Question

In: Statistics and Probability

Go to the Files section and download the AFE_Test file from the Datasets folder. We are...

Go to the Files section and download the AFE_Test file from the Datasets folder. We are interested in a onetail test described in the following fashion: Ho: u < or = to 200 CFM; H1: u > 200 CFM. At 5% significance level,

we can reject the null hypothesis given the sample information in AFE_Test1.

we can reject the null hypothesis given the sample information in AFE_Test2.

we cannot reject the null hypothesis.

we can reject the null hypothesis given the sample information in AFE_Test1 and AFE_Test2.

Engine

Number Air Flow Error

415.4743512

346.09016

-8.596266867

532.7726337

283.0702257

189.3824034

105.5314484

334.2984697

184.4056484

256.663138

-104.457736

467.3703235

258.2404746

-2.074507734

-40.79538064

6.80056894

90.27253748

338.7010153

331.0819963

649.4146936

313.2307526

503.0561507

403.6624912

223.0000044

214.666741

-201.8286026

181.0092996

405.3486368

78.97865964

105.2625563

-20.7479801

305.6369038

-25.34042812

290.649136

160.2956032

581.8788331

87.38225426

236.7845624

461.8920726

-1.765506909

-1.139319807

155.2220552

172.5709168

324.958664

137.8738076

431.9881597

294.0063993

370.8186838

-268.3206689

377.8966444

15.56555128

400.5351499

237.0884473

-114.9040103

-1.842450161

258.17274

306.9807505

199.0776624

-159.7275472

90.7100499

50.57186593

-235.5723112

239.8702733

252.8041065

66.01740517

139.3463705

157.9240731

398.7363967

349.8917564

157.576396

108.0615717

246.23303

284.4200068

416.5905053

39.32832863

188.2311195

218.0355792

198.1066631

399.3020115

158.6990307

404.0966402

Engine Number Air Flow Error

-5.910636889

454.3618488

248.8773234

280.4565938

353.3013896

-218.8510862

94.68939062

332.6359874

425.0468792

299.2086466

-5.854975813

304.4174591

539.9043275

408.6882527

-130.7805305

712.8235887

110.8343768

41.30960043

293.1632807

219.859895

147.6832367

522.4087418

-22.70800359

102.6290747

518.992134

49.33093443

324.0126134

486.8666893

522.5290679

264.1219098

37.28276716

106.6241894

45.27340572

362.001093

110.5986357

335.748915

226.6452257

350.815877

275.9994104

195.8830681

391.2196789

439.131587

274.5389211

210.0179118

302.4803718

307.4775905

-112.4925929

463.7959919

204.2295691

371.1563168

86.33736435

68.51681611

262.680861

268.7462811

444.2777809

468.5967597

388.3007466

276.8384193

184.5206371

94.26292855

453.4004675

175.5802814

22.65986588

249.5145609

155.0875923

243.9447699

528.9350029

512.2006642

94.9378191

604.680723

240.4991037

399.2537004

194.1203309

197.734822

268.7425977

356.5817097

515.0917659

394.3821284

399.1902631

338.4149168

192.9782557

Solutions

Expert Solution

We can not reject the null hypothesis given the sample information in AFE_Test1 because the estimated p-value 0.453 is more than 0.05 significance level.

We can reject the null hypothesis given the sample information in AFE_Test2 because the estimated p-value 0.001 is less than 0.05 significance level.

We can reject the null hypothesis given the sample information in AFE_Test1 and AFE_Test2 because the estimated p-value 0.026 is less than 0.05 significance level.

orchestra answered 1 year ago

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