Question

A sample of 12 joint specimens of a particular type gave a sample mean proportional limit stress of 8.55 MPa and a sample standard deviation of 0.76 MPa. (a) Calculate and interpret a 95% lower confidence bound for the true average proportional limit stress of all such joints. (Round your answer to two decimal places.) MPa

Answer 1

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)

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

`(x^(-) - t_{(\alpha )/(2) } (S)/(√(n) ) , x^(-) + t_{(\alpha )/(2) } (S)/(√(n) ))`

Degrees of freedom = n-1 = 12-1 = 11

t₀.₀₂₅ = 3.4966

`(8.55 - 3.4966(0.76)/(√(12) ) , 8.55 + 3.4966 (0.76)/(√(12) ))`

(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)