DA small amount of the trace element selenium, 50-200 micrograms (ug) per day, is considered essential to good health. Suppose that random samples of50 adults were selected from two regions of the United States and that each person's daily selenium intake was recorded. The means and standarddeviations of the selenium daily intakes for the two groups are shown in the table.Sample Mean (ug)Sample Standard Deviation (ug)Find a 95% confidence interval for the difference (₁-₂) in the mean selenium intakes for the two regions. (Round your answers to three decimalplaces.)49RegionPhase One-Tw_docx1167.422.8ug toInterpret this interval,O The probability that the difference in the average daily selenium intake for the populations in Region 1 and Region 2 is exactly in the middle of theInterval is 95%.A2140.517.3O The probability that the difference in the average daily selenium intake for the populations in Region 1 and Region 2 is within the interval is95%.We are 95% confident that the difference in the average daily selenium intake for the populations in Region 1 and Region 2 is within the interval.O We are 95% confident that the difference in the average dally selenium intake for the samples taken from Region 1 and Region 2 is within theinterval.O The daily selenium intake is within the interval for 95% of the populations in Region 1 and Region 2.howtolose(1).docphase3eval.docPhasethreefall20.docShow AllScre2-14Scre2-10scree2-10Screes2-10.

The provided information is as follows:The sample size for the sample 1 is .The sample mean for the…

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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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A sample of 20 items provides a sample standard deviation of 4.
A coin-operated drink machine was designed to discharge a mean of 7 fluid ounces of coffee per cup. In a test of the machine, the discharge amounts in 16 randomly chosen cups of coffee from the machine were recorded. The sample mean and sample standard deviation were 6.86 fluid ounces and 0.2 fluid ounces, respectively. If we assume that the discharge amounts are approximately normally distributed, is there enough evidence, to conclude that the population mean discharge, u, differs from 7 fluid ounces? Use the 0.05 level of significance. Perform a two-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.) (a) State the null hypothesis H, and the alternative hypothesis H1. p H, :0 H, :0 (b) Determine the type of test statistic to use. (Choose one) ▼ D=0 OSO (c) Find the value of the test statistic. (Round to three or more decimal places.) (d) Find the two critical values. (Round to…
The concentration of hexane (a common solvent) was measured in units of micrograms per liter for a simple random sample of twelve specimens of untreated ground water taken near a municipal landfill. The sample mean was 720.2 with a sample standard deviation of 8.8. Eleven specimens of treated ground water had an average hexane concentration of 695.1 with a standard deviation of 9.1.It is reasonable to assume that both samples come from populations that are approximately normal. Construct a 90% confidence interval for the reduction of hexane concentration after treatment. Interpret the results.
An ecologist is studying the impact of local polluted waters on the growth of alligators. The length of adult male alligators typically follows a normal distribution with a standard deviation of 2 feet. The ecologist wants to estimate the mean length of this population of alligators. Suppose she samples n alligators at random and uses the sample mean, X to as an estimator for u. a. What is the bias and variance of the estimator? (Note, these may be a function of n.) b. If n = 4, what is the probability that the estimator is within one foot of the true mean? (I.e. find P(|X – µ| < 1). c. What sample size, n, is required for the estimator to be within one foot of the true mean with 95% probability? (I.e. find the value of n that satisfies P(|X – µ| < 1) = 0.95.) d. Suppose the ecologist ends up sampling n = 9 alligators and calculates a sample mean of ī = 10.4 feet. Construct a 95% confidence interval for the population mean. e. Give an interpretation for the interval obtained in (d).
Ex. Derive the MLE for a random sample from Exp(A) distribution that is subject to left truncation and right-censoring (LTRC); be sure to provide its variance estimate as well. Compare to the case where there is only right-censoring that we derived in the lecture notes earlier.
Table 1: Summary statistics of the extra sodium content (%) in the thirty samples Min. : 0.14 1st Qu.: 1.17 Median : 2.13 Mean : 2.4813 3rd Qu.: 3.86 Max. : 5.22 SD = 1.6156 Researchers believe that the average percentage of sodium content is 3%. Test the appropriate hypotheses using a 5% level of significance. Specify the following: random variable of interest parameter of interest (Symbol for the parameter) statistic you would use to estimate parameter
Suppose, household color TVs are replaced at an average age of ? = 7.8 years after purchase, and the (95% of data) range was from 3.8 to 11.8 years. Thus, the range was 11.8 − 3.8 = 8.0 years. Let x be the age (in years) at which a color TV is replaced. Assume that x has a distribution that is approximately normal. (a) The empirical rule indicates that for a symmetric and bell-shaped distribution, approximately 95% of the data lies within two standard deviations of the mean. Therefore, a 95% range of data values extending from ? − 2? to ? + 2? is often used for "commonly occurring" data values. Note that the interval from ? − 2? to ? + 2? is 4? in length. This leads to a "rule of thumb" for estimating the standard deviation from a 95% range of data values. Estimating the standard deviationFor a symmetric, bell-shaped distribution, standard deviation ≈ range 4 ≈ high value − low value 4 where it is estimated that about 95% of the commonly occurring data…
Question 6 of 6 - /2 E View Policies Current Attempt in Progress Fast Auto Service provides oil and lube service for cars. It is known that the mean time taken for oil and lube service at this garage is 15 minutes per car and the standard deviation is 2.4 minutes. The management wants to promote the business by guaranteeing a maximum waiting time for its customers. If a customer's car is not serviced within that period, the customer will receive a 50 % discount on the charges. The company wants to limit this discount to at most 4% of the customers. What should the maximum guaranteed waiting time be? Assume that the times taken for oil and lube service for all cars have a normal distribution. Round your answer to the nearest minute. The maximum guaranteed waiting time should be approximately i minutes. eTextbook and Media Save for Later Attempts: 0 of 5 used Submit Answer 50 e Type here to search
In a random sample of 75 eighth grade students scores on a national mathematics assessment test has a mean score of 266. The test result prompts a state school administration to declare that the mean score for the stated eighth graders on this exam is more than 260. Assume the population standard deviation is 34. At a=0.09, is there enough evidence to support the administrations claim? Write the claim mathematically and identify Ho and Ha Find the standardized test statistic z, and its corresponding area
Suppose, household color TVs are replaced at an average age of μ = 9.0 years after purchase, and the (95% of data) range was from 5.8 to 12.2 years. Thus, the range was 12.2 − 5.8 = 6.4 years. Let x be the age (in years) at which a color TV is replaced. Assume that x has a distribution that is approximately normal. (a) The empirical rule indicates that for a symmetric and bell-shaped distribution, approximately 95% of the data lies within two standard deviations of the mean. Therefore, a 95% range of data values extending from μ − 2σ to μ + 2σ is often used for "commonly occurring" data values. Note that the interval from μ − 2σ to μ + 2σ is 4σ in length. This leads to a "rule of thumb" for estimating the standard deviation from a 95% range of data values. Estimating the standard deviationFor a symmetric, bell-shaped distribution, standard deviation ≈ range 4 ≈ high value − low value 4 where it is estimated that about 95% of the commonly occurring data…
A simple random sample of 25 has been collected from a normally distributed population for which it is known the standard deviation is 17. The sample mean has been calculated as 342 and the sample standard deviation of 14.9. Construct a 95% confidential interval for the population mean
A sample of 10 from population 1 has a mean of 32 and a standard deviation of 6. A sample of 13 from population 2 has a mean of 48 and a standard deviation of 8. The population standard deviations are unknown but assumed to be equal. Test the hypothesis, at the 1% level of significance, that the population means are the same. State your conclusion as a sentence and include the test statistic and the critical value.

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1 2 Sample Mean (ug) 167.4 140.5 Sample Standard Deviation (pg) 22.8 17.3 5% confidence interval for the difference (₁-₂) in the mean selenium intakes for the tw pa Pg to at this interval, probability that the difference in the average daily selenium intake for the populations in Re