Respuesta :
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
[tex]t = \frac{x^{-} - mean}{\frac{S}{\sqrt{n-1} } }[/tex]
[tex]t = \frac{5311.4 - 500}{\frac{220.7}{\sqrt{15-1} } }[/tex]
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
[tex](x^{-} - t_{0.05} \frac{S}{\sqrt{n} } ,x^{-} + t_{0.05}\frac{S}{\sqrt{n} } )[/tex]
substitute values and simplification , we get
[tex](5311.4 - 1.761 \frac{220.7}{\sqrt{15} } ,5311.4 +1.761\frac{220.7}{\sqrt{15} } )[/tex]
(5211.05 , 5411.7)
95% lower confidence bound on the mean.
