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 5311.4 hours and a sample standard deviation of 220.7 hours.
a. Test the hypothesis that the true mean life of a biomedical device is greater than 500 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.

a)Null hypothesis:- H₀: μ> 500

  Alternative hypothesis:-H₁ : μ< 500

b) (5211.05 , 5411.7)

95% lower confidence bound on the mean.

c) The test of hypothesis t = 5.826 >1.761 From 't' distribution table at 14 degrees of freedom at 95% level of significance.

Step-by-step explanation:

Step :-1

Given  a random sample of 15 devices is selected in the laboratory.

size of the small sample 'n' = 15

An average life of 5311.4 hours and a sample standard deviation of 220.7 hours.

Average of sample mean (x⁻) =  5311.4 hours

sample standard deviation (S) = 220.7 hours.

Step :- 2

a) Null hypothesis:- H₀: μ> 500

Alternative hypothesis:-H₁ : μ< 500

Level of significance :- α = 0.95 or 0.05

b) The test statistic

t = 5.826

The degrees of freedom γ= n-1 = 15-1 =14

tabulated value t =1.761 From 't' distribution table at 14 degrees of freedom at 95% level of significance.

calculated value t = 5.826 >1.761 From 't' distribution table at 14 degrees of freedom at 95% level of significance.

Null hypothesis is rejected at  95% confidence on the mean.

C) The 95% of confidence limits

substitute values and simplification , we get

(5211.05 , 5411.7)

95% lower confidence bound on the mean.

an

y = 6x -3y -12

y +3y +12 = 6x

=> . x = { 2y / 3 +2 } this is equation of line in terms of x

=> . if line meets the x-axis the co-ordinates of meeting point would be ( x , 0 )

=> . x = 2 ( 0 ) / 3 + 2

=> . x = 2

so, the point is ( 2 , 0 )