Solve the problems using R 25.4siological parameters in patients with end-stage renal diseasewithheBP0.974.431.6836.50ngISd25.1Baseline (n=102)Meansd0.220.640.4716.0818-month follow-up (n =69)Rimm19.2Mean1.004.491.5723.27sd0.190.710.4013.79bePharmacologySuppose we want to estimate the concentration (ug/mL)of a specific dose of ampicillin in the urine after various pe-riods of time. We recruit 25 volunteers who have receivedampicillin and find they have a mean concentration of 7.0ug/mL with a standard deviation of 2.0 µg/mL. Assumethe underlying population distribution of concentrations isnormally distributed.ModA*6.30 Find a 95% CI for the population mean concentration.*6.31 Find a 99% Cl for the population variance of theconcentrations.*6.32 How large a sample would be needed to ensurethat the length of the CI in Problem 6.30 is 0.5 ug/mLassuming the sample standard deviation remains at2.0 µg/mL?WY S6 S din

Answered: Image /qna-images/answer/ec0630f8-29f4-4e23-ad86-ae27b9fdcffd.jpg

Answered: Pharmacology Suppose we want to…

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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Assume that last year in a particular state with 700000 children there were 1471 children out of 32500 in a random sample who were diagnosed with Autism Spectrum Disorder. Nationally, 1 out of 88 children are diagnosed with ASD. Is there sufficient data to show that the incident rate of ASD is higher in that state than nationally? Test at the 1% level.
Cardiovascular DiseaseSuppose the incidence rate of myocardial infarction (MI)was 5 per 1000 among 45- to 54-year-old men in 2000.To look at changes in incidence over time, 5000 men in thisage group were followed for 1 year starting in 2010. Fifteennew cases of MI were found. 7.14 Of the 15 new MI cases in the preceding study, 5 died within 24 hours. Test whether the 24-hour case-fatality rate changed from 2000 to 2010.
“Passive and Active Smoke” in Appendix B includes cotinine levels measured in a group of nonsmokers exposed to tobacco smoke (n = 40, Mean = 60.58 ng>mL, s = 138.08 ng>mL) and a group of nonsmokers not exposed to tobacco smoke (n = 40, Mean = 16.35 ng>mL, s = 62.53 ng>mL). Cotinine is a metabolite of nicotine, meaning that when nicotine is absorbed by the body, cotinine is produced. Use a 0.05 significance level to test the claim that nonsmokers exposed to tobacco smoke have a higher mean cotinine level than nonsmokers not exposed to tobacco smoke. 1. Construct a confidence interval estimate of the difference betwen the mean continen levels fo the two groups of nonsmokers. What confidence level would be appropriate? 2. Find the margin of error E using the formula. 3. What is the confidence interval? Explain the meaning of the confidence interal and what the limit represents.
I NEED D AND E BE ANSWERED I answered a,b,c (a) First-come-first-serve ANS: Mean flow time =17.4 Mean Process time=5,2 The number of jobs in the system =Mean flow time /Mean Process time=17.4/5.2=3.34 days (b) Earliest due date ANS Mean flow time =12.2 Mean days past due date=0 No. of jobs in the system=61/26=3.24 (c) Shortest processing time ANS Mean flow time =11.2 Mean days past due date=1.2 No. of jobs in system=56/26=2.153 (d) Which rule should be used, in your opinion? Why?(e) The head mechanic at the production facility wants to perform maintenanceon the bottleneck workstation used to process the jobs above. Themechanic insists the maintenance needs to be performed before day 45,and has indicated that the maintenance will take 3 days to complete andwill take the workstation offline. Considering it is still day 13, when isthe best time to perform the maintenance? (Hint: Treat the maintenanceas though it is a new job.)
Cardiovascular DiseaseSuppose the incidence rate of myocardial infarction (MI)was 5 per 1000 among 45- to 54-year-old men in 2000.To look at changes in incidence over time, 5000 men in thisage group were followed for 1 year starting in 2010. Fifteennew cases of MI were found. 7.14 Of the 15 new MI cases in the preceding study,5 died within 24 hours. Test whether the 24-hour casefatality rate changed from 2000 to 2010.7.15 Suppose we eventually plan to accumulate 50 MIcases during the period 2010–2015. Assume that the24-hour case-fatality rate is truly 20% during this period.How much power would such a study have in distinguishingbetween case-fatality rates in 2000 and 2010–2015 if atwo-sided test with significance level .05 is planned?7.16 How large a sample is needed in Problem 7.15 toachieve 90% power?
“Passive and Active Smoke” in Appendix B includes cotinine levels measured in a group of nonsmokers exposed to tobacco smoke (n = 40, Mean = 60.58 ng>mL, s = 138.08 ng>mL) and a group of nonsmokers not exposed to tobacco smoke (n = 40, Mean = 16.35 ng>mL, s = 62.53 ng>mL). Cotinine is a metabolite of nicotine, meaning that when nicotine is absorbed by the body, cotinine is produced. Use a 0.05 significance level to test the claim that nonsmokers exposed to tobacco smoke have a higher mean cotinine level than nonsmokers not exposed to tobacco smoke. Based on your hypothesis tests and confidence intervals, what do you conclude about the effects of second hand smoke? Why?
A survey recently reported that the mean national annual expenditure for inpatient andoutpatient services of all persons over 64 years of age was $5,423 with a standarddeviation of $979. A random sample of 352 persons over age 64 living in Sudbury had anaverage expense of $5,516. We want to test whether the mean inpatient and outpatient expense of all Sudbury residents over age 64 is higher than the national average of $5,423. Calculate the P-value for this test.
Arsenic is a compound that occurs naturally in very low concentrations. Arsenic blood concentrations in healthy individuals are Normally distributed with a mean of 3.2 mg/dl. Some areas of the United States have naturally elevated arsenic levels in the ground and water supplies. Researchers took a sample 20 water samples from a particular area and measured the arsenic levels in each of the 20 water samples. They want to determine if there is sufficient evidence to conclude that the arsenic level in that area is higher than 3.2 mg/dl. hypothesis test for one population mean (unknown population standard deviation) confidence interval estimate for one population mean (unknown population standard deviation) hypothesis test for population mean from paired differences confidence interval estimate for population mean from paired differences hypothesis test for difference in population means from two independent samples confidence interval estimate for difference in population means from two…
Specific Motors of Detroit has developed a new automobile known as the M car. 24 M cars and 28 J cars (from Japan) were road tested to compare miles-per-gallon (mpg) performance. sample 1 (M cars) : n1 = 24, X ̄1 = 29.8, s1 = 2.56 sample 2 (J cars) : n2 = 28, X ̄2 = 27.3, s2 = 1.81 (b) Can we conclude, using a .05 level of significance, that the miles-per-gallon (mpg) performance of M cars is greater than the miles-per-gallon performance of J cars? (d) Find the critical values. (e) Rejection region and conclusion. (f) Interpret the results.
Overproduction of uric acid in the body can be an indication of cell breakdown. This may be an advance indication of illness such as gout, leukemia, or lymphoma. Over a period of months, anadult male patient has taken eight blood tests for uric acid. The mean concentration was ?̅= 5.35mg/dl. The distribution of uric acid in healthy adult males can be assumed to be normal, withσ = 1.85 mg/dl.a. Find a 90% confidence interval for the population mean concentration of uric acid in thispatient’s blood. What is the margin of error?b. Find a 99% confidence interval for the population mean concentration of uric acid in thispatient’s blood. What is the margin of error?c. Compare the margins of error for parts (a) and (b). As confidence levels increase, do marginsof error increase?
Citrus Rental is a popular car rental agency that has a history of having too few cars available, so that its available cars are overdriven. The mean monthly mileage over the years for Citrus cars has been about 1550 miles per month. Recently, though, Citrus purchased thousands of new cars, and the company claims that the average mileage of its cars is now less than in the past. To test this, a random sample of 20 recent mileages of Citrus cars was taken. The mean of these 20 mileages was 1438 miles per month, and the standard deviation was 240 miles per month. Assume that the population of recent monthly mileages of Citrus cars is normally distributed. At the 0.1 level of significance, can it be concluded that the mean recent monthly mileage, μ , of Citrus cars is less than 1550 miles per month? Perform a one-tailed test. Then fill in the table below. Carry your intermediate computations to at least three decimal places and round your answers as…
Citrus Rental is a popular car rental agency that has a history of having too few cars available, so that its available cars are overdriven. The mean monthly mileage over the years for Citrus cars has been about 1550 miles per month. Recently, though, Citrus purchased thousands of new cars, and the company claims that the average mileage of its cars is now less than in the past. To test this, a random sample of 19 recent mileages of Citrus cars was taken. The mean of these 19 mileages was 1450 miles per month, and the standard deviation was 202 miles per month. Assume that the population of recent monthly mileages of Citrus cars is normally distributed. At the 0.05 level of significance, can it be concluded that the mean recent monthly mileage, u, of Citrus cars is less than 1550 miles per month? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places and round your answers as specified in the table. (If necessary,…

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