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ID AREA GENDER BW WDI TDI RISK 101 2 2 78 4 14 4,58E-04 102 2...

ID

AREA

GENDER

BW

WDI

TDI

RISK

101

2

2

78

4

14

4,58E-04

102

2

2

65

6

10

9,23E-06

201

1

1

58

6

10

2,07E-04

202

1

2

67

10

22

4,93E-04

301

2

1

47

6

16

4,09E-05

302

2

2

112

15

15

8,04E-06

401

2

1

75

0

3

1,80E-05

403

2

1

70

0

8

4,11E-05

501

1

1

59

5

10

6,36E-04

502

1

1

68

2

4

4,76E-04

503

1

1

48

4

9

1,18E-03

510

1

2

84

7

10

1,64E-03

511

1

1

48

0

2

3,66E-04

602

1

1

50

2

6

1,08E-05

603

1

1

52

3

13

4,50E-05

604

1

2

38

0

6

3,79E-05

605

1

2

75

0

5

3,40E-04

607

1

2

85

0

3

6,35E-06

608

1

1

55

2

7

1,91E-05

609

1

2

79

3

18

3,42E-05

801

1

1

63

1

3

2,14E-04

802

1

1

68

0

8

1,04E-03

901

2

2

64

10

20

2,34E-04

1001

2

2

85

4

11

2,76E-04

1002

2

1

55

0

4

1,16E-04

1101

1

1

58

0

5

1,29E-05

1102

1

2

72

0

13

2,71E-05

1304

1

1

52

4

8

1,48E-03

1306

1

2

80

2

6

7,88E-04

1307

1

1

70

0

3

5,91E-04

1309

1

1

68

7

14

2,59E-03

1310

1

2

70

4

8

3,09E-05

1313

1

2

65

4

7

1,29E-03

1401

2

1

54

0

2

1,44E-05

1402

2

1

63

0

5

3,10E-05

Carcinogenic risk level of 1×10-3 is of critical importance because risks above this level require special environmental management actions.  Assuming the risks are distributed normally, the risk manager is going to be required to take special action if more than 16% of the population has risk >1×10-3.  Decide what is the case by testing the variance.

math   Statistics-And-Probability   
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Answer #1

From the data, we can see that out of n 35, x 6 has Carcinogenic risk level more than 1 x 10-3 ID AREA GENDER BW WDITDIRISK 101 102 201 202 301 302 401 403 501 502 503 510 511 602 603 604 605 607 608 78 65 58 67 147 2 112 75 70 59 14 4.58E-04 10 9.23E-06 10 2.07E-04 22| 4.93Е-04 16 4.09E-05 15 8.04E-06 3 1.80E-05 8 4.11E-05 106.36E-04 41 4.76Е-04 10 1 48 10 1 48 50 52 38 75 85 2 3.66E-04 6 1.08E-05 13 4.50E-05 6 3.79E-05 5 3.40E-04 3 6.35E-06 7 1.91E-05 18 3.42E-05 3 2.14E-04 5 79 801 802 901 1001 1002 1101 1102 1304 1306 1307 1309 1310 1313 1401 1402 63 10 20 2.34E-04 11 2.76E-04 4 1.16E-04 5 1.29E-05 13 2.71E-05 85 58 72 1 52 80 70 68 70 2 65 6 7.88E-04 3 5.91E-04 8 3.09E-05 2 1.44E-05 1 63 5 3.10E-05This would like to test, whether more than 16% of the population has risk > 1 x 10% (p will be the population proportion with risk>1x 103) i.e. p>0.16 Null hypothesis and alternative hypothesis would be Ho: p s 0.16 VS H:p> 0.16 P>0.16 will go to alternative hypothesis as the hypothesis with signs >, <, should be at alternative p = 0.1714 n 35 Test statistic is given by p-p 0.1714 - 0.16 = 0.0738 p(l-p 0.16(1-0.16 Since alternative is with >sign p-value P(z> 0.0738) 0.5294 Using -1-NORM.S.DIST(0.0738,TRUE) excel function we can calculate standard normal probabilities, we have subtracted it from 1 because this is right (> tailed test, and excel gives left tailed probabilities, hence after subtracting it from 1 we get right tailed probability. Since, p-value > 0.05 (assuming 5% level of significance) we fail to reject null hypothesis. That means (p s 0.16) 16% or less of the population has risk > 1 x 10%

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