An article in the Journal of Aircraft (Vol. 23, 1986, 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: Frequency Finite Element Equivalent Plate
1 14.48 14.79
2 48.45 49.08
3 97.13 99.98
4 113.97 117.43
5 174.75 181.18
6 212.54 220.06
7 277.40 294.79
(a) Do the data suggest that the two methods provide the same mean value for natural vibration frequency? Use
α = 0.05.
(b) Find a 95% confidence interval on the mean difference between the two methods. Round your answers to three decimal places (e. g. 98.765).

(a) No, the data suggests it is possible to reject the hypothesis which states that the 2 methods provide same mean value.

(b) (-10.957,-0.063)

Step-by-step explanation:

(a) Hypotheses for this data are:

:

First find the differences in values:

1 => -0.31

2 => -0.63

3 => -2.85

4 => -3.46

5 => -6.43

6 => -7.52

7 => -17.39

Now, find the mean and standard deviation of the differences:

mean = = - 5.51

sd = = 5.89

The value of test statistics is:

t = = = - 2.4750

Analysing Student's T distribution, at a df = 7-1 = 6:

p-value = 0.025*2 = 0.05

To reject the null hypothesis, p-value must be less or equal than α. Since they are equal, reject the null hypothesis, i.e., reject the claim suggesting the 2 methods provide the same mean value for natural vibration frequency.

(b) For a CI = 95%:

t-score for α = 0.025 and df = 6 is 2.447.

mean ±

-5.51 ± 2.447.

-5.51 ± 5.4471

lower limit: -5.51 - 5.4471 = - 10.957

upper limit: -5.51 + 5.4471 = - 0.063

The interval on the mean difference is (-10.957,-0.063)

10

step-by-step explanation:

the answer is c hope this