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Q.1 In a study to determine the effect of a certain gene carried out by pregnant mothers on infant weight at a later stage of their lives two samples were selected,  sample (A) for mothers with the gene and sample (B) for mothers without the gene. Each sample contained 35 subjects.

Group A-Birth weights of children born for mothers with the gene

27

49

59

67

75

36

49

60

68

76

37

51

60

72

76

40

52

63

73

84

42

52

64

73

102

44

56

66

74

103

48

59

66

75

117

 

The table above shows weights in pounds for group A. Use the table to answer the following questions:

For the ungrouped data what are the value of the 25th and the 75th percentiles?

Answer;

P= p/100*n+1/2

Therefore, 25/100*35+1/2

0.25*35+0.5

N=9, therefore 25th percentile =49

0.95*35+1/2

N=33.75=34

95th percentile =103

Use the data to construct a frequency distribution and a cumulative relative frequency for grouped data? (Use classes of width of 10).

answer;

group A weights

no of children

cumulative weight

27

1

27

36

1

36

37

1

37

40

1

40

42

1

42

44

1

44

48

1

48

49

2

98

51

1

51

52

2

104

56

1

56

59

2

118

60

2

120

63

1

63

64

1

64

66

2

132

67

1

67

68

1

68

72

1

72

73

2

146

74

1

74

75

2

150

76

2

152

84

1

84

102

1

102

103

1

103

117

1

117

For the grouped data calculate the 75th percentile

Answer;

75/100*27+1/2

0.75*27+0.5

=20.75=21

75th percentile =74

Construct a polygon for the grouped data for Group A, and use it to describe the distribution of weights (Shape of the graph, skew ness, distribution of weights, concentration of weights, etc.. ?

Answer;

6. The table below shows data for group B weights. Plot this data after grouping it by a polygon on the same graph with the grouped data for group A, then use the resulting graph to compare between the two groups (Which has a better weight distribution? justify your answer). Note: consider an optimal weight to be between 50 and 69.

Group B has better weight distribution since its curve is significantly higher than that of group A.

Group B-Birth weights of children born for mothers without the gene

30

53

76

87

97

36

54

78

87

100

37

55

79

91

101

40

62

79

91

101

42

66

82

91

102

44

66

83

92

115

48

68

85

97

117

Q. 2.  Dr. Green (a pediatrician) wanted to test if there is a correlation between the number of meals consumed by a child per day (X) and the child weight (Y). Included you will find a table containing the information on 6 of the children. Use the table to answer the following:

Child

Number of meals consumed per day (X)

child weight (Y)

XY

Ahmad

11

8

121

64

88

Ali

16

11

256

121

176

Osama

12

9

144

81

108

Husien

19

13

361

169

247

Lala

6

8

36

64

48

Noor

 

 

 

 

 

1.  Find the average and standard deviation for the child number of meals and weight.

Answer:

Average no of meals= (11+16+12+19+6+0)/6 =64/6= 10.67

Average weight = (8+11+9+13+8+0)/6 = 49/6 = 8.17

Standard deviation of meals= square root of 1/6*236 = 6.27

Standard deviation of weights = square root of 1/6*99 = 4.06

2.  Dose Ahmad has a better position among the 6 children in number of meals he consumed or in his weight? (Hint: use Z score formula to make the comparisons), justify your answer.

Answer;

Z score = (x –mean)/standard deviation

Meals = (11-10.6)/6.27 = 0.638

Weights = (8-8.16)/4.06 = -0.039 = -0.04

Therefore, Ahmad has a better position in number of meals than in weights because the z score of meals is positive while that of weights is negative.

 3.  If these six children represent a sample from a population of 100 children what will be the number of children who will have a number of meals between 10 and 14.

Z score = (10-10.6)/6.27 =-0.0956 =-0.1

Therefore (14-10.6)/6.27 = 0.54

 -0.1*100 =-10

0.54*100 = 54, therefore 54—10=64 children will have meals between 10 and 14

4.  Is there a correlation between the two variables? Find the value of correlation coefficient.

Answer;

Yes there is correlation

R= sum (x-mean of x) (y-mean of y)/square root (x-mean of x) ^2 *square root(y-mean of y) ^2

=144/square root 236 * square root of 99

=144/15.36*9.95

=144/152.83 =0.94

R=0.94

5.  Find out the prediction equation then use it to calculate the exact expected weight of a child that consumed 10 meals per day.

Answer;

8.16 weight = 10.6 meals

Therefore, 10 meals =?

= (8.16*10)/10.6 =81.6/10.6

=7.69

Approximate weight is 8

Q.3. A hospital supervisor has found that the medical stuff of the laboratory, on the average, complete a certain task in 10 minutes. If the times required to complete the task are approximately normally distributed with a standard deviation of 3 minutes, find:

1. The proportion of workers that will require more than 8 minutes to complete the task.

Answer:

(8-10)/3= -2/3 = 0.67 therefore 67%

2. If the supervisor has a total of 160 workers in his lab and any worker that completes the work in more than 15 minutes will be dismissed what is the total number of workers to be dismissed.

Answer:

Average time = 10, standard deviation 3

Expected time for all workers to finish with a standard deviation of 3

=10+-3 minutes, 10+3=13, therefore no employees are expected to finish past 15 minutes

3. If a new machine was inserted which reduces the average time required from the original average by 2 minutes whereas the standard deviation is reduced by 1 minute, what will be the reduction in the proportion of workers that takes 6 minutes to finish the work. (Hint: find the proportion based on the old average and standard deviation then based on the new ones and then complete your answer).

Answer:

New average=8, new standard deviation = 2 minutes

(6-8)/2= -1

Q.4. A population of 1000 children have been surveyed for their hemoglobin level. The population average hemoglobin was 12 and their standard deviation was 3. Based on this information find out the following:

1. If a child that has a hemoglobin level below 11 will need iron supplementation what proportion will need iron?

Z11=(x-12)/3=11, 33=x-12, 33+12=x =45

Proportion = 45/1000= 0.045

2. What propotion of children will have a hemoglobin between 15.5 and 11?

Z 15.5= (x-12)/3, =46.5=x-12, =46.5+12=58.5=59

Proportion = 59/1000-0.045=0.059-0.045 =0.014

3. What is the Z score for a child that has a hemoglobin level of 13

Z score = (13-12)/3=1/3=0.33

4. What is the number of children that will have a hemoglobin level above 13?

Answer;

= (1-0.33)*1000= 670 children

5. What proportion of children will have a Z score below -2.5?

Answer;

-2.5= (x-12)/3=-7.5+12=x, = 4.5=5children

Q.5. Answer the following:

  1. The least square prediction line is called so because it minimizes the total predictive errors. Answer by true or false______true______
  2. Presence of a strong correlation between two variables means that the pairs of the variables tend to occupy similar relative positions in their respective distributions. Answer by true or false__________true________
  3. The standard normal curve has an average of zero (0) and a standard deviation of 1.96. Answer by true or false______true______
  4. When correlation is found between two variables this mean that there is a causative relationship between them. Answer by true or false and justify your answer.

Answer;

 True because for there to be a correlation between two variables a relationship must exist which can be either positive or negative.

  1. The graph that is used to represent quantitative data and at the same time keeps the individual observations is the frequency polygon. True or False? true
  2. Bar charts are not suitable to describe continuous quantitative variables, why?

answer;

Bar charts cannot illustrate a trend in continuous quantitative variables but rather reflect final definitive data.

Q.6 For each situation described by the statements below, what is the variable being measured? and what type of scale is used for its measurement? (Nominal, ordinal, interval or ratio).

Statement

Variable measured

Scale of measurement

 

1. In order to study the effect of different types of treatments on the recovery status the researcher classified the patients recovery status as quick, moderate, slow and late.

time

interval

2. In a survey for health management purposes a researcher measured the time required by doctors to finish certain surgical operations.

time

ordinal

3. In a study on type of disease the researcher has to classify the disease as genetic or not genetic

disease

nominal

4.  In a study of the effect of certain drugs of temperature the mean temperature of children in each treatment group were compared with the others.

temperature

ratio

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