Question

A sample of 18 joint specimens of a particular type gave a sample mean proportional limit stress of 8.58 MPa and a sample standard deviation of 0.73 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

Interpret this bound.

With 95% confidence, we can say that the value of the true mean proportional limit stress of all such joints is centered around this value.With 95% confidence, we can say that the value of the true mean proportional limit stress of all such joints is less than this value.    With 95% confidence, we can say that the value of the true mean proportional limit stress of all such joints is greater than this value.

What, if any, assumptions did you make about the distribution of proportional limit stress?

We must assume that the sample observations were taken from a uniformly distributed population.We must assume that the sample observations were taken from a normally distributed population.    We do not need to make any assumptions.We must assume that the sample observations were taken from a chi-square distributed population.

(b) Calculate and interpret a 95% lower prediction bound for proportional limit stress of a single joint of this type. (Round your answer to two decimal places.)
MPa

Interpret this bound.

If this bound is calculated for sample after sample, in the long run 95% of these bounds will provide a higher bound for the corresponding future values of the proportional limit stress of a single joint of this type.If this bound is calculated for sample after sample, in the long run, 95% of these bounds will provide a lower bound for the corresponding future values of the proportional limit stress of a single joint of this type.    If this bound is calculated for sample after sample, in the long run 95% of these bounds will be centered around this value for the corresponding future values of the proportional limit stress of a single joint of this type.

Answer 1

solution- briven data:- Sample Size, n=18. límit stress, st = 8.58 Mpa. A sample mean proportional Sariple standard deviation, $=0-73. Mpa. Level of Significance, A=0.05. degree of freedom = n-t=124 = 1) At 95•% confidence interval, tential = 1.796. Margin of vroute, Ez tex GC) = 1-4768 Margin oly ergien, E = 0.3090. cm) hower confidence lesno 5-E 58.98 -0.3090 hower confidence Lesund = 8:27] Cat 95%) With 95% confidence, we can say that the value of the true mean proportional límit stress of all such joints - which is always greater than this valve. I and option is correct] What assumptions, aliol you make about the Allstalleuition of proportional dimet stress? sample oleservations were taken We assume that the

from a normally distributed population. (b) piesele = - taide (av ltcom). 5 = 8.58, thitht atoos, 11 =1.796, n = 1$. 145% = 8:98- 1.746 (073 V 17 ()). = 8.58 - 80796 x 0.75000 3703 = 8:58-4-347006652 -7.23 At 95%, I chower prediction limit = 7-23 | of this bound is caterelated for sample after sample, in the long run, 95% of these bounds will provide a lower bound to the future values of a single point of this type.