Prepared by:

Raveen Ismael Abdullah

Master student

College of Nursing

Hawler Medical University

Biostatic Assignment

3/15/2017

Exercise 1:

Simple random sampling using random number table.

Requirements: Random digit table.

Question:

- Use the random digit table to select 50 students out of 250

students.

- What do you call this type of sampling?

Answer:

It is simple random sampling.

Using coding system each student will have his/her own code for example student number one will have code 001 and so on

Selecting a random number between (1-250) through Excel 2007 by using random in between.

Starting number randomly is 87.

By using random digit table 50 students been selected out of 250 students.

Student No.

Code Random Digit Number

1 087 87

2 144 144

3 012 12

4 082 82

5 092 92

6 087 87

7 240 240

8 157 157

9 056 098

10 048 56

11 153 48

12 075 153

13 142 190

14 067 246

15 203 203

16 223 223

17 016 16

18 173 173

19 027 27

20 088 88

21 029 29

22 235 235

23 136 136

24 096 96

25 177 177

26 005 5

27 229 229

28 188 188

29 095 95

30 086 86

31 167 167

32 053 53

33 185 185

34 152 152

35 231 231

36 211 211

37 068 68

38 248 248

39 038 38

40 066 66

41 123 123

42 191 191

43 249 249

44 153 153

45 118 118

46 023 23

47 122 122

48 165 165

49 075 75

50 196 196

Exercise 2.

Systematic random sampling

Question:

Choose 50 students out of 200 students. Mention all steps in details.

Answer:

Making intervals by dividing No. of population by the required

sample 200/50=4

Coding the sample.

Choosing starting Number randomly by (excel 2007)

Random start was 11

For example :

11+4=15(student number fifteen will be selected)

14+4=19

No.

Student code

Systematic sampling

1. 15 15

2. 19 19

3. 23 23

4. 27 27

5. 31 31

6. 35 35

7. 39 39

8. 43 43

9. 47 47

10. 51 51

11. 55 55

12. 59 59

13. 63 63

14. 67 67

15. 71 71

16. 75 75

17. 79 79

18. 83 83

19. 87 87

20. 91 91

21. 95 95

22. 99 99

23. 103 103

24. 107 107

25. 111 111

26. 115 115

27. 119 119

28. 123 123

29. 127 127

30. 131 131

31. 135 135

32. 139 139

33. 143 143

34. 147 147

35. 151 151

36. 155 155

37. 159 159

38. 163 163

39. 167 167

40. 171 171

41. 175 175

42. 179 179

43. 183 183

44. 187 187

45. 191 191

46. 195 195

47. 199 199

48. 3 3

49. 7 7

Exercise 3.

Stratified sampling

Question

One thousand employees (office and manual workers) of an enterprise

were composed of 600 males and 400 females. Use a stratified random

sampling method to collect a sample of 50 males and 50 females.

Answer :

Dividing population into subgroups based on mutually exclusive

criteria

600 males

400females

Using systematic sampling for choosing 50 males out of 600 males

and choosing 50 females out of 400 females.

Female employees:

Sample interval =400/50 =8

The random starting number is 385 (by excel 2007) 385,393, 1, 9, 17, 25, 33, 41, 49, 57, 65, 73, 81, 89, 97, 5, 13, 21, 29, 37,

45,53,61,69,77,85,93,2,10,18,26,34,42,50,58,66,74,82,90,98,6,14,22,30,

38, 46, 54,62,70,78

Male employees:

Sample interval =600/50 =12

The random starting number is 5 (by excel 2007) 17,29,41,53,65,77,89,101,113,125,137,149,161,173,185,197,209,221,233,245,257,269,281,29

3,305,317,329,341,353,365,377,389,401,413,425,437,449,461,473,485,497,509,521,533,545,5

57,569,581,593,605

Exercise 4. Cluster sampling

Question

Use the cluster sampling method to choose a representative sample of

Iraqi primary school students.

Answer:

Iraqi primary school students will be divided into groups called

clusters based on geographic areas

Using Random sampling for selecting subsets of clusters and

completing the required number of sample size .

Exercise 5

The following are body weight in Kg for 50 students.

Q) Present the data in an order array.

50 82 99 78 59 77 67 79 72 93

55 68 78 99 78 98 56 72 75 94

88 95 78 66 68 92 95 74 74 95

67 71 56 67 89 78 94 59 73 77

45 70 55 78 56 45 56 78 72 83

By using excel 2007

Answer

45

45

50

55

55

56

56

56

56

59

59

66

67

67

67

68

68

70

71

72

72

72

73

74

74

75

77

77

78

78

78

78

78

78

79

82

83

88

89

92

93

94

94

95

95

95

98

99

99

Exercise 6:

The frequency distribution table

Q/Present the following data (marks of students) in a frequency

distribution table:

50 82 99 78 59 77 67 79 72 93

55 68 78 99 78 98 56 72 75 94

88 95 78 66 68 92 95 74 74 95

67 71 56 67 89 78 94 59 73 77

45 70 55 78 56 45 56 78 72 83

Answer:

Putting data in order array from smallest to largest number

Determine the actual range : X max-X min =99-45=54

Decide number of classes=6

Determine class interval CI=R/NC=54/6=9

Relative frequency fi/(n)*100

Class interval Frequency Cumulative frequency

Relative frequency

Cumulative R. Frequency

≥49 2 2 0.04 0.04

50-59 9 11 0.18 0.22

60-69 6 17 0.12 0.34

70-79 19 36 0.38 0.72

80-89 4 40 0.08 0.8

≤90 10 50 0.2 1.0000

Total 50 1.0000

Exercise 7:

Measures of central tendency and dispersion

Q/For the following data (Blood urea in mg/dl), calculate measures of

central tendency, and measures of dispersion, then present the data in

a proper way or ways.

42 44 30 78 56 30 60 40

41 67 50 35 65 64 49 73

78 56 56 37 72 66 45 49

Answer:

Putting data in order array from smallest to largest number

by using Excel 2007

Choosing data analysis

Descriptive statistics

Summary statistics

measures of central tendency, and measures

of dispersion

Mean 53.45833333

Standard Error 3.000591226

Median 53

Mode 56

Standard Deviation 14.69983486

Sample Variance 216.0851449

Kurtosis -1.060919758

Skewness 0.101284958

Range 48

Minimum 30

Maximum 78

Sum 1283

Count 24

Exercise 8:

The normal distribution

Question 1

Suppose that the mean (+sd) of the marks of students is 70 + 2 and

their distribution is normal.

1.What is the percentage of students whose marks are more than 70?

2. What is the percentage of students whose marks are less than 70?

3. What is the percentage of students whose marks are between 68 and 72?

4. What is the percentage of students whose marks are between 70 and 72?

5. What is the percentage of students whose marks are more than 72?

6. What is the percentage of students whose marks are less than 72?

7. What is the percentage of students whose marks are more than 74?

8. What is the percentage of students whose marks are less than 66?

9. What is the percentage of students whose marks are more than 76?

10. What is the percentage of students whose marks are less than 64?

Answer: Drawing normal distribution chart(symmetrical and bell shaped )

Using 68,95,99.7 rule

64 66 68 70 72 74 76

-2SD +2SD

2.1

5%

13

.59%

34

.13%

34

.13%

13

.59%

2.1

5%

68.26%

of area 95.44% of area

99.74% of area

0.1

3

%

0.1

3

%

Answers

1. The percentage of students marks more than 70 is 50%.

2. The percentage of students marks less than 70 is 50%.

3.The percentage of students marks between 68-72 is 68.26%

4.The percentage of students whose marks between 70-72 is 34.13%.

5.The Percentage of students whose marks more than 72 is 15.87%.

6.The percentage of students whose marks less than 72 is 84.13%.

7.The percentage of students whose marks more than 74 is 2.28%.

8.The percentage of students whose marks are less than 66 is 2.28%.

9.The percentage of students whose marks are more than 76 is 0.13%.

10.The percentage of students whose marks are less than 64 is 0.13%.

Exercise 9:

The t test

Q1: The following values represent SGOT readings (in IU/Liter) of 20

patients affected with liver disease. If you know that the mean SGOT

of the population from which the sample of patients was drawn is 42

IU/Liter. Is there significant difference from the mean of normal

population?

Let alpha = 0.05

34 35 37 40 42 46 46 47 48 49

50 50 55 56 56 56 62 63 65 70

Answer

SPSS VERSION 18

One-Sample Statistics

N Mean Std. Deviation Std. Error Mean

SGOT IU/Liter 20 50.35 10.075 2.253

One-Sample Test

T-test

t df Sig. (2-tailed)

Mean

Difference

95% Confidence Interval of the

Difference

Lower Upper

SGOT IU/Liter 22.328 19 .000 50.300 45.58 55.02

The above table shows that the SGOT mean value for patients affected

with liver disease significantly differ from the SGOT mean value of

population p value is 0.001.

Q2: The weight in gm of two groups of new-born infants was

measured. Mothers of the first group were regular attendants of

antenatal care clinics, while mothers of the second group had no

history of such attendance.

Do these data provide evidence that antenatal care affect the weight

of new-born infants.

Let alpha = 0.01

Group1 (Antenatal care)

3000 3500 4000 3200 3150 2950 4100 3750 3450 3200

Group2 (No Antenatal care)

2700 2450 2500 2000 2200 2100 2300 2450 2600 2250

Answer :

Using excel 2007

Two independent groups

Using t-Test: Two-Sample Assuming Unequal Variances because we have two

sample Two independent variables ,two groups t-Test: Two-Sample Assuming Unequal Variances

3000 2700

Mean 3477.777778 2316.666667

Variance 159444.4444 39375

Observations 9 9

Hypothesized Mean Difference 0 df 12 t Stat 7.812060676 P(T<=t) one-tail 2.39594E-06 t Critical one-tail 2.680997992 P(T<=t) two-tail 4.79188E-06 t Critical two-tail 3.054539586

Assumption: Data normally distributed

o HO= M1 is equal to M2

o HA=M1 is not equal to M2

Calculated t for alpha 0.01 =2.68

Tabulated T for alpha 0.01 =3.05

There is no significant difference between two groups because calculated t is less

than tabulated T and p value is 4.79

Hence calculated t less than tabulated T

Here we accept HO and reject HA, the difference between two mean due to chance

The antenatal care have no significant affect on the weight of new born infant .

.

Q3. Eight jaundiced newborns were subjected to phototherapy. Do

these data provide evidence that phototherapy affect bilirubin level?

Bilirubin mg/dl before phototherapy Bilirubin mg/dl after phototherapy

15 6

18 9

14 14

25 26

17 5

20 6

17 4

16.5 3.5

Answer : t-Test: Paired Two Sample for Means Excel 2007

Variable 1 Variable 2

Mean 17.8125 9.1875

Variance 11.70982143 57.56696429

Observations 8 8

Pearson Correlation 0.657690579 Hypothesized Mean

Difference 0 df 7 t Stat 4.116194905 P(T<=t) one-tail 0.002240457 t Critical one-tail 1.894578604 P(T<=t) two-tail 0.004480914 t Critical two-tail 2.364624251

The table shows the significant difference between the mean of two

groups p value is equal to 0.004

Assumption: Data normally distributed

o HO= M1 is equal to M2

o HA=M1 is not equal to M2

Calculated t for alpha 0.01 =4.116

Tabulated T for alpha 0.01 =2.36

P value 0.004

There is significant difference between two groups because

calculated t is more than tabulated T and p value is 0.004

Hence calculated t more than tabulated T ,Here we accept HA and

reject HO

the difference between two mean is significant Phototherapy affect

bilirubin level .