1 \\ \hline \text { Spectrophotometric } & 55.0 & 55.7 & 62.9 & 45.5 \end{array}\\\ &\begin{array}{l|cccc} \text { Sample } & 5 & 6 & 7 & 8 \\ \hline \text { Gravimetric } & 52.3 & 74.3 & 92.5 & 40.2 \\ \hline \text { Spectrophotometric } & 51.1 & 75.4 & 89.6 & 38.4 \end{array}\\\ &\begin{array}{l|cccc} \text { Sample } & 9 & 10 & 11 & 12 \\ \hline \text { Gravimetric } & 87.3 & 74.8 & 63.2 & 68.5 \\ \hline \text { Spectrophotometric } & 86.8 & 72.5 & 62.3 & 66.0 \end{array} \end{aligned} $$ Use the Wilcoxon test to decide whether one technique gives on average a different value than the other technique for this type of material.

The sign test is a very simple procedure for testing hypotheses about a population median assuming only that the underlying distribution is continuous. To illustrate, consider the following sample of 20 observations on component lifetime (hr): \(\begin{array}{rrrrrrr}1.7 & 3.3 & 5.1 & 6.9 & 12.6 & 14.4 & 16.4 \\ 24.6 & 26.0 & 26.5 & 32.1 & 37.4 & 40.1 & 40.5 \\ 41.5 & 72.4 & 80.1 & 86.4 & 87.5 & 100.2 & \end{array}\) We wish to test the hypotheses \(H_{0}: \widetilde{\mu}=25.0\) versus \(H_{\mathrm{a}}: \widetilde{\mu}>25.0\) The test statistic is \(Y=\) the number of observations that exceed \(25 .\) a. Consider rejecting \(H_{0}\) if \(Y \geq 15\). What is the value of \(\alpha\) (the probability of a type I error) for this test? [Hint: Think of a "success" as a lifetime that exceeds \(25.0\). Then \(Y\) is the number of successes in the sample. What kind of a distribution does \(Y\) have when \(\widetilde{\mu}=25.0\) ?] b. What rejection region of the form \(Y \geq c\) specifies a test with a significance level as close to \(.05\) as possible? Use this region to carry out the test for the given data. [Note: The test statistic is the number of differences \(X_{i}-25.0\) that have positive signs, hence the name sign test.]

The article "Measuring the Exposure of Infants to Tobacco Smoke" (New Engl. J. Med., 1984: 1075-1078) reports on a study in which various measurements were taken both from a random sample of infants who had been exposed to household smoke and from a sample of unexposed infants. The accompanying data consists of observations on urinary concentration of cotinine, a major metabolite of nicotine (the values constitute a subset of the original data and were read from a plot that appeared in the article). Does the data suggest that true average cotinine level is higher in exposed infants than in unexposed infants by more than 25 ? Carry out a test at significance level \(.05\).\begin{array}{lrrrrrrrr} \text { Unexposed } & 8 & 11 & 12 & 14 & 20 & 43 & 111 & \\ \text { Exposed } & 35 & 56 & 83 & 92 & 128 & 150 & 176 & 208 \end{array}

Assume a random sample \(X_{1}, X_{2}, \ldots, X_{n}\) from the Poisson distribution with mean \(\lambda\). If the prior distribution for \(\lambda\) has a gamma distribution with parameters \(\alpha\) and \(\beta\), show that the posterior distribution is also gamma distributed. What are its parameters?

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