A sample of 15 joint specimens of a particular type gave a sample mean proportio

Question

A sample of 15 joint specimens of a particular type gave a sample mean proportional limit stress of 8.47 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. 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 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. 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. 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 chi-square distributed population. 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. (b) Calculate and interpret a 95% lower prediction bound for proportional limit stress of a single joint of this type. MPa Interpret this bound. 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. If this bound is calculated for sample after sample, in the long run 95% of these bounds will provide a lower 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 higher bound the corresponding future values of the proportional limit stress of a single joint of this type. You may need to use the appropriate table in the Appendix of Tables to answer this question.

Explanation / Answer

a)

sample mean =8.47

sample standard devaition=0.73

significan level alpha = 1-0.95 =0.05

df=15-1 =14

n=15

t critical value at df=14 and alpha =0.05 is 1.76131014

margin of error = t*s/sqrt(n)

=0.331980875

95% lower bound = x-e

=8.47-0.331980875

=8.138019124

b)

mean(x)-(t*s*sqrt(1+(1/n))

me=(t*s*sqrt(1+(1/n)

=1.327923502

lower bound =x-Me

=8.47-1.327923502

=7.142076498

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.