Solution Found!
An article in the Journal of Aircraft (1986, Vol. 23, pp.
Chapter 10, Problem 54E(choose chapter or problem)
An article in the Journal of Aircraft (1986, Vol. 23, pp. 859–864) described a new equivalent plate analysis method formulation that is capable of modeling aircraft structures such as cranked wing boxes and that produces results similar to the more computationally intensive finite element analysis method. Natural vibration frequencies for the cranked wing box structure are calculated using both methods, and results for the first seven natural frequencies follow:
|
Freq. |
Finite Element Cycle/s |
Equivalent Plate, Cycle/s |
1 |
14.58 |
14.76 |
2 |
48.52 |
49.10 |
3 |
97.22 |
99.99 |
4 |
113.99 |
117.53 |
5 |
174.73 |
181.22 |
6 |
212.72 |
220.14 |
7 |
277.38 |
294.80 |
(a) Do the data suggest that the two methods provide the same mean value for natural vibration frequency? Use \(\alpha=0.05\). Find the P-value.
(b) Find a 95% confidence interval on the mean difference between the two methods.
Equation Transcription:
Text Transcription:
alpha=0.05
Questions & Answers
QUESTION:
An article in the Journal of Aircraft (1986, Vol. 23, pp. 859–864) described a new equivalent plate analysis method formulation that is capable of modeling aircraft structures such as cranked wing boxes and that produces results similar to the more computationally intensive finite element analysis method. Natural vibration frequencies for the cranked wing box structure are calculated using both methods, and results for the first seven natural frequencies follow:
|
Freq. |
Finite Element Cycle/s |
Equivalent Plate, Cycle/s |
1 |
14.58 |
14.76 |
2 |
48.52 |
49.10 |
3 |
97.22 |
99.99 |
4 |
113.99 |
117.53 |
5 |
174.73 |
181.22 |
6 |
212.72 |
220.14 |
7 |
277.38 |
294.80 |
(a) Do the data suggest that the two methods provide the same mean value for natural vibration frequency? Use \(\alpha=0.05\). Find the P-value.
(b) Find a 95% confidence interval on the mean difference between the two methods.
Equation Transcription:
Text Transcription:
alpha=0.05
ANSWER:
Step 1 of 4
Part a)
The quantity of interest is the difference in the mean and values of natural vibration frequency for two methods, 
Test 
Find the difference in frequencies in the table given below:
Frequency |
Difference |
1 |
-0.18 |
2 |
-0.58 |
3 |
-2.77 |
4 |
-3.54 |
5 |
-6.49 |
6 |
-7.42 |
7 |
-17.42 |