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

Answer:

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

$t = \frac{x^{-} - mean}{\frac{S}{\sqrt{n-1} } }$

$t = \frac{5311.4 - 500}{\frac{220.7}{\sqrt{15-1} } }$

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

$(x^{-} - t_{0.05} \frac{S}{\sqrt{n} } ,x^{-} + t_{0.05}\frac{S}{\sqrt{n} } )$

substitute values and simplification , we get

$(5311.4 - 1.761 \frac{220.7}{\sqrt{15} } ,5311.4 +1.761\frac{220.7}{\sqrt{15} } )$

(5211.05 , 5411.7)

95% lower confidence bound on the mean.