A sample of 16 joint specimens of a particular type gave a sample mean proportio

Question

A sample of 16 joint specimens of a particular type gave a sample mean proportional limit stress of 8.56 MPa and a sample standard deviation of 0.76 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.) 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 chi-square distributed population. We must assume that the sample observations were taken from a uniformly distributed population. We do not need to make any assumptions. We must assume that the sample observations were taken from a normally distributed population. 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.)

Explanation / Answer

here n=16, mean=8.56, standard deviation=sd=0.76

95% confidence interval for sample mean=mean± z(.025)*sd/sqrt(n)

=8.56±1.96*0.76/sqrt(16)=8.56 ±0.3724=(8.8176,8.9324)

(here z(0.025)=1.96 for two tailed )

the lower limit is 8.8176

answer b) 95% prediction interval =mean±z(0.025)*sd*sqrt(1+1/n)

=8.56±1.96*0.76*sqrt(1+1/16)=8.56±1.54=(7.02,10.1)

95% lower prediction bound=7.02

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