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In a study of hypnosis, breathing patterns were observed in an experimental group of subjects and in a control group. The measurements of total ventilation (liters of air per minute per square meter of body area) are shown. \({ }^{67}\) (These are the same data that were summarized in Exercise \(7.5 .6 .\) ) Use a Wilcoxon-Mann-Whitney test to compare the two groups at \(\alpha=0.10 .\) Use a nondirectional alternative. $$ \begin{array}{|cc|} \hline \text { Experimental } & \text { Control } \\ \hline 5.32 & 4.50 \\ 5.60 & 4.78 \\ 5.74 & 4.79 \\ 6.06 & 4.86 \\ 6.32 & 5.41 \\ 6.34 & 5.70 \\ 6.79 & 6.08 \\ 7.18 & 6.21 \\ \hline \end{array} $

Short Answer

Expert verified
Calculate U for both groups, find the critical value of U from the table, and compare to determine if there is a statistically significant difference at the \(\alpha=0.10\) level.

Step by step solution

01

- Rank All Data

Combine both the experimental and control groups' data and rank them from smallest to largest, regardless of which group they belong to. Assign ranks to the measurements, with the smallest value assigned rank 1. In case of ties (identical values), assign the average rank that these values would have received if they were slightly different.
02

- Calculate Sum of Ranks for Each Group

Once all measurements have been ranked, add up the ranks for each group separately. Let the sum of the ranks for the experimental group be denoted as T_e and the sum for the control group as T_c.
03

- Use Test Statistic for Wilcoxon-Mann-Whitney Test

The test statistic for the Wilcoxon-Mann-Whitney test is usually denoted as U. Calculate U for each group using the following formulas: For the experimental group, U = T_e - (n_e * (n_e + 1))/2, where n_e is the number of observations in the experimental group. For the control group, U = T_c - (n_c * (n_c + 1))/2, where n_c is the number of observations in the control group. Choose the smaller of the two U values as the test statistic.
04

- Determine the Critical Value of U

Consult the Wilcoxon-Mann-Whitney critical values table for two samples of the given sizes at the desired alpha level. Compare your calculated test statistic to the critical value to determine whether the null hypothesis can be rejected.
05

- Make a Decision Based on the Critical Value

If the calculated U is less than or equal to the critical value from the table, reject the null hypothesis and conclude that there is a statistically significant difference between the two groups. If the calculated U is greater, do not reject the null hypothesis.

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Key Concepts

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

Nonparametric Statistics
Nonparametric statistics offer powerful tools for analyzing data without making assumptions about its distribution. Unlike parametric tests, which require the data to follow certain distributional criteria (e.g., normal distribution), nonparametric methods are distribution-free. These are particularly useful when working with small sample sizes, ordinal data, or when the presence of outliers would significantly skew the results of parametric tests.
Rank Sum Test
The rank sum test, also known as the Wilcoxon-Mann-Whitney test, is a nonparametric method used to determine whether there is a statistically significant difference between two independent groups. It is effective when comparing median values of two samples. To perform the test, all observations from both groups are ranked together. The ranks are then summed for each group separately, and the test statistic is calculated. This test does not rely on any assumptions about the populations from which the samples were drawn, making it a versatile tool in statistical analysis.
Hypothesis Testing
Hypothesis testing is a cornerstone of statistical analysis used to determine the evidence against a null hypothesis. The null hypothesis typically posits no effect or no difference, whereas the alternative hypothesis suggests there is an effect or a difference. In the context of the Wilcoxon-Mann-Whitney test, the null hypothesis would state that there is no difference in the median values between the two groups. By comparing a calculated test statistic to a critical value, researchers can decide whether to reject the null hypothesis or not, thus assessing if the observed data provide sufficient evidence to support the alternative hypothesis.
Statistical Significance
Statistical significance is a determination about the non-randomness of the results obtained from a statistical test. In simpler terms, it assesses whether the observed effect or difference is likely due to chance or if it reflects a true pattern in the data. A significance level, denoted as alpha (α), is predetermined by the researcher, often set at 0.05 or 0.10. If the probability of the observed test statistic is less than or equal to this alpha level, the result is considered statistically significant. This means that the null hypothesis is rejected in favor of the alternative hypothesis. It is essential to select an appropriate alpha level before conducting the test to avoid misuse of statistical analysis.

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Most popular questions from this chapter

Researchers in Norway examined the records of 85,176 children to look for a possible association between use of folic acid during pregnancy and development of autism. The null hypothesis of interest is that use of folic acid is not related to risk of autism. The researchers set \(\alpha\) at 0.05 and stated that they had \(93 \%\) power to detect a certain difference in autism rates. Explain what "93\% power" means in this context. In particular, what is the probability of a Type II error? \(^{58}\)

A researcher performed a \(t\) test of the null hypothesis that two means are equal. He stated that he chose an alternative hypothesis of \(H_{A}: \mu_{1}>\mu_{2}\) because he observed \(\bar{y}_{1}>\bar{y}_{2}\). (a) Explain what is wrong with this procedure and why it is wrong. (b) Suppose he reported \(t=1.97\) on 25 degrees of freedom and a \(P\) -value of \(0.03 .\) What is the proper \(P\) -value? Why?

Postoperative ileus (POI) is a form of gastrointestinal dysfunction that commonly occurs after abdominal surgery and results in absent or delayed gastrointestinal motility. Does rocking in a chair after abdominal surgery reduce postoperative ileus (POI) duration? Sixty-six postoperative abdominal surgery patients were randomly divided into two groups. The experimental group \((n=34)\) received standard care plus the use of a rocking chair while the control group \((n=32)\) received only standard care. For each patient, the postoperative time until first flatus (days) (an indication that the POI has ended) was measured. The results are tabulated below. \(^{43}\) Here is computer output for a \(t\) test. \(\mathrm{t}=-3.524, \mathrm{df}=63.99, \mathrm{p}\) -value \(=0.000396\) alternative hypothesis: true difference in means is less than 0 (a) State the null and alternative hypotheses in context. (b) Is there evidence at the level of \(\alpha=0.05\) that use of the rocking chair reduces POI duration (i.e., the time until first flatus)? (c) Although the researchers hypothesized that the use of a rocking chair could reduce POI duration, it is not unreasonable to hypothesize that the use of a rocking chair could increase POI duration. Based on this possibility, discuss the appropriateness of using a directional versus nondirectional test. (Hint: Consider what medical recommendations might be made based on this research.)

In a study of the nutritional requirements of cattle, researchers measured the weight gains of cows during a 78 -day period. For two breeds of cows, Hereford (HH) and Brown Swiss/Hereford (SH), the results are summarized in the following table. \({ }^{6}\) $$ \begin{array}{|lll|} \hline & \text { HH } & \text { SH } \\ \hline n & 33 & 51 \\ \bar{y} & 18.3 & 13.9 \\ s & 17.8 & 19.1 \\ \hline \end{array} $$ (a) What is the value of the \(t\) test statistic for comparing the means? (b) In the context of this study, state the null and alternative hypotheses. (c) The \(P\) -value for the \(t\) test is \(0.29 .\) If \(\alpha=0.10\), what is your conclusion regarding the hypotheses in (b)?

In a study of 1,040 subjects, researchers found that the prevalence of coronary heart disease increased as the number of cups of coffee consumed per day increased. \({ }^{31}\) (a) What is the explanatory variable? (b) What is the response variable? (c) What are the observational units?

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