Question

10-54. An article in the journal of Aircraft 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: Finite Element Cycle/s 14.6 48.5 97.0 113.8 174.7 210.7 276.3 Equivalent Plate Frequency cle/s 14.7 49.1 99.9 4 180.2 220.2 290.4 7 (a) Do the data suggest that the two methods provide the same mean value for natural vibration frequency? Use α = 0.01 (b) Find a 99% confidence interval on the mean difference between the two methods 10-65. Consider the hypothesis test H0: σ-σ against Hlof < σ2. Suppose that the sample sizes are ni-6 and n,-13, and that sf-23.8 and sỈ-27.8. Use -0.05. Test the hypothesis and explain how the test could be conducted with a confidence interval on ơy

Answer 1

a) The step-by-step instructions to obtain the paired T- test output using Minitab18 K Type the data into the Minitab18 work sheet *Go to Stat -Basic Statistics -Paired -t Choose Each sample in its own column *Choose Two-Tailed test. * Choose the confidence level is 1-α: 0 99 * Choose the Hypothesized Differene is zero. * Choose the Alternative hypothesis is Difference # *Choose ok. pothesized Difference. The Paired!-test output using Minitab 18.

Paired T-Test and Cl: Finite Element, Equivalent plate Descriptive Statistics Sample Finite Element 7 133.7 92.3 Equivalent plate 7 138.9 97.2 36.8 N Mean StDev SE Mean 34.9 Estimation for Paired Difference 99% CI for μ difference Mean StDev SE Mean 5.20 5.04 1.91 (-12.26, 1.86) -difference: mean of (Finite Element-Equivalent plate) Test Ho: μ-difference-o H1: difference # 0 Null hypothesis Alternative hypothesis T-Value P-Value -2.73 0.034

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Step1: To, test the two methods provi de the same mean value for natural vibration frequency at the 1% level of significance The null and alternative hypotheses are as follows: Ho Ap 0 Step2: Deci de the test statistic and the level of significance The level of significance is α-001 The critical value of t is os -3.707428 Step3: State the decision Rule Reject Ho if t> 3.707428, otherwise do not reject Ho Step4: Calculate the value of test stati stic The paired difference experiment produced the following results 7,133.6571, 138.8571,5.2,s 5.041494 2 5.041494 5.041494 2.645751 1.905506 Sp1905506 Step5: The formula for calculating the test stati stic is SE 5.2 1.905506 -2.72893 t -2.72893 Step6: The P- value approach: P-value = 2P(t > 2.72893) 2(1-Pe <272893) 2(1-0.982883) -2(0.017117) 0.034234 P-value034234>a-001. We do not reject the null hypothesis H Step7: Conclusion Compare with the critical value and make a decision Since, the P- value-0.0342> 0.01. We do not reject the null hypothesis Ho Since, the t-test statistic is 272893 oos 3.707428. We do not reject the null hypothesis Ho There is not sufficient evidence to reject the null hypothesis Ho Hence, we can conclude that the two methods provide the same mean value for natural frequency at the 1% level of significance

b) The sample size isn-7 The mean difference is d 52. The formula for calculating the 99% mean difference between the methods is d -t 5.041494 5.041494 -52-(3.707428) 5041494 S H S-5.2+(3.707428) (0423455335 5.041494 2645751-32+(3707428) 5 041494 2.645751 -5.2 (3.707428) (1.905506)SDS-5.2+(3.707428 (1.905506) -5.2-7.064526 S A-5.2+7.064526 →一12.2645 H0s1 .864526 Interpretat on we can be 99% confident that the true mean difference between the two methods (4) is between -12.2645 and 1.8645