Question

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.

Answer 1

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.

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 = (x^(-) - mean)/((S)/(√(n-1) ) )`

`t = (5311.4 - 500)/((220.7)/(√(15-1) ) )`

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

`(x^(-) - t_(0.05) (S)/(√(n) ) ,x^(-) + t_(0.05)(S)/(√(n) ) )`

substitute values and simplification , we get

`(5311.4 - 1.761 (220.7)/(√(15) ) ,5311.4 +1.761(220.7)/(√(15) ) )`

(5211.05 , 5411.7)

95% lower confidence bound on the mean.