We model the time until we catch a fish (in minutes), assuming it is an exponential distribution with an unknown parameter A. Theobserved times until a fish is caught are provided below:3, 3+1, 3+2, , 16.Find the value of the method of moments estimator for the parameter A for this dataset, based on the VARIANCE.For the same data, the method of quantiles estimator for A, based on the median, amounts to

Ans: Let X : Time until catch a fish ( in minutes) 1/ : Average time for to catch a fish X ~…

Answered: We model the time until we catch a fish…

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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2. Let X - tp. Verify the mean and variance formulas. (Hint: Let X = p/2UV-1/2.)
The weights (in pounds) of 6 vehicles and the variability of their braking distances (in feet) when stopping on a dry surface are shown in the table. Can you conclude that there is a significant linear correlation between vehicle weight and variability in braking distance on a dry surface? Use α=0.01. Weight, x 5920 5370 6500 5100 5890 4800 Variability in 1.76 1.91 1.50 braking distance, y EEE Click here to view a table of critical values for Student's t-distribution. Setup the hypothesis for the test. Ho: P Ha: P 0 0 1.86 1.62 1.67 M
J 1
I have asked this question twice and both times it has been wrong. Answer given F=.82 and .77 both these are incorrect. A bakery is considering buying one of two gas ovens. The bakery requires that the temperature remain constant during a baking operation. A study was conducted to measure the variance in temperature of the ovens during the baking process. The variance in temperature before the thermostat restarted the flame for the Monarch oven was 3.3 for 22 measurements. The variance for the Kraft oven was 4 for 25 measurements. Does this information provide sufficient reason to conclude that there is a difference in the variances for the two ovens? Assume measurements are normally distributed and use a 0.02 level of significance.
You are testing the claim that having lights on at night increases weight gain (abstract). A sample of 10 mice lived in an environment with bright light on all of the time and 8 mice who lived in an environment with a normal light/dark cycle is given below. Test the claim using a 10% level of significance. Assume the population variances are unequal and that the weight changes are normally distributed. Give answers to 3 decimal places. Data available at StatKey, choose Mice Wgt Gain-2e data set Light (x₁) 1.71 4.67 4.99 5.33 5.43 6.94 7.15 9.17 10.26 11.67 What are the correct hypotheses? Note this may view better in full screen mode. Select the correct symbols in the order they appear in the problem. Ho: Select an answer ? V Select an answer V Ha: Select an answer ✓ Based on the hypotheses, find the following: Test Statistic = Dark (₂) 2.27 2.53 2.83 4 4.21 4.6 5.95 6.52 p-value = ? V Select an answer V (Hint: difference in means from Ha) The correct decision is to Select an answer…
The table below lists measured amounts of redshift and the distances (billions of light-years) to randomly selected astronomical objects. There is sufficient evidence to support a claim of a linear correlation, so it is reasonable to use the regression equation when making predictions. For the prediction interval, use a 90% confidence level with a redshift of 0.0126. Find the explained variation. Redshift Distance OA, 0.157904 OB. 0.444904 OC. 0.000582 OD. 0.219582 0.0231 0.34 0.0536 0.76 0.0715 0.99 0.0391 0.56 0.0438 0.63 0.0109 0.15
Parts b, c, and d only please.
The durability of the hacksaw is measured in a quality control laboratory. A standard saw needs 2500 cuts. The cutting average of 28 saws selected randomly from a company was found to be 2600 and the variance as 17500. I wonder if the company's production is above the standard? (α = 0.01)
Actuaries use various parameters when evaluating the cost of a life insurance policy. The variance of the life spans of a population is one of the parameters used for the evaluation. Each year, the actuaries at a particular insurance company randomly sample 30 people who died during the year (with the samples chosen independently from year to year) to see whether the variance of life spans has changed. The life span data from this year and from last year are summarized below. Current Last Year Year x1 = 75.8 | x,= 76.2 si= 47.61 s3= = 92.16 (The first row gives the sample means and the second row gives the sample variances.) Assume that life spans are approximately normally distributed for each of the populations of people who died this year and people who died last year. Can we conclude, at the 0.05 significance level, that the variance of the life span for the current year, of, differs from the variance of the life span for last year, o,? Perform a two-tailed test. Then complete the…

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