The titanium content in an aircraft-grade alloy is an important determinant of strength. A sample of 20 test coupons reveals the following titanium content (in percent): 8.32, 8.05, 8.93, 8.65, 8.25, 8.46, 8.52, 8.35, 8.36, 8.41, 8.42, 8.30, 8.71, 8.75, 8.60, 8.83, 8.50, 8.38, 8.29, 8.46. The median titanium content should be 8.5%. Use the sign test with a = 0.05 to investigate this hypothesis. Match the answers. Parameter of Interest Null and Alternative Hypothesis: Test Statistic: Reject the null hypothesis if: v Conclusion:

MATLAB: An Introduction with Applications
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The titanium content in an aircraft-grade alloy is an important determinant of strength. A
sample of 20 test coupons reveals the following titanium content (in percent):
8.32, 8.05, 8.93, 8.65, 8.25, 8.46, 8.52, 8.35, 8.36, 8.41, 8.42, 8.30, 8.71, 8.75, 8.60, 8.83, 8.50,
8.38, 8.29, 8.46. The median titanium content should be 8.5%.
Use the sign test with a = 0.05 to investigate this hypothesis. Match the answers.
v Parameter of Interest:
a. Median Titanium content
v Null and Alternative Hypothesis:
b. Mean of the Titanium content
- v Test Statistic:
c. Hi u=8.5
H:u+ 8.5
v Reject the null hypothesis if:
d. Fail to reject the null hypothesis. There is not enough evidence to conclude that the median of the titanium
content differs from 8.5%
v Conclusion
e. Hi u =8.5
H: i >8.5
f. Reject the null hypothesis. The median of the titanium content differs from 8.5%.
g.r- = 12
h.r <r.
i. H: u =8.5
H:ũ + 8.5
j. r = min (R+,R¯)
Activate Windows
Transcribed Image Text:The titanium content in an aircraft-grade alloy is an important determinant of strength. A sample of 20 test coupons reveals the following titanium content (in percent): 8.32, 8.05, 8.93, 8.65, 8.25, 8.46, 8.52, 8.35, 8.36, 8.41, 8.42, 8.30, 8.71, 8.75, 8.60, 8.83, 8.50, 8.38, 8.29, 8.46. The median titanium content should be 8.5%. Use the sign test with a = 0.05 to investigate this hypothesis. Match the answers. v Parameter of Interest: a. Median Titanium content v Null and Alternative Hypothesis: b. Mean of the Titanium content - v Test Statistic: c. Hi u=8.5 H:u+ 8.5 v Reject the null hypothesis if: d. Fail to reject the null hypothesis. There is not enough evidence to conclude that the median of the titanium content differs from 8.5% v Conclusion e. Hi u =8.5 H: i >8.5 f. Reject the null hypothesis. The median of the titanium content differs from 8.5%. g.r- = 12 h.r <r. i. H: u =8.5 H:ũ + 8.5 j. r = min (R+,R¯) Activate Windows
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