The life in hours of a biomedical device under development in the laboratory is known to be approximately normally distributed. A random sample of 15 devices is selected and found to have an average life of 5625.1 hours and a sample standard deviation of 226.1 hours. (a) Test the hypothesis that the true mean life of a biomedical device is greater than 5500 using the P-value approach.
(b) Construct a 95% lower confidence bound on the mean.
(c) Use the confidence bound found in part (b) to test the hypothesis

Kindly check explanation

Step-by-step explanation:

H0 : μ = 5500

H1 : μ > 5500

The test statistic assume normal distribution :

Test statistic :

(Xbar - μ) ÷ s/sqrt(n)

(5625.1 - 5500) ÷ 226.1/sqrt(15) = 2.1429 = 2.143

Pvalue from test statistic :

The Pvalue obtained using the calculator at df = 15 - 1 = 14 is 0.025083

α = 0.05

Since ;

Pvalue < α

0.025083 < 0.05 ; Reject H0

The confidence interval :

Xbar ± Tcritical * s/sqrt(n)

Tcritical at 95% = 1.761 ;

margin of error = 1.761 * 226.1/sqrt(15) = 102.805

Lower boundary : (5625.1 - 102.805) = 5522.295

(5522.295 ; ∞)

The hypothesized mean does not occur within the constructed confidence boundary. HENCE, there is significant eveidbce to support the claim that the true mean life of a biomedical device is greater than 5500

irrational, cannot

the 8 in 8,420 is in the thousands place so it's value is 8 thousand.

the 8 in 6,870 is in the hundreds place so it's value is 8 hundred.

divide 8000 by 800 to find the answer:

8000 / 800 = 10

the 8 in 8,420 is 10 times greater than the 8 in 6,870