Question

A sample of 16 joint specimens of a particular type gave a sample mean proportional limit stress of 8.55 MPa and a sample standard deviation of 0.79 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.)

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.)

Answer 1

SOLUTION:

From given,

A sample of 16 joint specimens of a particular type gave a sample mean proportional limit stress of 8.55 MPa and a sample standard deviation of 0.79 MPa.

The total number of sample = n = 16

Sample mean = = 8.55 M Pa

Standard deviation = = 0.79 MPa

Degree of freedom = n-1 = 16-1 = 15

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.)

The corresponding t statistic for 95% significance is : 1.761

95% = 95/100 = 0.95

=1-0.95 = 0.05

/ 2 = 0.05/2 = 0.025

t _{/ 2,df} = t_0.025,15 = 1.761

95% lower confidence bound

( - t _{/ 2,df} * (/sqrt(n))) = ( 8.55 - 1.761* (0.79 /sqrt(16))) = 8.2022

It means that with 95% confidence we can say that the value of the true mean is greater than 8.2022 .We have assumed that the sample observations were obtained from a population that is normally distributed.

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.)

95% lower prediction bound is given by:

( - t _{/ 2,df} * *sqrt(1+1/n))) = ( 8.55 - 1.761* 0.79 *sqrt(1+1/16))) = 7.11599

If this bound is calculated for many samples , 95% of these bounds will provide a lower bound.

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