How do you answer #48 using ANOVA? ### Study on Rehabilitation Program for Pulmonary Disease**Objective:**A study by Strijbos et al. aimed to compare the results of a 12-week hospital-based outpatient rehabilitation program (Group 1) with a 12-week home-care rehabilitation program (Group 2) in patients with chronic obstructive pulmonary disease and airflow limitation. A control group (Group 3) did not receive rehabilitation therapy. The data collected included breathing frequency scores of subjects 18 months after rehabilitation.**Breathing Frequency Data:**| Group | 1 | 2 | 3 ||-------|-------|-------|-------|| | 12 | 12 | 16 || | 16 | 16 | 24 || | 12 | 12 | 12 || | 16 | 12 | 12 || | 14 | 14 | 12 || | 12 | 15 | 16 || | 14 | 12 | 16 || | 12 | 14 | 16 || | 15 | 18 | 12 || | 12 | 12 | 12 || | 12 | 14 | 18 || | 10 | 10 | 15 || | 16 | 16 | 10 || | 18 | 20 | 16 |**Source:**Data provided by Dr. Jaap H. Strijbos.### Development of GAD Self-Report MeasureRoemer et al. developed a self-report measure for generalized anxiety disorder (GAD) targeting undergraduate populations. Reliability studies involved subjects completing the GAD questionnaire (GAD-Q) and the Penn State Worry Questionnaire (PSWQ). PSWQ scores were made by four groups based on their GAD status:- **Study II (Group 1):** GAD by questionnaire- **Non-GAD by Questionnaire (Study II

Anova is the measure of parametric test. It compare the mean significance between the three or more…

Answered: 48. The objective of a study by…

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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You take a survey that is supposed to measure whether you are an extrovert or not. The score you get on the extrovert survey is a z-score of 2.0. What does that mean about your level of extraversion compared to the average score?
You are testing the claim that the proportion of men who own cats is smaller than the proportion of women who own cats.You sample 110 men, and 75% own cats.You sample 150 women, and 65% own cats.Find the proportion of the pooled samples, (pcpc), as a decimal, rounded to two decimal places.
Quit smoking: A recent study reported that in a sample of 234 European-American HIV-positive smokers, 105 of them had used a nicotine patch to try to quit smoking, and in a sample of 81 Hispanic-American HIV-positive smokers, 25 had used a nicotine patch. Can you conclude that the proportion of patch users is greater among European-Americans? Let p₁ denote the proportion of European-Americans who used the patch. Use the a=0.01 level of significance. Part 1 of 4 (a) State the appropriate null and alternate hypotheses. Ho: P₁ (Choose one) ▼ P₂ H₁: P₁ (Choose one) P2 This is a (Choose one) test. Part 2 of 4 (b) Find the P-value. Round the answer to four decimal places. P-value= X S Ś
You are testing the claim that the proportion of men who own cats is larger than the proportion of women who own cats. You sample 110 men, and 25% own cats. You sample 150 women, and 85% own cats. Find p (p-bar), as a decimal, rounded to two decimal places.
You are testing the claim that the proportion of men who own cats is larger than the proportion of women who own cats.You sample 60 men, and 50% own cats.You sample 140 women, and 20% own cats.Find the test statistic, rounded to two decimal places.
You want to obtain a sample to estimate the proportion of a population that possess a particular genetic marker. Based on previous evidence, you believe approximately P∗=15% of the population have the genetic marker. You would like to be 98% confident that your estimate is within 3% of the true population proportion. How large of a sample size is required?n =
← The results of a survey on the belief that stem cell research has merit are shown in the table below. % X 1524 SEP 28 95% CI (0.7052, 0.7434) Here, X denotes the number who believed that stem cell research has merit. Complete parts a and b below. N 2104 a. Explain how to interpret "Sample p" and "95% CI" in this table. The "Sample p" is the proportion of all respondents in the sample who believe stem cell research has merit, 1524/2104 0.72. The "95% CI" is the 95% confidence interval and it means that we can be 95% confident that the interval 0.7052 to 0.7434 contains the population proportion. (Round to four decimal places as needed.) b. What is the 95% margin of error associated with the poll? The 95% margin of error is. (Round to four decimal places as needed.) F5 6 Sample p 0.724335 tv F6 Clear all & F7 Check answer Ndi 8 DIL 9 elp students like you learn core concep for 10 level of significance and two (0-65) (0-35) 1000 Chegg Study F9 A O
A medical researcher tested a new treatment for poison ivy against the traditional ointment. He concluded that the new treatment is more effective. Explain what the P-value of 0.037 means in this context. Explain what the P-value of 0.037 means in this context. O A. There is a 3.7% difference between the effectiveness of the new treatment and the effectiveness of the traditional ointment. O B. If there is no difference in effectiveness, the chance of seeing an observed difference this large or larger is 3.7% by natural sampling variation. O C. If there is a difference in effectiveness, the chance of seeing an observed difference this large or larger is 3.7% by natural sampling variation. D. There is a 3.7% chance that the new treatment is effective.

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