Answer:
95% of the lower confidence interval for the true average proportional limit stress of all such joints
7.7829
95% of the confidence interval for the true average proportional limit stress of all such joints
(7.7829, 9.3171)
Step-by-step explanation:
Step(i):-
Given that the sample size 'n' = 12
Mean of the sample = 8.55
The standard deviation of the sample (S) = 0.76
Step(ii):-
95% of the confidence interval is determined by
[tex](x^{-} - t_{\frac{\alpha }{2} } \frac{S}{\sqrt{n} } , x^{-} + t_{\frac{\alpha }{2} } \frac{S}{\sqrt{n} })[/tex]
Degrees of freedom = n-1 = 12-1 = 11
t₀.₀₂₅ = 3.4966
[tex](8.55 - 3.4966\frac{0.76}{\sqrt{12} } , 8.55 + 3.4966 \frac{0.76}{\sqrt{12} })[/tex]
(8.55 - 0.7671 , 8.55+0.7671)
(7.7829, 9.3171)
Final answer:-
95% of the confidence interval for the true average proportional limit stress of all such joints
(7.7829, 9.3171)
