In: Math
Ultra high performance concrete (UHPC) is a relatively new construction material that is characterized by strong adhesive properties with other materials. The article "Adhesive Power of Ultra High Performance Concrete from a Thermodynamic Point of View"+ described an investigation of the intermolecular forces for UHPC connected to various substrates. The following work of adhesion measurements (in ml/m2) for UHPC specimens adhered to steel appeared in the article. 107.1 109.5 107.4 106.8 109.1
(a) Is it plausible that the given sample observations were selected from a nomal distribution?
It's plausible that the distribution could be normal.
It's not plausible that the distribution could be normal.
(b) Calculate a two-sided 95% confidence interval for the true average work of adhesion for UHPC adhered to steel. (Round your answers to two decimal places.)
Does the interval suggest that 107 is a plausible value for the true average work adhesion for UHPC adhered to steel?
The interval suggests that 107 is a possible value for the true average.
The interval doesn't suggest that 107 is a possible value for the true average.
Does the interval suggest that 111 is a plausible value for the true average work adhesion for UHPC adhered to steel?
The interval suggests that 111 is a possible value for the true average.
The Interval doesn't suggest that 111 is a possible value for the true average
(c) Predict the resulting work of adhesion value resulting from a single future replication of the experiment by calculating a 95% prediction interval. (Round your answers to two decimal places.)
Compare the width this interval to the width of the CI from (b).
The two intervals are the same width.
The PI is wider than the CI.
The Cl is wider than the PI.
(d) Calculate interval for which you can have a high degree of confidence that at least 95% of all UHPC specimens adhered to steel will have work of adhesion values between the limits of the interval. (Round your answers to two decimal places.)
Sol:
(a).
Yes, it is plausible that the distribution could be normal as there is no outlier or unusual value is here.
(b).
Here n= 5
Degree of freedom dF = 5 - 1
df = 4
Critical value for 95% confidence interval = t0.95, 4
= 2.7764
sample standard deviation s = 1.0756
stanadrd error of sample mean = s/sqrt(n)
= 1.0756/sqrt(5)
stanadrd error of sample mean = 0.481
sample mean = 107.78
Here 95% confidence interval = +- tcritical * s/sqrt(n)
= 107.78 + 2.7764 * 0.4810
95% confidence interval = (106.44, 109.12)
The interval suggest that 107 is a plausible value for the true average.
The interval also suggest that 111 is not a plausible value for the true average.
(c).
Here
95% prediction interval for a single value = +- tcritical * s *sqrt(1+ 1/n)
= 107.78 +- 2.7764 * 1.0756 * sqrt(1 + 1/5)
= 107.78 +- 2.9865 * 1.0954
95% prediction interval for a single value =(104.51, 111.05)
d).
The formula for the 95% tolerance interval is,
k =95 with n = 5 , the critical value is 5.079
95% tolerance interval = 107.78 5.079 * 1.0756
= 107.78 5.463
95% tolerance interval = ( 102.31 , 113.24 )