Statement: A battery manufacturer guarantees that his batteries will last, on av

Question

Statement: A battery manufacturer guarantees that his batteries will last, on aver upon a year. His claims are based u age, 3 years, with a standard deviation of 1 very large population of his 'good' batteries. A consumer watch group decides to test his guarantee. Their small sample of his 'good batteries indicates bat- tery lifetimes (in years) of 1.9, 2.4, 3.0, 3.5, and 4.2. Determine (a) the percent confidence that the difference between the watch group's sample variance and manufacturer's true variance is due solely to random effects. Next, based upon the manufacturer's battery population average life time and standard deviation (b) determine the probabilities that a battery lifetime will be less than 1.9 years and (c) between 3 and 4 years. Below is breif solution with correct final answers for referance, please provide detailed solution step by step arriving at the same answers Solution: (a) First, estimate the sample variance: z = 3.0 S = 0.815. So = 3.26. For v 4, this gives = 51.4 %. (b)21.9 years-(1.9-3)/1=-1.1. Using the z-table, the one-sided area is 0.3643. (e) 24 years-(4-3)/1 = 1, which gives a one-sided area of 0.3413, 23 years Thus, Pr|s

Explanation / Answer

Part a

We are given a sample as

1.9, 2.4, 3.0, 3.5, 4.2

From this sample, we have

Sample mean = Xbar = 3.0

Sample variance = S2 = 0.815

Sample standard deviation = S = sqrt(0.815) = 0.902774

Sample size = n = 5

Degrees of freedom = n – 1 = 4

Sum of squares = X2 = df*S2 = 4*0.815 = 3.26

Sum of squares follows Chi square distribution. So, using Chi square table with value 3.26 and df = 4,

We get = 0.515295 or 51.5%

Correct Answer = 51.5%

Part b

We are given

Population standard deviation = = 1

Population mean = µ = 3

We have to find P(X<1.9)

Z = (X - µ) /

Z = (1.9 – 3) / 1 = -1.1

P(Z<-1.1) = P(X<1.9) = 0.135666 (By using z-table or excel)

Required probability = 0.135666 or 13.57%

Part c

We have to find P(3<X<4)

P(3<X<4) = P(X<4) – P(X<3)

First we have to find P(X<4)

Z = (X - µ) /

We are given

Population standard deviation = = 1

Population mean = µ = 3

Z = (4 – 3) / 1 = 1

P(Z<1) = P(X<4) = 0.841345 (By using z-table or excel)

Now, we have to find P(X<3)

Z = (3 – 3) / 1 = 0

P(Z<0) = P(X<3) = 0.5000 (By using z-table or excel)

P(3<X<4) = P(X<4) – P(X<3)

P(3<X<4) = 0.841345 - 0.5000

P(3<X<4) =0.341345

Required probability = 0.341345 or 34.13%

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