Question

A sample of 18 joint specimens of a particular type gave a sample mean proportional limit stress of 8.54 MPa and a sample standard deviation of 0.71 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.)

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 do not need to make any assumptions.

We must assume that the sample observations were taken from a normally distributed population.   

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.

(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.)

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 higher bound 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 bound for the corresponding future values of the proportional limit stress of a single joint of this type.

Answer 1

a)

sample std dev ,    s =    0.7100
Sample Size ,   n =    18
Sample Mean,    x̅ =   8.5400

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   17          
't value='   tα/2= 1.740 [Excel formula =t.inv(α,df) ]      
                  
Standard Error , SE = s/√n =   0.710   / √   18   =   0.1673
margin of error , E=t*SE = 1.740 *   0.167   = 0.291
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    8.54   - 0.291 = 8.25

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.

We must assume that the sample observations were taken from a normally distributed population.   

b)

margin of error for prediction interval,E=   t*s*√(1+1/n)= 1.269
prediction interval is       
Interval Lower Limit=   x̅ - E = 7.27

   If this bound is calculated for sample after sample, in the long run, 95% of these bounds will provide a lower bound for the corresponding future values of the proportional limit stress of a single joint of this type.