EXAMPLE 8.5 | Alloy Adhesion An article in the Journal of Materials Engineering ["Instrumented Tensile Adhesion Tests on Plasma Sprayed Thermal Barrier Coatings" (1989, Vol. 11(4), pp. 275-282)] describes the results of tensile adhesion tests on 22 U-700 alloy specimens. The load at specimen failure is as follows (in megapascals): 19.8 15.4 11.4 19.5 10.1 18.5 14.1 8.8 14.9 7.9 17.6 13.6 O 0.01 O 0.025 O 0.05 O 0.95 O 0.975 7.5 12.7 16.7 11.9 15.4 11.9 15.8 11.4 The sample mean is x = 13.71, and the sample standard deviation is s = 3.55. Figures 8.6 and 8.7 show a box plot and a normal probability plot of the tensile adhesion test data, respectively. These displays provide good support for the assumption that the population is normally distributed. We want to find a 95% CI on μ. Since n = 22, we have n - 1 = 21 degrees of freedom for t, so to.025,21 = 2.080. The resulting CI is X-1/2-1/√x+1a/2n-1³/√n 13.71-2.080 (3.55)/√/22 ≤ ≤ 13.71 +2.080 (3.55)/√22 13.71-1.57 ≤ ≤ 13.71 +1.57 12.14 ≤ ≤ 15.28 Referring to Example 8-5 above, the probability that the true mean, u, lies below 12.14 or above 15.28 is 15.4 11.4
EXAMPLE 8.5 | Alloy Adhesion An article in the Journal of Materials Engineering ["Instrumented Tensile Adhesion Tests on Plasma Sprayed Thermal Barrier Coatings" (1989, Vol. 11(4), pp. 275-282)] describes the results of tensile adhesion tests on 22 U-700 alloy specimens. The load at specimen failure is as follows (in megapascals): 19.8 15.4 11.4 19.5 10.1 18.5 14.1 8.8 14.9 7.9 17.6 13.6 O 0.01 O 0.025 O 0.05 O 0.95 O 0.975 7.5 12.7 16.7 11.9 15.4 11.9 15.8 11.4 The sample mean is x = 13.71, and the sample standard deviation is s = 3.55. Figures 8.6 and 8.7 show a box plot and a normal probability plot of the tensile adhesion test data, respectively. These displays provide good support for the assumption that the population is normally distributed. We want to find a 95% CI on μ. Since n = 22, we have n - 1 = 21 degrees of freedom for t, so to.025,21 = 2.080. The resulting CI is X-1/2-1/√x+1a/2n-1³/√n 13.71-2.080 (3.55)/√/22 ≤ ≤ 13.71 +2.080 (3.55)/√22 13.71-1.57 ≤ ≤ 13.71 +1.57 12.14 ≤ ≤ 15.28 Referring to Example 8-5 above, the probability that the true mean, u, lies below 12.14 or above 15.28 is 15.4 11.4
MATLAB: An Introduction with Applications
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Author:Amos Gilat
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Chapter1: Starting With Matlab
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![**Example 8.5 | Alloy Adhesion**
An article in the *Journal of Materials Engineering* ["Instrumented Tensile Adhesion Tests on Plasma Sprayed Thermal Barrier Coatings" (1989, Vol. 11(4), pp. 275-282)] describes the results of tensile adhesion tests on 22 U-700 alloy specimens. The load at specimen failure is as follows (in megapascals):
\[
\begin{array}{cccccc}
19.8 & 10.1 & 14.9 & 7.5 & 15.4 & 15.4 \\
15.4 & 18.5 & 7.9 & 12.7 & 11.9 & 11.4 \\
11.4 & 14.1 & 17.6 & 16.7 & 15.8 & 11.4 \\
19.5 & 8.8 & 13.6 & 11.9 & 11.4 \\
\end{array}
\]
The sample mean is \( \bar{x} = 13.71 \), and the sample standard deviation is \( s = 3.55 \). Figures 8.6 and 8.7 show a box plot and a normal probability plot of the tensile adhesion test data, respectively. These displays provide good support for the assumption that the population is normally distributed. We want to find a 95% CI on \( \mu \). Since \( n = 22 \), we have \( n - 1 = 21 \) degrees of freedom for \( t \), so \( t_{0.025,21} = 2.080 \). The resulting CI is
\[
\bar{x} - t_{ \alpha/2, n-1} s / \sqrt{n} \leq \mu \leq \bar{x} + t_{ \alpha/2, n-1} s / \sqrt{n}
\]
\[
13.71 - 2.080 (3.55) / \sqrt{22} \leq \mu \leq 13.71 + 2.080 (3.55) / \sqrt{22}
\]
\[
13.71 - 1.57 \leq \mu \leq 13.71 + 1.57
\]
\[
12.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F032f9f2f-4a86-4f7c-abbd-a161821f35fd%2F4863c73c-64cb-4605-922f-2a6602c25011%2Fxl1nn7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Example 8.5 | Alloy Adhesion**
An article in the *Journal of Materials Engineering* ["Instrumented Tensile Adhesion Tests on Plasma Sprayed Thermal Barrier Coatings" (1989, Vol. 11(4), pp. 275-282)] describes the results of tensile adhesion tests on 22 U-700 alloy specimens. The load at specimen failure is as follows (in megapascals):
\[
\begin{array}{cccccc}
19.8 & 10.1 & 14.9 & 7.5 & 15.4 & 15.4 \\
15.4 & 18.5 & 7.9 & 12.7 & 11.9 & 11.4 \\
11.4 & 14.1 & 17.6 & 16.7 & 15.8 & 11.4 \\
19.5 & 8.8 & 13.6 & 11.9 & 11.4 \\
\end{array}
\]
The sample mean is \( \bar{x} = 13.71 \), and the sample standard deviation is \( s = 3.55 \). Figures 8.6 and 8.7 show a box plot and a normal probability plot of the tensile adhesion test data, respectively. These displays provide good support for the assumption that the population is normally distributed. We want to find a 95% CI on \( \mu \). Since \( n = 22 \), we have \( n - 1 = 21 \) degrees of freedom for \( t \), so \( t_{0.025,21} = 2.080 \). The resulting CI is
\[
\bar{x} - t_{ \alpha/2, n-1} s / \sqrt{n} \leq \mu \leq \bar{x} + t_{ \alpha/2, n-1} s / \sqrt{n}
\]
\[
13.71 - 2.080 (3.55) / \sqrt{22} \leq \mu \leq 13.71 + 2.080 (3.55) / \sqrt{22}
\]
\[
13.71 - 1.57 \leq \mu \leq 13.71 + 1.57
\]
\[
12.
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