Answer asap An article in the American Journal of Emergency Medicine compared the ability to detect acute traumatic aortic injury (ATAI) on cervical x-ray images. The data consisted of 13 cases of ATAI, 19 cases with negative aortography (NAO) and 18 cases with multiple trauma (MT)without aortography. Measurements of the cervical soft-tissue width at the third cervical vertebrae generated the following results.Averages were 9.2,8.1 and 7.8 mm and standard deviations were 2.2, 3.3 and 3 mm for the ATAI, NAO, and MT groups, respectively.Consider an ANOVA for these data.a) Use the sample sizes and group averages to compute the grand mean width and the mean square for treatments.Round your answer to two decimal places (e.g. 9.87).μ =MSTreatments =b) Use the sample sizes and standard deviations in the groups to compute the mean square error.Round your answer to two decimal places (e.g. 9.87).MSError =ic) Use the results in the previous parts to conduct the ANOVA test and prove that the mean widths do not differ. Calculate the P-valuefor the test. c) Use the results in the previous parts to conduct the ANOVA test and prove that the mean widths do not differ. Calculate the P-valuefor the test.SourceTreatmentsErrorTotalDFiiiSS{to twodecimal places)iiiMS{to twodecimal places)iiF{to twodecimal places)iP(to threedecimal places)i

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MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Patients with rheumatoid arthritis are at a greater risk of developing osteoporosis. The reasons are not well understood due to the difficulty in qualitatively assessing bone metabolism and mineral content. A researcher measures human calcitonin (HCT) levels in men with rheumatoid arthritis. The researcher randomly selects 43 men with rheumatoid arthritis and measured the mean HCT level of 33.8 pg/mL. Assume that HCT for individuals with rheumatoid arthritis follows a normal distribution with standard deviation σ = 23.6 pg/mL. 5. What is the effect on the interval width if we decreased or increase the confidence level to 90% or 99% respectively? (Hint, create the intervals)
Patients with rheumatoid arthritis are at a greater risk of developing osteoporosis. The reasons are not well understood due to the difficulty in qualitatively assessing bone metabolism and mineral content. A researcher measures human calcitonin (HCT) levels in men with rheumatoidarthritis. The researcher randomly selects 43 men with rheumatoid arthritis and measured the mean HCT level of 33.8 pg/mL. Assume that HCT for individuals with rheumatoid arthritis follows a normal distribution with standard deviation σ = 23.6 pg/mL. 3. Create a 95% confidence interval for the population mean HCT.
Patients with rheumatoid arthritis are at a greater risk of developing osteoporosis. The reasons are not well understood due to the difficulty in qualitatively assessing bone metabolism and mineral content. A researcher measures human calcitonin (HCT) levels in men with rheumatoidarthritis. The researcher randomly selects 43 men with rheumatoid arthritis and measured the mean HCT level of 33.8 pg/mL. Assume that HCT for individuals with rheumatoid arthritis follows a normal distribution with standard deviation σ = 23.6 pg/mL. 6. Describe the effect on your confidence interval if you increase or decrease the samplesize
Patients with rheumatoid arthritis are at a greater risk of developing osteoporosis. The reasons are not well understood due to the difficulty in qualitatively assessing bone metabolism and mineral content. A researcher measures human calcitonin (HCT) levels in men with rheumatoidarthritis. The researcher randomly selects 43 men with rheumatoid arthritis and measured the mean HCT level of 33.8 pg/mL. Assume that HCT for individuals with rheumatoid arthritis follows a normal distribution with standard deviation σ = 23.6 pg/mL. 4. Explain in detail what the 95% Confidence Interval for the mean HCT represents within this context.
The heart rate of humans are normally distributed with a mean of 90.3 bpm and std. dev. of 7.3 bpm. What are the heartbeats that separate the lower 25% from the rest and the upper 20% from the rest?
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How heavy a load (pounds) is needed to pull apart pieces of Douglas fir 4 inches long and 1.5 inches square? Students doing a laboratory exercise sample 20 pieces of wood and find an average load capacity of 30,841 pounds. We are willing to regard the wood pieces prepared for the lab session as an SRS of all similar pieces of Douglas fir. Engineers also commonly assume the characteristics of materials vary normally. Suppose that the strength of pieces of wood like these follow a normal distribution with a population standard deviation of 3,000 pounds (As you can see all three necessary assumptions are met). a. Assuming that all evergreen wood has a known "load" capacity average of 30,000 pounds. Make a two-sided hypothesis (null and alternative statement) about Douglas Fir "load" capacity compared to the overall average. b. Apply the formula for finding our test statistic (show your work or describe the process) Specifically, use the formula for the z-statistic (pasted below and on…
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Scoliosis is a condition involving curvature of the spine. In a study, 286 girls age 10 to 15 years, were followed to determine the effects of an underarm plastic brace (111 patients), nighttime surface electrical stimulation (46 patients) and observation only (129 patients). A treatment was deemed to have failed if the curvature of the spine increased by 6° on two successive examinations. The preceding table summarizes the results obtained by researchers. At the 1% significance level, does the data provide sufficient evidence to conclude that a difference in failure rate exists among the three types of treatments? Use the Chi-Square Homogeneity Test and interpret your result. Success Failure Total Brace 94 17 111 Stimulation 24 22 46 Observation 71 58 129 Total 189 97 286

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