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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In a recent year, a sample of the South American Redeye Piranha was collected by fishermen on a particular river. Wildlife biologists regard this sample as a random sample of all Redeye Piranha in that river. The mean length of the 60 fish in the sample was 12.88 cm, and a 99% confidence level for the true mean length of all Redeye Piranha in that river is 12.09 cm to 13.67 cm. Explain what would happen to the length of the interval if the confidence level were decreased to 95%.
In a test of the effectiveness of garlic for lowering cholesterol, 48 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (before - after) in their levels of LDL cholesterol (in mg/dL) have a mean of 5.5 and a standard deviation of 19.8. 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 in reducing LDL cholesterol? Click here to view at distribution table. Click here to view page 1 of the standard normal distribution table, Click here to view page 2 of the standard normal distribution table. What is the confidence interval estimate of the population mean u? O mg/dL < µ
In a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (before - after) in their levels of LDL cholesterol (in mg/dL) have a mean of 3.9 and a standard deviation of 15.3. 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 in reducing LDL cholesterol? Click here to view at distribution table. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. What is the confidence interval estimate of the population mean u? | mg/dL < µ
In a test of the effectiveness of garlic for lowering cholesterol, 48 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (before - after) in their levels of LDL cholesterol (in mg/dL) have a mean of 3.6 and a standard deviation of 17.5. Construct a 95% 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 in reducing LDL cholesterol? Click here to view a t distribution table. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. What is the confidence interval estimate of the population mean u? mg/dL <µ< mg/dL (Round to two decimal places as needed.)
the variance in waiting times(in minutes) of 10 attendees at bethel woods, where attendees enter a single waiting line that feeds three checkpoint windows is 0.227 minutes. A 95% confidence interval for the population variance would be
In a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (before - after) in their levels of LDL cholesterol (in mg/dL) have a mean of 2.0 and a standard deviation of 16.5. 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 in reducing LDL cholesterol?
In a test of the effectiveness of garlic for lowering cholesterol,44 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes(beforeminus after) in their levels of LDL cholesterol (in mg/dL) have a mean of 5.6 and a standard deviation of 17.5. 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 in reducing LDL cholesterol? What is the confidence interval estimate of the population mean mu? mg/dLless than muless than mg/dL (Round to two decimal places as needed.)
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.
A random sample of 180 men indicated that 50 were smokers, and a random sample of 150 women surveyed indicated that 39 smoked. Construct a 98% confidence interval for the true difference in proportions of male and female smokers.
In a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (before - after) in their levels of LDL cholesterol (in mg/dL) have a mean of 4.1 and a standard deviation of 19.1. 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 in reducing LDL cholesterol? Click here to view at distribution table. Click here to view page 1 of the standard normal distribution table, Click here to view page 2 of the standard normal distribution table. What is the confidence interval estimate of the population mean u? mg/dL < µ
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?

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