A random sample of 15 health maintenance organizations (HMOS) was selected. For each HMO, the co-payment (In dollars) for aEspañoldoctor's office visit was recorded. The results are as follows.7,6, 7, 10, 9, 10, 11, 6, 10, 9, 5, 6, 5, 11, 8Send data to calculatorUnder the assumption that co-payment amounts are normally distributed, find a 99% confidence interval for the mean co-paymentamount in dollars. Then find the lower limit and upper limit of the 99% confidence interval.00Carry your Intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consulta list of formulas.)Lower limit:Upper limit:Check

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Answered: A random sample of 15 health…

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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An internet service provider sampled 540 customers and found that 55 of them experienced an interruption in their service during the previous month. Construct a 95% confidence interval for the proportion of all the customers who have experienced a service interruption. Interpret the results.
n June 2012, a Gallup Poll reported that 1503 out of a total of 2004 U.S. adults said they had a “great deal of confidence” or “quite a lot of confidence” in the military. Assume the conditions for using the CLT are met. PLEASE ONLY DO B,C, D please and thank you (a) Find a 95% confidence interval for the proportion that has a great deal of confidence in the military and interpret this interval. (b) Find an 80% confidence interval and interpret it. (c) Which interval is wider? (d) How would a 90% interval compare with the others in width?
A store randomly samples 603 shoppers over the course of a year and finds that 142 of them made their visit because of a coupon they’d received in the mail. Construct a 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they’d received in the mail.
According to the February 2008 Federal Trade Commission report on consumer fraud and identity theft, 23% of all complaints in 2007 were for identity theft ('Consumer fraud and,' 2008). Assume a state had 373 complaints of identity theft out of 1460 consumer complaints . Calculate a 91% confidence interval for the proportion of all complaints that are complaints of identity theft in that state.
A second-semester junior at a large high school wants to estimate the number of colleges a senior applies to on average before deciding where they will attend. The junior randomly selects 19 seniors and asks them how many colleges they applied to and records the number. Use the data below to construct a 99% confidence interval to estimate the true mean number of colleges seniors at this school apply to before deciding about which college to attend. 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 7, 7, 8, 9 3. Do. Use technology to find the mean and standard deviation of the sample. Round each answer to 2 decimal places. Use those values and t* from the table to calculate the confidence interval. Round the endpoints to 1 decimal place. 99% confidence interval is (, )
In a random sample of 250 high school students, it was found that 85.2% (213) of them later graduated. Find the 95% and 98% confidence interval for the true proportion of all high school students who graduate. Compare the two proportions.
A pharmaceutical company is running tests to see how well its new drug lowers cholesterol. Fourteen adults volunteer to participate in the study. The total cholesterol level of each participant (in mg/dL) is recorded once at the start of the study and then again after three months of taking the drug. The results are given in the following table. Construct a 90% confidence interval for the true mean difference between the cholesterol levels for people who take the new drug. Let Population 1 be the initial cholesterol level and Population 2 be the cholesterol level after three months. Round the endpoints of the interval to one decimal place, if necessary. Total Cholesterol Levels (in mg/dL) Initial Level Level after Three Months 190 192 230 207 206 182 229 199 224 181 195 209 226 214 190 189 224 196 219 206 185 195 208 204 194 198 225 216 Please highlight the answer
Assume a nationwide random survey of 800 adults asked about attitudes toward “alternative medicine” such as acupuncture, massage therapy, and herbal therapy. Among the respondents, 448 said they would use alternative medicine if traditional medicine was not producing the results they wanted. Give a 99% confidence interval for the proportion of all adults who would use alternative medicine.
6. In the past, 70% of a garage's business was with former patrons. Clarence Wade, the owner of the garage, samples 600 repair invoices and finds that for 432 of them, the patron was a repeat customer. a. Develop a 90% confidence interval for the proportion of all current business that is with repeat customers.
A study of 1907 traffic fatalities found that 725 were alcohol related. Suppose you want your 99% confidence interval to be accurate within 1% of the population portion, how many traffic fatalities would need to be in the study?
The ability to find a job after graduation is very important to GSU students as it is to the students at most colleges and universities. Suppose we take a poll (random sample) of 3878 students classified as Juniors and find that 3241 of them believe that they will find a job immediately after graduation. Find the 95% confidence interval for the proportion of GSU Juniors who believe that they will, immediately, be employed after graduation. Use Z = 1.96. Use the unrounded values in Excel to find the answers to #2, #3 and #4. 1. Find the sample proportion of GSU Juniors who believe that they will, immediately, be graduation. 2. The standard error for sample proportion has a value of 3. The margin of error has a value of --
Of 130 adults selected randomly from one town, 30 of them smoke. Construct a 90% confidence interval for the true percentage of all adults in the town that smoke.

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