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Chapter 15: Problem 4

A random sample of 15 automobile mechanics certified to work on a certain type of car was selected, and the time (in minutes) necessary for each one to diagnose a particular problem was determined, resulting in the following data: $$ \begin{array}{rrrrrrrr} 30.6 & 30.1 & 15.6 & 26.7 & 27.1 & 25.4 & 35.0 & 30.8 \\ 31.9 & 53.2 & 12.5 & 23.2 & 8.8 & 24.9 & 30.2 & \end{array} $$ Use the Wilcoxon test at significance level .10 to decide whether the data suggests that true average diagnostic time is less than 30 minutes.

Short Answer

Expert verified
The data suggests that the true average diagnostic time is less than 30 minutes.

Step by step solution

01

Organize Data

List the sample data in ascending order: \(\{8.8, 12.5, 15.6, 23.2, 24.9, 25.4, 26.7, 27.1, 30.1, 30.2, 30.6, 30.8, 31.9, 35.0, 53.2\}\).
02

Calculate the Ranks

Assign ranks to the data, starting from 1 for the lowest value: \[\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15\}\]. In this case, the data values are unique.
03

Identify Observations Below Hypothesized Mean

Identify the values less than 30. These values and their ranks are: \(\{8.8, 12.5, 15.6, 23.2, 24.9, 25.4, 26.7, 27.1\}\) with ranks \(\{1, 2, 3, 4, 5, 6, 7, 8\}\).
04

Sum the Ranks of Values Below Hypothesized Mean

Sum the ranks of the values less than the hypothesized mean (30): \(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36\).
05

Determine Wilcoxon Test Statistic

The Wilcoxon test statistic for this sample is the sum of the ranks of values less than the hypothesized mean, which is 36.
06

Find Critical Value

For a significance level of 0.10 and \(n = 15\), we use a Wilcoxon rank-sum table. The critical value for a one-sided test is approximately 60.
07

Make Decision

Compare the test statistic (36) to the critical value (60). Since 36 is less than 60, we reject the null hypothesis.

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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 statistical method used to make decisions about a population based on sample data. It begins by stating a null hypothesis ( H_0 ) and an alternative hypothesis ( H_a ). The null hypothesis represents the default or initial assumption, while the alternative hypothesis represents the conclusion you aim to establish.
For example, in the problem at hand, we test whether the average diagnostic time is less than 30 minutes. Here, the null hypothesis might be that the average time is 30 minutes or more, while the alternative hypothesis states that it is less than 30 minutes.
Steps in hypothesis testing often include:
  • Formulating the null and alternative hypotheses.
  • Choosing a significance level, like 0.10, to determine how strict the test should be.
  • Calculating a test statistic from the sample data.
  • Comparing this statistic to a critical value from a statistical table to make a final decision.

Understanding these steps can help you critically evaluate the reliability of your results, and make more informed conclusions.
Non-Parametric Statistics
Non-parametric statistics are types of statistical methods designed to analyze data that doesn't necessarily fit a normal distribution. These methods do not assume that the underlying dataset is distributed in any particular form.
The Wilcoxon test is a non-parametric test often used when analyzing small samples or when the data doesn鈥檛 meet the assumptions necessary for parametric tests. In our example, the Wilcoxon rank-sum test is employed. It ranks the data points instead of relying on the mean and standard deviation. This makes it ideal when you have ordinal data or your data deviates significantly from normality.
Some advantages of non-parametric tests include:
  • They are more flexible and robust than their parametric counterparts.
  • They can be used with small sample sizes.
  • They are resistant to outliers that can affect parametric tests severely.

However, they may be less powerful than parametric tests when the assumptions for parametric tests are met. Non-parametric statistics can thus be a valuable tool in your analytical toolbox, especially in diverse real-world scenarios where data doesn鈥檛 follow ideal conditions.
Significance Level
The significance level, often denoted as \(\alpha\), is a threshold used to decide whether or not to reject the null hypothesis in a hypothesis test. It represents the probability of making a Type I error, which is mistakenly rejecting a true null hypothesis.
In the given exercise, we used a significance level of 0.10. This means that there's a 10% risk of concluding that the mean diagnostic time is less than 30 minutes when it actually is not. Researchers choose significance levels based on how much risk they are willing to accept.

A few common choices include:
  • 0.05: A standard level indicating a 5% risk of error.
  • 0.01: A more stringent level for a 1% risk of error, often used in critical fields.
  • 0.10: A more lenient level allowing a 10% risk of error, which could be used in exploratory analyses.

If the test result is statistically significant at the chosen level, it means the result is unlikely due to random chance, leading us to reject the null hypothesis. Fully understanding the significance level helps you interpret the results of your hypothesis test efficiently and determine the reliability of your decisions based on sample data.

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Most popular questions from this chapter

Suppose we wish to test\(H_{0}\) : the \(X\) and \(Y\) distributions are identicalversus \(H_{\mathrm{a}}\) : the \(X\) distribution is less spread out than the \(Y\) distributionThe 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. 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 \(I I_{0}\) is true. Thus \(c\) can be chosen from Appendix Table A.14 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. $$ \begin{array}{llllll} \text { SIDS } & 4.0 & 4.4 & 4.8 & 4.9 & \\ \text { Control } & 3.7 & 4.1 & 4.3 & 5.1 & 5.6 \end{array} $$ Consult the Lehmann book (in the chapter bibliography) for more information on this test, called the Siegel-Tukey test.

The accompanying data on cortisol level was reported in the article "Cortisol, Cortisone, and 11-Deoxycortisol Levels in Human Umbilical and Maternal Plasma in Relation to the Onset of Labor" (J. of Obstetric Gynaecology of the British Commonwealth, 1974: 737-745). Experimental subjects were pregnant women whose babies were deliverec between 38 and 42 weeks gestation. Group 1 individuals elected to deliver by Caesarean section before labor onset group 2 delivered by emergency Caesarean during inducec labor, and group 3 individuals experienced spontaneous labor. Use the Kruskal-Wallis test at level \(.05\) to test for equality of the three population means. $$ \begin{array}{lrrrrrr} \text { Group 1 } & 262 & 307 & 211 & 323 & 454 & 339 \\ & 304 & 154 & 287 & 356 & & \\ \text { Group 2 } & 465 & 501 & 455 & 355 & 468 & 362 \\ \text { Group 3 } & 343 & 772 & 207 & 1048 & 838 & 687 \end{array} $$

Suppose that observations \(X_{1}, X_{2}, \ldots, X_{n}\) are made on a process at times \(1,2, \ldots, n\). On the basis of this data, we wish to test \(H_{0}\) : the \(X_{i}\) 's constitute an independent and identically distributed sequence versus \(H_{\mathrm{a}}: X_{i+1}\) tends to be larger than \(X_{i}\) for \(i=1, \ldots, n\) (an increasing trend) Suppose the \(X_{i}\) 's are ranked from 1 to \(n\). Then when \(H_{\mathrm{a}}\) is true, larger ranks tend to occur later in the sequence, whereas if \(H_{0}\) is true, large and small ranks tend to be mixed together. Let \(R_{i}\) be the rank of \(X_{i}\) and consider the test statistic \(D=\sum_{i=1}^{n}\left(R_{i}-i\right)^{2}\). Then small values of \(D\) give support to \(H_{\mathrm{a}}\) (e.g., the smallest value is 0 for \(R_{1}=10, R_{2}=2, \ldots, R_{n}=n\) ), so \(H_{0}\) should be rejected in favor of \(H_{\mathrm{a}}\) if \(d \leq c\). When \(H_{0}\) is true, any sequence of ranks has probability \(1 / n\) !. Use this to find \(c\) for which the test has a level as close to 10 as possible in the case \(n=4\).

The article "Measuring the Exposure of Infants to Tobacco Smoke" (New Engl. J. of 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 cotanine, 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 cotanine level is higher in exposed infants than in unexposed infants by more than 25 ? Carry out a test at significance level \(.05\). $$ \begin{array}{lrrrrrrrr} \text { Unexposed } & 8 & 11 & 12 & 14 & 20 & 43 & 111 & \\ \text { Exposed } & 35 & 56 & 83 & 92 & 128 & 150 & 176 & 208 \end{array} $$

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 \(H_{0}: \tilde{\mu}=25.0\) versus \(H_{\mathrm{a}}: \tilde{\mu}>25.0\). 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 \(\tilde{\mu}=25.0\) ? 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.
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