Suppose that insurance companies did a survey. They randomly surveyed 450 drivers and found that 300 claimed to always buckle up. We are interested in the population proportion of driverswho claim to always buckle up.NOTE: If you are using a Student's t-distribution, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)O Part (a) A Part (d)Construct a 95% confidence interval for the population proportion that claim to always buckle up.(i) State the confidence interval. (Round your answers to four decimal places.)(ii) Sketch the graph.aC.L. =22P'(ii) Calculate the error bound. (Round your answer to four decimal places.)O Part (e)

The sample proportion of driver who claim to buckle up can be calculated as:The critical value from the standard normal distribution at 95% confidence level will be: The 95% confidence interval for the population proportion whole claims to buckle up can be calculated as: The C.L means central line which is the point estimate of the population proportion which is 0.6667 in this case. The confidence level is 95% . Thus , The graph can be completed as:Error bound can be calculated as: Thus, error bound is 0.0436

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Construct a 95% confidence interval for the population proportion that claim to always buckle up. (1) State the confidence interval. (Round your answers to four decimal places.) (ii) Sketch the graph. a C.L. = 2 2 P' (ii) Calculate the error bound. (Round your answer to four decimal places.)

Construct a 95% confidence interval for the population proportion that claim to always buckle up. (1) State the confidence interval. (Round your answers to four decimal places.) (ii) Sketch the graph. a C.L. = 2 2 P' (ii) Calculate the error bound. (Round your answer to four decimal places.)

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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