MATLAB: An Introduction with Applications
6th Edition
ISBN: 9781119256830
Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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- 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 *1 = 75.8 x2 = 76.2 s3 = 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…arrow_forwardPart 1. .If b = 0.523, then for every 1 unit increase in the independent variable, there is a 0.523 unit increase in the dependent variable. True or false? 2. .If b = .704, then 70.4% of the variance in the dependent variable can be explained by the variance in the independent variable. In contrast, 29.6% of the variance in the dependent variable can be explained by outside factors. True or false? 3. .If r = 0.444, then for every 1 unit increase in the independent variable, there is a 0.444 unit increase in the dependent variable. True or false?arrow_forwardOnce the correlation coefficient is known, how can we find the amount of the shared variance?arrow_forward
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