i. Losses on an automobile liability coverage follow a lognormal distribution with parameters and o = 3.μlii. Thirty-six (n = 36) losses have been observed.iii.ln x; = 108Estimate the lower bound of the 95% normal confidence interval of u.Possible AnswersABCDLess than 2.1At least 2.1, but less than 2.3At least 2.3, but less than 2.5At least 2.5, but less than 2.7E At least 2.74

i. Losses on automobile liability coverage follow a lognormal distribution with parameters μ and…

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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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If the random sample X1,..., X, is taken from a normal distribution with mean value u and standard deviation o, then regardless of the sample size n, the sample mean X is distributed with expected value and standard deviation If a confidence level of 95% is used to construct a confidence interval for the mean u of a normal distribution when the value of standard deviation o is known, the z critical value is Standard error of an estimator ô is the of ê. error consists of rejecting the null hypothesis H, when H, is true. The area under a chi-squared curve with 10 degrees of freedom, which is captured between the two critical values X/2 and x-a/2 is
A weight loss program claims that program participants have a mean weight loss of at least 12 pounds after 1 month. You work for a medical association and are asked to test this claim. A random sample of 30 program participants and their weight losses (in pounds) after 1 month is listed in the table below. Assume the population standard deviation 4. At α = 0.03, do you have enough evidence to reject the program's claim? Complete parts (a) through (e). 5.1 5.2 6.5 6.9 7.2 8.7 9.2 9.4 9.5 9.7 11.1 11.2 11.6 11.6 11.9 (a) Identify Ho and H₂. Choose the correct answer below. O A. Ho: μ ≥ 12 H₂: 12 7.3 7.4 9.7 10.1 12.4 12.5 O E. Ho: μ 12 H₂:μ≤ 12 O D. Ho: μ#12 H₂:μ = 12 OF. Ho: H=12 H₂:μ#12 8.5 10.6 15.2
Q1. Using 625 trading days of data, you estimated the daily log return follows a normal distribution with a mean of 5 bps and and a stdev of 125 bps. Q1b. what is the 90, 95, and 99% confidence interval for your mean return estimate?
H. Let X1,..., X, be a random sample from some distribution with unknown u. It's known that X 100 and X = 1783. %3D %3D (40) Supposen = 144 and o? = 12. The value of a, corresponding to the confidence interval [0.7, 1.3] is (a) 0.212 (b) 0.122 (c) 0.050 (d) 0.298
4. A 95% confidence inter val for a population mean µ, based on a sample of size 49, is (12.5, 20.1). (a) The sample mean, , is (i) 3.8. (ii) 16.3. (iii) 7.6. (iv) 32.6. (v) 1.96. (b) The maximum error for this confidence interval, E, is (i) 3.8. (ii) 16.3. (iii) 7.6. (iv) 32.6. (v) 1.96. (c) Compared to the 95% confidence interval, (12.5, 20.1), a 90% confidence interval based on the same sample will be (i) wider. (ii) narrower. (iii) of the same width.
The minimum sample size needed to estimate a population mean within 2 units with a 95% confidence when the population standard deviation equals 8 is A. 62 OB. 9 C. 8 O D. 61 A random sample of 64 observations has a mean of 30. The population variance is assumed to be 9. The 85.3% confidence interval estimate for the population mean (to the third decimal place) is O A. (28.369, 31.631) O B. (29.383, 30.617) O C. (29.456, 30.544) O D. (28.560, 31.440)
e this K 23-2....xlsx ^ Assume that you have a sample of n₁ = 7, with the sample mean X₁ = 43, and a sample standard deviation of S₁ = 5, and you have an independent sample of n₂ = 16 from another population with a sample mean of X₂ = 31, and the sample standard deviation S₂ = 8. Construct a 90% confidence interval estimate of the population mean difference between ₁ and μ2. Assume that the two population variances are equal. ₁-₂ (Round to two decimal places as needed.) View an example Get more help. data-2_13_2023-2....xlsx ^ ... (0,0) Clear all More- Check answer Show all X ►
Suppose that a random sample X₁, X2, X20 follows an exponential distribution with parameter B. Check whether or not a pivotal quantity exixts, if it exists, find a 100(1 - α)% confidence interval for B. www.
Suppose, household color TVs are replaced at an average age of μ = 8.2 years after purchase, and the (95% of data) range was from 6.2 to 10.2 years. Thus, the range was 10.2 – 6.2 = 4.0 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 symmetrical 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 values…
E. Non-parametric statistics does not use the values of μμ or σσ . True False F. When data points are widely spaced around the regression line, the coefficient of correlation is close to 1. True False G. If the value of the y intercept of the regression line is positive, then the coefficient of correlation could be negative. True False H. ANOVA can tell which mean is different. True False I. If a confidence interval for the difference of 2 proportions is found and the interval contains 0, it suggests the proportions are the same. True False
For a one-tailed test (lower tail) with n = 10 at 90% confidence (i.e. α = 0.10), the appropriate tα (critical value) is a. -1.383 b. 2.821 c. 1.383 d. -2.821
Independent samples of size n1 = 25 and n2 = 36 are taken from two normal populations with knownstandard deviations of σ1 = 5.5 and σ2 = 4.2. e sample means are x¯1 = 13.6 and x¯2 = 19.2. Find a95% confidence interval for µ1 − µ2.

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