3. A trial compares a new drug to that considered standard care in patients undergoing kneereplacement surgery. The outcome of interest is whether patients report a reduction in pain by3+ scale points or not. Construct a 95% confidence interval for the difference in proportions ofpatients between treatments who report a pain reduction of 3+ points.Reduction of 3+ PointsTreatmentnNumberProportionNew Drug50400.8Standard50150.3

The objective of this question is to construct a 95% confidence interval for the difference in…

Answered: 3. A trial compares a new drug to that…

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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According to a report on sleep deprivation by the Centers for Disease Control and Prevention, the proportion of California residents who reported insufficient rest or sleep during each of the preceding 30 days is 8.0%, while this proportion is 8.8% for Oregon residents. These data are based on simple random samples of 11,545 California and 4,691 Oregon residents. Calculate a 95% confidence interval for the difference between the proportions of Californians and Oregonians who are sleep deprived and interpret it in context of the data.
Suppose an All Greens store in Sonoma, California, wants to estimate a range of advertising costs appropriate to its store. If it spends too little on advertising, it will not reach enough customers. However, it does not want to overspend on advertising for this type and size of store. At this store, x1 = 163, x2 = 2.4, x3 = 188, x5 = 6.6, and x6 = 10. Use these data to predict x4 (advertising costs) and find an 80% confidence interval for your prediction. (Use 2 decimal places.) prediction lower limit upper limit At the 80% confidence level, what range of advertising costs do you think is appropriate for this store? (Round to nearest integer.) lower limit $ upper limit $
A researcher surveys 50 individuals in Smithville and 40 in Amherst, finding that 30% of smithville residents moved last year, while only 22% of Amherst residents did. Is there enough evidence to conclude that mobility rates in the two communities differ? Use a two-tailored alternative, and a=0.10. Again, find the p-value and a 90% confidence interval for the difference in proportions
In a trial of 275 patients who received 10-mg doses of a drug daily, 44 reported headache as a side effect. From the information above, construct and interpret a 95% confidence interval or the population proportion of patients who received the drug and report a headache as a side effect. One can be ____ % confident that the proportion of patients who receive the drug and report a headache as a side effect is between ___ and ___ .
In a study of the accuracy of fast food drive-through orders, restaurant a had 346 accurate orders and 61 that were not accurate. Construct a 90% confidence interval estimate of the percentage of orders that are not accurate. Compare the results from part A to this 90% confidence interval for the percentage of orders that are not accurate at restaurant B: 0.132< P < 0.189. What do you conclude?
According to a report on sleep deprivation by the Centers for Disease Control and Prevention, the proportion of California residents who reported insufficient rest or sleep during each of the preceding 30 days is 8.0%, while this proportion is 8.8% for Oregon residents. These data are based on simple random samples of 11,545 California and 4,691 Oregon residents. Calculate a 95% confidence interval for the difference between the proportions of Californians and Oregonians who are sleep deprived and interpret it in context of the data. Also, conduct a hypothesis, with alpha = 0.05, test to determine if these data provide strong evidence the rate of sleep deprivation is different for the two states. Construct a 95% confidence interval Which of the following correctly displays the standard error? 0.0046 0.009 0.001 0.00002 0.008
A healthcare information company is interested in estimating the average charge for a standard patient visit to a chiropractor in Maryland, after applying the discount negotiated with a large HMO plan. Data is collected from 16 randomly selected chiropractic practices in Maryland, and the following are some summary statistics:- Mean charge: 25.50 USD- SD of charges: 2.10 USDAssuming the charge data is normally distributed for all chiropractic practices in Maryland, estimate a 95% confidence interval for the mean amount charged by Maryland chiropractors (t (15, 0.05) = 2.13):
According to a report on sleep deprivation by the Centers for Disease Control and Prevention, the proportion of California residents who reported insufficient rest or sleep during each of the preceding 30 days is 8.0%, while this percent is 8.8% for Oregon residents. These data are based on simple random samples of 11,545 California and 4,691 Oregon residents. Suppose we want to calculate a 99% confidence interval for the difference between the proportions of Californians and Oregonians who are sleepdeprived. 1. State the critical value for the interval (3 decimal places): 2. State the difference between the sample proportions, POregon-PCalifornia (3 decimal places): 3. State the standard error of the difference between the proportions (3 decimal places): 4. State the confidence interval being sure to use proper notation (3 decimal places each):
A researcher finds that individuals with a high school education or less are more likely to have hypertension than individuals with at least some college education. They report a relative risk of 2.1 with a 95% confidence interval or (1.6, 2.5). They then stratify their analysis by smoking status and report a relative risk of hypertension in smokers with a high school education or less as compared to smokers with at least some college education of 1.4 with a 95% confidence interval of (0.9, 1.9), and a relative risk of hypertension in non-smokers with a high school education or less as compared to non-smokers with at least some college education of 2.6 with a 95% confidence interval of (2.1, 3.2), Which of the following is not true? A) there is effect modification by smoking status B) there is effect modification by hypertension status C) there is no statistically significant assocation between education and hypertension amoung smokers D) Education and hypertension are not…
It is believed that large doses of acetaminophen (the active ingredient in over the counter pain relievers like Tylenol) may cause damage to the liver. A researcher wants to conduct a study to estimate the proportion of acetaminophen users who have liver damage. For participating in this study, he will pay each subject $20 and provide a free medical consultation if the patient has liver damage. a) If he wants to limit the margin of error of his 95% confidence interval to 1%, what is the minimum amount of money he needs to set aside to pay his subjects?
In a controlled and scientifically valid study of sleeping pills, 40 people who took Pill A averaged 7.5 hours of sleep, with a standard deviation of 1.5 hours, while 40 people who took Pill B averaged 8 hours of sleep, with a standard deviation of 2 hours. Is there enough evidence to support the assertion that Pill B is more effective than Pill A to a 95% confidence interval? Explain.
5. A phase three clinical trial for a new medication proposed to treat depression has 1191 participants. If all these individuals are from the group of individuals who were given the medication and 759 individuals who received the medication experienced improvement in their symptoms of depression, construct a 90% confidence interval for the true proportion of individuals on the medication who experienced improvement in their symptoms of depression.A) (.363, .417)B) (.335, .363)C) (.623, .651)D) (.614, .660)

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