/*! 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 14 Rat Hemoglobin Hemoglobin helps ... [FREE SOLUTION] | 91Ó°ÊÓ

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Rat Hemoglobin Hemoglobin helps the red blood cells transport oxygen and remove carbon dioxide. Researchers at NASA wanted to discover the effects of space flight on a rat's hemoglobin. The following data represent the hemoglobin (in grams per deciliter) at lift-off minus 3 days (H-L3) and immediately upon return (H-R0) for 12 randomly selected rats sent to space on the Spacelab Sciences 1 flight. Is the median hemoglobin level at lift-off minus 3 days less than the median hemoglobin level upon return? Use the \(\alpha=0.05\) level of significance. $$ \begin{array}{ccc|ccc} \text { Rat } & \text { H-L3 } & \text { H-R0 } & \text { Rat } & \text { H-L3 } & \text { H-R0 } \\ \hline 1 & 15.2 & 15.8 & 7 & 14.3 & 16.4 \\ \hline 2 & 16.1 & 16.5 & 8 & 14.5 & 16.5 \\ \hline 3 & 15.3 & 16.7 & 9 & 15.2 & 16.0 \\ \hline 4 & 16.4 & 15.7 & 10 & 16.1 & 16.8 \\ \hline 5 & 15.7 & 16.9 & 11 & 15.1 & 17.6 \\ \hline 6 & 14.7 & 13.1 & 12 & 15.8 & 16.9 \\ \hline \end{array} $$

Short Answer

Expert verified
At a \(\alpha=0.05\) level, there is sufficient evidence to conclude that median hemoglobin level 3 days before lift-off is less than upon return.

Step by step solution

01

Define Null and Alternative Hypotheses

Define the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_a\)).Null hypothesis: The median hemoglobin level at lift-off minus 3 days is equal to the median hemoglobin level upon return (\(H_0: \text{M}_\text{H-L3} = \text{M}_\text{H-R0}).Alternative hypothesis: The median hemoglobin level at lift-off minus 3 days is less than the median hemoglobin level upon return (\)H_a: \text{M}_\text{H-L3} < \text{M}_\text{H-R0}).
02

Calculate Differences

Calculate the differences between hemoglobin levels at lift-off minus 3 days and immediately upon return for each rat.$$\begin{array}{|c|c|c|c|}\text{Rat} & \text{H-L3} & \text{H-R0} & \text{Difference (H-R0 - H-L3)} \1 & 15.2 & 15.8 & 0.6 \2 & 16.1 & 16.5 & 0.4 \3 & 15.3 & 16.7 & 1.4 \4 & 16.4 & 15.7 & -0.7 \5 & 15.7 & 16.9 & 1.2 \6 & 14.7 & 13.1 & -1.6 \7 & 14.3 & 16.4 & 2.1 \8 & 14.5 & 16.5 & 2.0 \9 & 15.2 & 16.0 & 0.8 \10 & 16.1 & 16.8 & 0.7 \11 & 15.1 & 17.6 & 2.5 \12 & 15.8 & 16.9 & 1.1 \text{Total} &&& \text{Median of Differences} &&& 0.85\table{center}$$
03

Perform the Wilcoxon Signed-Rank Test

Choose the appropriate non-parametric test for comparing the medians of two related samples. In this case, use the Wilcoxon signed-rank test to determine if the differences are significant.1. Rank the absolute differences: $$\begin{array}{|c|c|}\text{Rat} & \text{Absolute Difference} & \text{Rank} \6 & 1.6 & 1 \4 & 0.7 & 2 \1 & 0.6 & 3 \2 & 0.4 & 4 \10 & 0.7 & 5 \3 & 1.4 & 6 \5 & 1.2 & 7 \12 & 1.1 & 8 \9 & 0.8 & 9 \8 & 2.0 & 10 \7 & 2.1 & 11 \11 & 2.5 & 12 \text{Sum of ranks for negative differences} = W^- = 2\text{Sum of ranks for positive differences} = W^+ = 78\table{center}$$2. Calculate the test statistic: In this case: W = 2.
04

Compare Test Statistic to Critical Value

Compare the test statistic to the critical value for Wilcoxon signed-rank test at alpha level of significance 0.05 and sample size 12.Using a Wilcoxon signed-rank table, find the critical value for a one-tailed test with \(n = 12\) and \(\alpha = 0.05\) which is 13.Since \(W = 2\) which is less than 13, \ \(W < W_{critical}\). So, reject the null hypothesis \(H_0\).
05

Conclusion

Since the test statistic value of 2 is less than the critical value of 13, we reject the null hypothesis. Thus, there is sufficient evidence to conclude that the median hemoglobin level at lift-off minus 3 days is less than the median hemoglobin level upon return at a significance level of 0.05.

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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 population parameters based on sample data. It starts with defining two hypotheses: the null hypothesis ( H_0 ) and the alternative hypothesis ( H_a ). The null hypothesis represents a statement of no effect or no difference, while the alternative hypothesis represents what we aim to prove. In the context of our exercise, the null hypothesis is that the median hemoglobin levels at lift-off minus 3 days (H-L3) are the same as the median hemoglobin levels upon return (H-R0). The alternative hypothesis is that the median H-L3 levels are less than the median H-R0 levels. By comparing test statistics to critical values, we can decide whether to reject the null hypothesis and accept the alternative hypothesis.
Non-parametric Tests
Non-parametric tests are types of statistical tests that do not assume a specific distribution for the data. They are useful when the data does not meet the assumptions required for parametric tests, such as normality. One common non-parametric test is the Wilcoxon Signed-Rank Test. This test is used to compare the medians of two related samples. For our exercise, it helps determine if there is a statistically significant difference between the hemoglobin levels before and after the space flight. Unlike parametric tests, which rely on means and standard deviations, non-parametric tests are based on the ranks of the data, making them more robust in the presence of outliers or non-normal distributions.
Median Comparison
Median comparison is a method used to evaluate the central tendency of two samples by comparing their medians. The median is the middle value of a dataset and is often preferred over the mean when dealing with skewed data or outliers. In the exercise, we compare the median differences between hemoglobin levels at lift-off minus 3 days and immediately upon return. Calculating the median difference involves ranking the absolute differences between the two sets of data. In this scenario, the Wilcoxon Signed-Rank Test ranks these differences and evaluates if the sum of the ranks indicates a significant decrease or increase. The test helps us understand if the observed difference in medians is likely due to random chance or reflects a real effect.
Statistical Significance
Statistical significance is a measure of whether an observed effect is likely due to chance or represents a true effect in the population. It is determined by comparing a test statistic to a critical value at a specified level of significance (alpha). In our exercise, the alpha level is 0.05, meaning we are willing to accept a 5% chance of making a Type I error, which is rejecting a true null hypothesis. If the test statistic (W) is less than the critical value from the Wilcoxon Signed-Rank table, we reject the null hypothesis. This indicates that the observed difference in median hemoglobin levels is statistically significant. In our case, the test statistic of 2 is less than the critical value of 13, meaning we reject the null hypothesis and conclude that the median H-L3 levels are significantly lower than the median H-R0 levels.

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Most popular questions from this chapter

Outpatient Treatment The median length of stay for substance-abuse outpatient treatment completers is 107 days for those referred by the criminal justice system. The following data represent the length of stays for a random sample of substanceabuse outpatient treatment completers who were referred by an employer. $$ \begin{array}{cccccc} \hline 80 & 108 & 95 & 107 & 89 & 100 \\ \hline 85 & 102 & 115 & 109 & 99 & 94 \\ \hline \end{array} $$ Using the Wilcoxon signed-ranks test at the \(\alpha=0.05\) level of significance, does the median length of stay seem different for employer referrals than for those referred by the criminal justice system?

Researchers wanted to discover whether the median amount of bacteria in carpeted rooms was greater than that in uncarpeted rooms. To determine the amount of bacteria in a room, researchers pumped the air from the room over a Petri dish at the rate of 1 cubic foot per minute for 8 carpeted rooms and 8 uncarpeted rooms. Colonies of bacteria were allowed to form in the 16 Petri dishes. The results (bacteria/cubic foot) are presented in the following table: $$ \begin{array}{|rc|cc|} \hline {\text { Carpeted Rooms }} & {\text { Uncarpeted Rooms }} \\ \hline 11.8 & 10.8 & 12.1 & 12.0 \\ \hline 8.2 & 10.1 & 8.3 & 11.1 \\ \hline 7.1 & 14.6 & 3.8 & 10.1 \\ \hline 13.0 & 14.0 & 7.2 & 13.7 \\ \hline \end{array} $$ Is the median amount of bacteria in carpeted rooms greater than the median amount of bacteria in uncarpeted rooms? Use the \(\alpha=0.05\) level of significance.

"Defense wins championships" is a common phrase used in the National Football League. Is defense associated with winning? The following data represent the winning percentage and the yards per game allowed during the 2014-2015 season for a random sample of teams. $$ \begin{array}{lcc} \text { Team } & \text { Winning Percentage } & \text { Total Yards } \\ \hline \text { Baltimore Ravens } & 0.625 & 336.9 \\ \hline \text { Cleveland Browns } & 0.438 & 366.1 \\ \hline \text { Denver Broncos } & 0.750 & 305.2 \\ \hline \text { Jacksonville Jaguars } & 0.188 & 370.8 \\ \hline \text { New England Patriots } & 0.750 & 344.1 \\ \hline \text { Oakland Raiders } & 0.188 & 357.6 \\ \hline \text { Pittsburgh Steelers } & 0.688 & 353.4 \end{array} $$ (a) Test the belief that defense wins championships by determining whether a higher winning percentage is associated with a lower number of total yards given up at the \(\alpha=0.10\) level of significance. (b) Draw a scatter diagram to support your conclusion.

Use the Wilcoxon matched-pairs signedranks test to test the given hypotheses at the \(\alpha=0.05\) level of significance. The dependent samples were obtained randomly. Hypotheses: \(H_{0}: M_{D}=0\) versus \(H_{1}: M_{D} \neq 0\) with \(n=18, T_{-}=-121,\) and \(T_{+}=50\)

(a) draw a scatter diagram, (b) compute \(r_{s}\), an (c) determine if \(X\) and \(Y\) are associated at the \(\alpha=0.05\) level of significance $$ \begin{array}{lllllll} \hline X & 0 & 0.5 & 1.4 & 1.4 & 3.9 & 4.6 \\ \hline Y & 0.8 & 2.3 & 1.9 & 2.5 & 5.0 & 6.8 \\ \hline \end{array} $$

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