A manufacturer produces piston rings for anautomobile engine. It is known that ring diameter isnormally distributed with standard deviation 0.001millimeters. A random sample of 35 rings has amean diameter of 74.036 millimeters.a. Construct a 99% two sided confidence interval onthe mean piston ring diameter.b. How many piston rings would have to bemeasured for the manufacturer to be 95% confidentthat the mean diameter from the sample is between74.0359 to 74.0361 of true population mean(upload the picture of solution and roundoffyour final answer into 5 decimal places)

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Answered: A manufacturer produces piston rings…

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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selected parts. The sample standard deviation is s =0.37. Construct a 95% two-sided confidence interval for the standard deviation 5. The operations manager of a large production plant would like to estimate the mean amount of time a worker takes to assemble a new electronic component. Assume that the standard deviation of this assembly time is 3.6 minutes. a. After observing 120 workers assembling similar devices, the manager noticed that their average time was 16.2 minutes. Construct a 90% confidence interval for the mean assembly time. b. How many workers should be involved in this study in order to have the mean assembly time estimated up to + 15 seconds with 90% confidence? 6. A new rocket-launching system is being considered for deployment of small, short-range rockets. The existing system has p = 0.8 as the probability of a successful launch. A sample of 40 experimental launches is made with the new system, and 34 are successful. Using 95% O Focus English (United States) search…
Health insurers and the federal government are both putting pressure on hospitals to shorten the average length of stay (LOS) of their patients. The average LOS in the United States is 4.6 days (Healthcare Cost and Utilization Project Statistical Brief, December 2018). A random sample of 22 hospitals in one state had a mean LOS of 3.6 days and a standard deviation of 2.1 days. Assume that the sampled population is approximately normal. Complete parts a through c. Question content area bottom Part 1 a. Use a 99% confidence interval to estimate the population mean LOS for the state's hospitals. enter your response here, enter your response here (Round to three decimal places as needed.)
Engineers must consider the breadths of male heads when designing helmets. The company researchers have determined that the population of potential clientele have head breadths that are normally distributed with a mean of 5.7-in and a standard deviation of 1.1-in.In what range would you expect to find the middle 50% of most head breadths? Between and .If you were to draw samples of size 46 from this population, in what range would you expect to find the middle 50% of most averages for the breadths of male heads in the sample? Between and .Enter your answers as numbers. Your answers should be accurate to 2 decimal places.
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Assume that a sample is used to estimate a population mean μ. Find the margin of error M.E. that corresponds to a sample of size 22 with a mean of 62.5 and a standard deviation of 9.6 at a confidence level of 80%.Report ME accurate to one decimal place because the sample statistics are presented with this accuracy. M.E. =
A data set includes 106 body temperatures of healthy adult humans having a mean of 98.9°F and a standard deviation of 0.63°F. Construct a 99% confidence interval estimate of the mean body temperature of all healthy humans. What does the sample suggest about the use of 98.6°F as the mean body temperature? Click here to view a t distribution table. LOADING... Click here to view page 1 of the standard normal distribution table. LOADING... Click here to view page 2 of the standard normal distribution table. LOADING... Question content area bottom Part 1 What is the confidence interval estimate of the population mean μ? enter your response here°F<μ<enter your response here°F (Round to three decimal places as needed.)
C. To assess the actual mean weight of apples, a farm manager selected a random sample of 75 apples. He found that the sample mean weight was 126.2 grams with standard deviation 8.7 grams. With what degree of confidence can we say that the confidence limits are (126.2±2.57) grams?
A medical researcher wishes to estimate the mean systolic blood pressure of heart surgery patients the day following surgery. She desires a margin of error of 2.2 mmHg. Past studies suggest that a population standard deviation of 41 mmHg is reasonable. Estimate the minimum sample size needed to estimate the population mean with the stated accuracy. OA. 1,389 OB. 37 OC. 695 OD. 3,056 $ 10 % 9: 17 [7] hp C 8 BICCE 9 ho ) fi num Ik + insert Next 5:04 PM 10/13/2022 prt scr backspace delete
In a random sample of ten people, the mean driving distance to work was 25.5 miles and the standard deviation was 7.3 miles. Assume the population is normally distributed and use the t-distribution to find the margin of error and construct a 90% confidence interval for the population mean μ. Interpret the results. Identify the margin of error. (Round to one decimal place as needed.) Construct a 90% confidence interval for the population mean. (...) (Round to one decimal place as needed.) Interpret the results. Select the correct choice below and fill in the answer box to complete your choice. (Type an integer or a decimal. Do not round.) OA. % of all random samples of ten people from the population will have a mean driving distance to work (in miles) that is between the interval's endpoints. OB. It can be said that % of the population has a driving distance to work (in miles) that is between the interval's endpoints. OC. With % confidence, it can be said that most driving distances to…
The teacher in a kindergarten school measures the height of her students. If the heights were normally distributed with the mean of 112 cm and a standard deviation 3.25 cm, find the heights x that corresponds to the following z-scores: z = 2.24 z = -1.52 z = 0.76

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