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

1

415.4743512

2

346.09016

3

-8.596266867

4

532.7726337

5

283.0702257

6

189.3824034

7

105.5314484

8

334.2984697

9

184.4056484

10

256.663138

11

-104.457736

12

467.3703235

13

258.2404746

14

-2.074507734

15

-40.79538064

16

6.80056894

17

90.27253748

18

338.7010153

19

331.0819963

20

649.4146936

21

313.2307526

22

503.0561507

23

403.6624912

24

223.0000044

25

214.666741

26

-201.8286026

27

181.0092996

28

405.3486368

29

78.97865964

30

105.2625563

31

-20.7479801

32

305.6369038

33

-25.34042812

34

290.649136

35

160.2956032

36

581.8788331

37

87.38225426

38

236.7845624

39

461.8920726

40

-1.765506909

41

-1.139319807

42

155.2220552

43

172.5709168

44

324.958664

45

137.8738076

46

431.9881597

47

294.0063993

48

370.8186838

49

-268.3206689

50

377.8966444

51

15.56555128

52

400.5351499

53

237.0884473

54

-114.9040103

55

-1.842450161

56

258.17274

57

306.9807505

58

199.0776624

59

-159.7275472

60

90.7100499

61

50.57186593

62

-235.5723112

63

239.8702733

64

252.8041065

65

66.01740517

66

139.3463705

67

157.9240731

68

398.7363967

69

349.8917564

70

157.576396

71

108.0615717

72

246.23303

73

284.4200068

74

416.5905053

75

39.32832863

76

188.2311195

77

218.0355792

78

198.1066631

79

399.3020115

80

158.6990307

81

404.0966402

Engine Number Air Flow Error

1

-5.910636889

2

454.3618488

3

248.8773234

4

280.4565938

5

353.3013896

6

-218.8510862

7

94.68939062

8

332.6359874

9

425.0468792

10

299.2086466

11

-5.854975813

12

304.4174591

13

539.9043275

14

408.6882527

15

-130.7805305

16

712.8235887

17

110.8343768

18

41.30960043

19

293.1632807

20

219.859895

21

147.6832367

22

522.4087418

23

-22.70800359

24

102.6290747

25

518.992134

26

49.33093443

27

324.0126134

28

486.8666893

29

522.5290679

30

264.1219098

31

37.28276716

32

106.6241894

33

45.27340572

34

362.001093

35

110.5986357

36

335.748915

37

226.6452257

38

350.815877

39

275.9994104

40

195.8830681

41

391.2196789

42

439.131587

43

274.5389211

44

210.0179118

45

302.4803718

46

307.4775905

47

-112.4925929

48

463.7959919

49

204.2295691

50

371.1563168

51

86.33736435

52

68.51681611

53

262.680861

54

268.7462811

55

444.2777809

56

468.5967597

57

388.3007466

58

276.8384193

59

184.5206371

60

94.26292855

61

453.4004675

62

175.5802814

63

22.65986588

64

249.5145609

65

155.0875923

66

243.9447699

67

528.9350029

68

512.2006642

69

94.9378191

70

604.680723

71

240.4991037

72

399.2537004

73

194.1203309

74

197.734822

75

268.7425977

76

356.5817097

77

515.0917659

78

394.3821284

79

399.1902631

80

338.4149168

81

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 2 years ago
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