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Conduct a Single factor ANOVA, within, among, f-stat, study, clinical, test, probability, statistics

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Attached is the actual problem and the chart needed to answer the questions. Below are the questions to be answered. Please show work I am trying to learn to work this type of problem for a test on Thursday.

A. What is the experimental design?

B. State the null and alternative hypothesis

C. Calculate the correction factor

D. Calculate SS total

E. Calculate SS within

F. Calculate SST among

g. Calculate SSE

H Calculate MST

I Calculate MSE

J. Calculate F-statistic

K. what is the critical F-statistic (alpah=0.05)

l. state the conclusion

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Solution Summary

The following outlines the process for conducting a single factor ANOVA experiement given a set of data. Work is computed in Excel.

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2. The J. Jenkins group recently conducted a study that investigated the effect of carbon monoxide exposure on patients with coronary artery disease. The subjects involved in the study were recruited from three different medical centers - the Johns Hopkins University School of Medicine (nl=21), the Rancho Los Amigos Medical Center (n2=16) and the St. Louis University School of Medicine (n3=23). Before combining the subjects into one large group to conduct the analysis, Dr. Jenkins first examined some of the baseline characteristics to ensure .that the patients from the various centers are in fact comparable.
One characteristic is pulmonary function before the start of the study; if patients from one medical center begin with measures of forced expiratory volume in 1 second (FEV 1) that are much larger or, much smaller - than those from the other centers, then the results of .the analysis may be affected. Therefore, given that the populations of patients in the three centers have mean baseline FEV ...

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  • BSc , Wuhan Univ. China
  • MA, Shandong Univ.
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  • "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
  • "excellent work"
  • "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
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