Answered: Defective alternator and corroded or…

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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NEED HELP PLS ANSWER IN 20 MINS CI on Variance Defective alternator and corroded or loose battery charges are a few of the main causes of battery degradation. A company producing automobile batteries is developing a prototype on their next product and that they claim their batteries will last on average, 3 years with a variance of 1 year. A random sample of 5 prototype batteries were taken and in the course of the company's product development, have measured their lifetimes as 1.9, 2.4, 3.0, 3.5, and 4.2 years. Construct a 95% confidence interval for o? and decide if the company's claim that 6? = 1 is valid. Assume the population of battery lives to be approximately normally distributed.

CI on Variance
Defective alternator and corroded or loose battery charges are a few of the main causes of battery
degradation. A company producing automobile batteries is developing a prototype on their next product
and that they claim their batteries will last on average, 3 years with a variance of 1 year. A random sample
of 5 prototype batteries were taken and in the course of the company's product development, have
measured their lifetimes as 1.9, 2.4, 3.0, 3.5, and 4.2
years.
Construct a 95% confidence interval for o and decide if the company's claim that o = 1 is valid. Assume
the population of battery lives to be approximately normally distributed.
O a. 0.29 < o2 < 6.74. 95% confidence interval for o2 is (0.29, 6.74)
O b. 0.34 < o2 < 4.59. Because 95% confidence interval for o is (0.34, 4.59), and the s2 = 0.815
is in that interval, we can conclude that the company's claim is valid
O c. 0.539 <o< 2.596. Because 95% confidence interval for o is (0.539, 2.596), and 1 is in that
interval, we can conclude that the company's claim that o =1 is valid
O d. 0.29 < o² < 6.74. Because 95% confidence interval for o is (0.29, 6.74), and 1 is in that
interval, we can conclude that the company's claim that o2 =1 is valid

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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