Question

2. The service life of a battery used in a cardiac pacemaker is assumed to be normally distributed. A random sample of ten batteries is subjected to an accelerated life test by running them continuously at an elevated temperature until failure, and the follow­ ing lifetimes (in hours) are obtained: 25.5, 26.1, 26.8, 23.2, 24.2, 28.4, 25.0, 27.8, 27.3, and 25.7. a. The manufacturer wants to be certain that the mean battery life exceeds 25 h. What conclusions can be drawn from these data (use a= 0.05)? b. Construct a 90% two-sided confidence interval on mean life in the accelerated test.

Answer 1

Kindly check explanation

Explanation:

Assuming a normal distribution :

Sample (x) :25.5, 26.1, 26.8, 23.2, 24.2, 28.4, 25.0, 27.8, 27.3, 25.7

Null hypothesis : μ = 25

Alternative hypothesis : μ > 25

α = 0.05

Decision region :

Reject Null if :

t0 > t

t can be obtained from the t distribution table at : df = (n - 1) = 10 - 1 = 9 ; α = 0.05

Hence, t(0.05, 9) = 1.833

From the sample given ; t0

t0 = (m - μ) / (s /√n)

n = sample size = 10

s = sample standard deviation

m = sample mean

Using calculator :

25.5, 26.1, 26.8, 23.2, 24.2, 28.4, 25.0, 27.8, 27.3, 25.7

Sample mean(m) = 26

Sample standard deviation (s) = 1.624

t0 = (26 - 25) / (1.624 /√10) = 1.947

1.947 > 1.833

Hence, reject Null

B)

Confidence interval at 90% ; α = 0.1

m ± t(0.1/2 ; 9) * s/√n

m - t * s/√n ≤ μ ≤ m + t * s/√n

26 - 1.833 * (1.624/√10) ≤ μ ≤ 26 + 1.833 * (1.624/√10)

25.059 ≤ μ ≤ 26.941