Question

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

Test if cancer risks statistically differ between areas (urban & non-urban) and genders (females and males). urban : 1 nonurban :2 female :1 male:2

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Answer #1

Save the excel file in your working directory

Using R


risk <- read.csv("risk.csv",header=TRUE)
risk$RISK <- as.numeric(risk$RISK)

## for comparison between areas
t.test(risk$RISK[risk$AREA==1], risk$RISK[risk$AREA==2] )

Welch Two Sample t-test

data:  risk$RISK[risk$AREA == 1] and risk$RISK[risk$AREA == 2]
t = -0.59192, df = 18.991, p-value = 0.5609
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -10.223452   5.715876
sample estimates:
mean of x mean of y 
 17.29167  19.54545 

p-value = 0.5609 > 0.05

hence we fail to reject the null hypothesis

we conclude that there is not significant difference between area.

## for gender


t.test(risk$RISK[risk$GENDER==1], risk$RISK[risk$GENDER==2] )

        Welch Two Sample t-test

data:  risk$RISK[risk$GENDER == 1] and risk$RISK[risk$GENDER == 2]
t = -2.2054, df = 31.596, p-value = 0.03483
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -13.9175761  -0.5490906
sample estimates:
mean of x mean of y 
 14.90000  22.13333 

P-VALUE = 0.0348 < 0.05

hence we reject the null hypothesis

we conclude that there is significant difference between genders

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