Solution Found!
Answer: The value of Young’s modulus (GPa) was determined
Chapter 1, Problem 45E(choose chapter or problem)
The value of Young's modulus (GPa) was determined for cast plates consisting of certain intermetallic substrates, resulting in the following sample observations ("Strength and Modulus of a Molybdenum-Coated Ti-25Al10Nb-3U-1Mo Intermetallic," J. of Materials Engr. and Performance, 1997: 46-50):
\(\begin{array}{lllll} 116.4 & 115.9 & 114.6 & 115.2 & 115.8 \end{array}\)
a. Calculate \(\bar{x}\) and the deviations from the mean.
b. Use the deviations calculated in part (a) to obtain the sample variance and the sample standard deviation.
c. Calculate \(s^{2}\) by using the computational formula for the numerator \(S_{x x}\).
d. Subtract 100 from each observation to obtain a sample of transformed values. Now calculate the sample variance of these transformed values, and compare it to \(s^{2}\) for the original data.
Questions & Answers
QUESTION:
The value of Young's modulus (GPa) was determined for cast plates consisting of certain intermetallic substrates, resulting in the following sample observations ("Strength and Modulus of a Molybdenum-Coated Ti-25Al10Nb-3U-1Mo Intermetallic," J. of Materials Engr. and Performance, 1997: 46-50):
\(\begin{array}{lllll} 116.4 & 115.9 & 114.6 & 115.2 & 115.8 \end{array}\)
a. Calculate \(\bar{x}\) and the deviations from the mean.
b. Use the deviations calculated in part (a) to obtain the sample variance and the sample standard deviation.
c. Calculate \(s^{2}\) by using the computational formula for the numerator \(S_{x x}\).
d. Subtract 100 from each observation to obtain a sample of transformed values. Now calculate the sample variance of these transformed values, and compare it to \(s^{2}\) for the original data.
ANSWER:Step 1 of 5
Before we start the exercise, we have created the following table which we are going to use.
\(\begin{array}{|c|c|c|c|c|} \hline i & x_{i} & x_{i}-\bar{x} & \left(x_{i}-\bar{x}\right)^{2} & x_{i}^{2} \\ \hline 1 & 116.4 & 0.82 & 0.672 & 13548.96 \\ 2 & 115.9 & 0.32 & 0.102 & 13432.81 \\ 3 & 114.6 & -0.98 & 0.960 & 13133.16 \\ 4 & 115.2 & -0.38 & 0.144 & 13271.04 \\ 5 & 115.8 & 0.22 & 0.048 & 13409.64 \\ \hline \text { Total: } & 577.9 & 0.00 & 1.928 & 66795.61 \\ \hline \end{array}\)