(See Section 7.3.) A sample of 18 joint specimens of a particular type gave a sa
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Question
(See Section 7.3.) A sample of 18 joint specimens of a particular type gave a sample mean proportional limit stress of 8.45 MPa and a sample standard deviation of 0.74 MPa. 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.) 8.42 X 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 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 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. You may need to use the appropriate table in the Appendix of Tables (Table A.5) to answer this question.Explanation / Answer
Mean is 8.45 and s is 0.74
for sample size of 18, the standard error SE is s/sqrt(N)=0.74/sqrt(18)=0.1744
thus lower bound is mean-SE*z= 8.45-0.1744*0.74 =8.32
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