Parts a and b (b) Find a 95% CI on the mean difference between the twomethods and use it to answer the question in part (a). 5-43. An article in the Journal of Aircraft (Vol. 23, 1986,pp. 859-864) describes a new equivalent plate analysis methodformulation that is capable of modeling aircraft structures suchas cranked wing boxes and that produces results similar to themore computationally intensive finite element analysis method.Natural vibration frequencies for the cranked wing box struc-ture are calculated using both methods, and results for the firstseven natural frequencies are shown here.Element,Car Cycle/s14.5848.5297.22113.991234Finite EquivalentPlate,Finite EquivalentPlate,Element,Cycle/s Car Cycle/sCycle/s14.76 5 174.73181.2249.106 212.72220.1499.997277.38294.80117.53(a) Do the data suggest that the two methods provide the samemean value for natural vibration frequency? Use theP-value approach.

The given data is:CarFinite Element Cycle\sEquivalent Plate Cycle\s 114.5814.76248.5249.1397.2299.994113.99117.535174.73181.226212.72220.147277.38294.8In this case, each of the sampled cars's natural vibration frequency is measured in two different scenarios, i.e., with finite elements and with equivalent plate. Considering the sampled cars in both scenarios are indifferent, the sample is considered to be paired in nature, and the paired t-test is to be used.Part (a)Considering μd as the difference between the average natural vibration frequency for finite element and the average natural frequency for the equivalent plate cycle, it is hypothesized as follows:H0:μd=0; the average natural vibration frequencies for both methods are hypothesized to be indifferent.H0:μd=0; the average natural vibration frequencies for both methods are hypothesized to be not indifferent.Calculation Table:CarFinite Element Cycle\s; xEquivalent Plate Cycle\s; y diffrence, d=x−yd2114.5814.76-0.180.0324248.5249.1-0.580.3364397.2299.99-2.777.67294113.99117.53-3.5412.53165174.73181.22-6.4942.12016212.72220.14-7.4255.05647277.38294.8-17.42303.4564Total-38.4421.2062The sampled number of cars, n=7.The sample average of the paired differencesd¯=∑dn=−38.47=−5.4857The sample variance of the paired differences,sd2=1n−1[∑d2−(∑d)2n]=17−1×[421.206−(−38.4)27]≈35.09243The associated standard error for the mean of the paired differences is calculated as:SE(d¯)=sd2n=35.092437≈2.23902Now the test statistic is:t=d¯SE(d¯)=−5.48572.23902≈−2.450045The degrees of freedom, df=n−1=7−1=6The p-value (two-tailed) for the t-score of −2.450045 with the distribution degrees of freedom df=6 is obtained using the excel function "T.DIST.2T(ABS(t),df)" as follows:The p-value is 0.0498.The decision rule is: Reject H0, if p-value≤α.For α=0.05 (the standard significance level), p-value (=0.0498)<α.Thus H0 is rejected.The average natural vibration frequencies for both methods are considered to be not indifferent.Part (b)For 95% CI, α=1−95%=0.05, the t-critical value for df=6 is obtained as follows:Thus tα2=2.447.The 95% CI for the difference between the average natural vibration frequency for both models is estimated as:d¯±tα2SE(b¯)i.e., −5.4857±[2.447×2.23902]i.e., −5.4857±5.48i.e., (−5.4857−5.48, −5.4857+5.48)≈(−10.9657, −0.0057)With 95% CI for the mean difference in natural frequency vibration between the two methods is (−10.9657, −0.0057).As the estimated interval of (−10.9657, −0.0057) does not include the value of 0, which essentially implies indifference, a significant difference between the average frequency vibration by the two methods is claimed with a surety level of 95%.The conclusion in part (a) holds true in part (b).(a) A significant difference between the average frequency vibration by the two methods is concluded.(b) The CI for the mean difference between the two methods is (−10.9657, −0.0057). The conclusion in part (a) holds true in part (b).