Conduct calculation steps to determine the 95% plus fourconfidence interval for the difference in proportionsP₁ P2 of individuals developing a cold after viralexposure between the echinacea treatment and the placebo.Let p₁ represent the population proportion for the placebopatient population and p2 represent the populationproportion for the echinacea patient population.First provide the plus four sample proportions for theplacebo group ₁ and the echinacea group 2. Compute thestandard error SE of the difference of the proportionestimates. Provide all answers to four decimal places.PI =P2||SE= Echinacea is widely used as an herbal remedy for thecommon cold, but does it work? In a double-blindexperiment, healthy volunteers agreed to be exposed tocommon-cold-causing rhinovirus type 39 and have theirsymptoms monitored. The volunteers were randomlyassigned to take either a placebo or an echinaceasupplement daily for 5 days following viral exposure.Among the 103 subjects taking a placebo, 88 developed acold, whereas 44 of the 48 subjects taking echinacea1developed a cold.

The standard error is nothing but the estimate of standard error. It is the form of a sample…

Answered: Conduct calculation steps to determine…

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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Two types of flares are tested and their burning times are recorded. The summary statistics are given below. Brand X Brand Y n = 35 n = 40 x= 19.4 min x= 15.1 min s= 1.4 min s= 0.8 min Construct a 95% confidence interval for the differences between the mean burning time of the brand X flare and the mean burning time of the brand Y flare.
A repeated measures study is done to measure the change in IQ test scores taken on Monday versus those taken on Friday. There were n=9 participants in the study. The mean difference was MD =-5 points and the standard error for the mean difference was SMD=0.51. Construct a 95% confidence interval to estimate the size of the population mean difference.
A researcher wants to estimate the difference between the mean score points of home and visitor teams in a basketball game. For this purpose, researcher randomly selects 24 basketball games and obtains their scores. In these games, home teams' mean score is 80 points with a standard deviation of 35 points while visitor teams' mean score is 79 points with a standard deviation of 10 points. Assuming equal population variances, find the (93)% confidence interval for the difference between the mean score points of home and visitor teams.
Sales personnel for Skilling Distributors submit weekly reports listing the customer contacts made during the week. A sample of 15 weekly reports showed a sample mean of 19.5 customer contacts per week. The sample standard deviation was 5.2. Provide a 90% confidence interval for the population mean number of weekly customer contacts for the sales personnel.
A football coach claims that the average weight of all opposing teams’ members is 225 pounds. For a test of the claim, a simple random sample of 50 players is taken from all the opposing teams. The mean is found to be 230 pounds with a standard of 15 pounds. Construct a 95% confidence interval for the unknown average weight of the population of all the players of opposing teams. Interpret your result.
The mean score on the math portion of the ACT for the year 2000 is μ = 20.7. A sample of 20 students in a local school district had a mean of 22.95. Using the sample, a 95% confidence interval estimate for the mean is (20.76, 25.14). State the point estimate for the mean ACT score for students. Based on this sample information , do you believe that the local students have a mean different from the general population? Why? 22.95 is in the center of the interval. The means are definitely not different. 22.95 is just outside the 95% confidence interval. Can't say the means are really different. 22.95 is not equal to 20.7. The means are definitely different. 20.7 is just outside the 95% confidence interval. Can't say the means are really different.
In a sample of 10 randomly selected black bears from northeast Pennsylvania, the sample mean was 214.1 pounds and the sample standard deviation was 51.1 pounds. Construct a 99% confidence interval estimate for the population standard deviation of all black bears in this region.
12. A new manufacturing process of component parts is being considered. Samples were taken improves the proportion of defective components. 1425 of 1500 components from the existing process and 1920 of 2000 components from the new process were found to be non-defective. 75 80 P1 = = 0.05, P2 = 0.04 and Pi - P2 = 0.05 - 0.04 = 0.01 1500 2000 a. Find a 99% confidence interval for the true difference in the fraction of defectives between the existing and the new processes? b. Based on your answer in (a), is there any significant difference between the existing and the new processes? Justify?
Total plasma volume is important in determining the required plasma component in blood replacement therapy for a person undergoing surgery. Plasma volume is influenced by the overall health and physical activity of the individual. Suppose that a random sample of 45 male firefighters was tested and the sample mean was 37.5 ml/kg. Assuming that the population standard deviation for the distribution of blood plasma is 7.5 ml/mg. Find a 99% confidence interval for the mean blood plasma volume in male firefighters.
2. There was a sample of 14 adult wild mountain lions who were caught and released in the San Andres Mountains. The average weight of the sample was 82.55 pounds with a sample standard deviation of 13.45 pounds. Find a 95% confidence interval for the population mean weight of all the wild mountain lions in the San Andres Mountains.
in a test of the effectiveness of garlic for lowering cholesterol, 43 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes in their levels of LDL cholesterol have a mean of 5.7 and a standard deviation of 17.7. Construct a 99% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic and reducing LDL cholesterol?
A telephone poll of 500 adult Americans was reported in an issue of Time Magazine. One of the questions asked was “What is the main problem facing the country?” Thirty percent answered “crime”. We are interested in the population proportion of adult Americans who feel that crime is the main problem. Define the random variables X and P’ in words. Which distribution should be used for this problem? Explain your choice. Construct a 97% confidence interval for the population proportion of adult Americans who feel that crime is the main problem. i. State the confidence interval, ii. Sketch the graph, iii. Calculate the error bound 4. Suppose we want to lower the sampling error. What is one way to accomplish that?

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