Question

13. Question Details DevoreStan9 7.E.034. My Note A sample of 13 joint specimens of a particular type gave a sample mean proportional limit stress of 8.47 MPa and a sample standard deviation of 0.79 MPa (a Calculate and interpret a 95% lower confidence bound for the true average proportional mit 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 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 less 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 chi-square distributed population O we do not need to make any assumptions. We must assume that the sample observations were taken from a normally distributed population (b Calculate and interpret a 95% lower prediction bound for proportional mit stress of a single oint of his type. Round your answer to two decima places. MPa Interpret this bound. If this bound is calculated or sample after sample n the long run 95% of these bound will provide a higher bound for the corresponding future values of the proportional mi stress of as type. If this bound is calculated type. e int t is 番 or sample after sample n the long run, 95% of these bounds will provide a lower bound for the corresponding future values of the proportional limit stress of as n le joint of this ● 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. You may need to use the appropriate table in the Appendix of Tables to answer this question

Answer 1

Solution

given, n=13, x̅=8.47, s=0.79

degree of freedom=n-1=13-1=12

a.) Critical value at 95% confidence with degree of freedom 12: (t)=2.179 (use student t table for this)

95% Lower confidence bound is given by:

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