Researchers want to test a new hand exercise plan for arthritis patients to see if they have improvedflexibility. They recruited 110 patients with arthritis in their hands and randomly assigned them to one oftwo treatment groups: group i took their regular medicine and did the new hand exercises; and group 2only took medicine. After one month, 42 of the 55 in group 1 improved in their flexibility, and 37 of the55 in group 2 improved their flexibility.Construct a 95% confidence interval to estimate the true difference in proportion pex - Pnoex of patientslike these who would show improvement in flexibility if they did the new hand exercises along withtaking the medicine versus just taking the medicine.(0.764)(0.236). (0.673(0.327)550.091t1.645.55(0.764)(0.236)(0.673)(0.327)0.091t1.9605555(0.764)(0.236)(0.673)(0.327)0.045 +1.960,55550.091t1.960,(0.764)(0.236)110(0.673)(0.327)+110

Answered: Image /qna-images/answer/2137cfd3-ac4f-43f9-8639-449be30c474d.jpg

Answered: Researchers want to test a new hand…

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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Similar questions

Take an SRS where n equals 50 and calculate at 95% confidence interval. If we took a different SRS of the same size from the same population, and calculate a confidence interval with the same level of confidence, the margin of error would stay the same. Is this statement True or False and explain why it is true or false.
Ms. Rains believes more seniors from PLD plan to attend college than seniors at Lafayette. In a random sample of 35 PLD seniors, 21 responded they plan to attend college. In a random sample of 32 Lafayette seniors only 17 responded they plan to attend college. Using a 95% confidence interval, predict the true difference in the proportion of seniors at PLD and Lafayette who plan to attend college and comment on Ms. Rains' belief.
Males and females were asked what they would do if they received a $100 bill in the mail that was addressed to their neighbor, but had been incorrectly delivered to them. Of the 70 males sampled, 55 said yes they would return it to their neighbor. Of the 130 females sampled, 120 said yes. We want to construct a 90% confidence interval estimate for the difference between the true proportion of males who say they would return the money to their neighbor and the true proportion of females who say they would return the money to their neighbor. Choose the best option for the result of constructing that interval. O -226 < p1 -p2 < +.003 O-226 < p1 -p2 <.-.048 O -243 < p1 -p2 < -.031
A computer store near campus is stocking its shelves and wants to know the relative proportions of students who use PCs vs Macs. A random sample of 60 students is surveyed, and each student is asked whether they use a PC or Mac more often. 55% select ”PC.” (a) What is the population and what is the sample in this study? (b) Calculate a 95% confidence interval for the proportion of UCI students that are PC users. (c) Provide an interpretation of this confidence interval in the context of this problem. (d) The confidence interval is pretty wide and leaves a lot of uncertainty over the proportion of UCIundergarduates who use PCs. With the goal to estimate a narrower 95% confidence interval, what is a simple change to this study that you could suggest for the next time that a similar survey is conducted?
Suppose you are planning an experiment and a sample has yet been selected. For this experiment you plan on taking a SRS of 50 mice with pancreatic cancer measuring a particular hormone level. What would be the impact on a 95% confidence interval calculated from the experiment on these mice if instead of a SRS of 50 mice, a SRS of 200 mice were taken?
A study would like to investigate the risk of lung cancer in url. area and rural area. Assume the proportion of urban residents with lung cancer is p1, and the proportion of rural area residents with lung cancer is p2. If it is found that 20 of 200 adults in an urban community have lung cancer and 10 of 150 adults in the rural community have lung cancer. Find the 95% confidence interval for the difference in the proportions. Would you conclude that p1 is greater than p2? Select all correct answers. O "O" is not in the interval (C.I) O Yes, we can conclude p1 is greater than p2. O "0.1" is in the interval (C.I) O "O" is in the interval (C.I) Not enough information is given. O (-0.02431, 0.090972) O (21.48721, 22.90809) O No, we can't conclude p1 is greater than p2. O "0.1" is not in the interval (C.I) (-1.21908, 2.64790) O (-19.20998, -17.21458) Previous hp
An economist at the Florida Department of Labor and Employment needs to estimate the unemployment rate in Indian River county. The economist found that 3.09% of a random sample of 632 county residents were unemployed.Find three different confidence intervals for the true proportion of Indian River county residents who are unemployed. Calculate one confidence interval with a 95% confidence level, one with a 96% confidence level, and one with a 97% confidence level. Notice how the confidence level affects the margin of error and the width of the interval.Report confidence interval solutions using interval notation. Express solutions in percent form, rounded to two decimal places, if necessary.The margin of error for a 95% confidence interval is given by E =.A 95% confidence interval is given by:
Several years ago, 45% of parents who had children in grades K-12 were satisfied with the quality of education the students receive. A recent poll asked 1,025 parents who have children in grades K-12 if they were satisfied with the quality of education the students receive. Of the 1,025 surveyed, 484 indicated that they were satisfied. Construct a 99% confidence interval to assess whether this represents evidence that parents' attitudes toward the quality of education have changed.
Chronic liver disease is associated with poorer health outcomes, including renal disease and cardiovascular diseases. Given this, a group of public health consultants are interested in identifying risk factors for chronic liver disease among 72 men. They found that compared to men without hepatitis B, the relative risk of chronic liver disease among those with hepatitis B was 1.24 and the 95% confidence interval was 0.94–1.54. A) Interpret the results (the relative risk and the corresponding 95% confidence interval) in not more than 75 words B) The confidence interval for the best estimate appears to be wide. What could be the reason? Justify in not more than 75 words.
A recent provincial report states that 90% of the population is aware that piercings as well as tattoos can transmit infectious disease. A survey was conducted in the province and random subjects were asked if they knew that piercings as well as tattoos could transmit an infectious disease. Following this, 1450 people responded with "yes" and 48 responded with "no". Find the 95% confidence interval estimate of the proportion of those who know that piercings as well as tattoos can transmit an infectious disease.
Two researchers, Jaime and Mariya, are each constructing confidence intervals for the proportion of a population who is lef-handed. They find the point estimate is 0.23. Each independently constructed a confidence interval based on the point estimate, but Jaime's interval has a lower bound of 0.220 and an upper bound of 0.234, while Mariya's interval has a lower bound of 0.169 and an upper bound of 0.272. Which interval is wrong? Why? 28 Choose the correct answer below. OA Jaime's interval is wrong because it does not include the point estimate. OB. Mariya's interval is wrong because it is too wide. OC. Mariya's interval is wrong because it is not centered on the point estimate, OD. Jaime's interval is wrong because it is to narrow O of 36
For many years businesses have struggled with the rising cost of health care. But recently, the increases have slowed due to less inflation in health care prices and employees paying for a larger portion of health care benefits. A recent Mercer survey showed that 52% of U.S. employers were likely to require higher employee contributions for health care coverage. Suppose the survey was based on a sample of 600 companies. Compute the margin of error and a 95% confidence interval for the proportion of companies likely to require higher employee contributions for health care coverage. If required, round your answer to four decimal places. Round intermediate calculations to four decimal places. Margin of Error: 0304 Confidence Interval: 4854 to 5546

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