HS311: Epidemiology and Biostatistics I – 1503A / Unit 7 Assignment Unit 2, Topic 08, 9 & 10 Exercises

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HS311: Epidemiology and Biostatistics I –
1503A / Unit 7 Assignment

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Unit 2, Topic 08, Exercise 8-15

Population Growth

The population of
the United States grew by 9.1 % between 2000 and 2009. The following table reports the
percentage growth in population for each of the 50 states during this decade. States are classified by whether
they are east or west of the Mississippi River:

Western State % Eastern State %
Alaska 11.4 Alabama 5.9
Arizona 28.6 Connecticut 3.3
Arkansas 8.1 Delaware 13.0
California 9.1 Florida 16.0
Colorado 16.8 Georgia 20.1
Hawaii 6.9 Illinois 4.0
Idaho 19.5 Indiana 5.6
Iowa 2.8 Kentucky 6.7
Kansas 4.8 Maine 3.4
Louisiana 0.5 7.6
Minnesota 7.0 Massachusetts 3.9
Missouri 7.0 Michigan 0.3
Montana 8.1 Mississippi 3.8
Nebraska 5.0 New Hampshire 7.2
Nevada 32.3 New Jersey 3.5
New Mexico 10.5 New York 3.0
North Dakota 0.7 North Carolina 16.6
Oklahoma 6.9 Ohio 1.7
Oregon 11.8 Pennsylvania 2.6
South Dakota 7.6 Rhode Island 0.5
Texas 18.8 South Carolina 13.7
Utah 24.7 Tennessee 10.7
Washington 13.1 Vermont 2.1
Wyoming 10.2 Virginia 11.4
West Virginia 0.6
Wisconsin 5.4

a.Calculate the
median percentage change for the eastern states and for the western states.
Comment on how the medians compare.

Enter your answers to one decimal
place.

For the western states, the median is *1%.

For the eastern states, the median is *2%.

These medians indicate that the population in the
western states is growing, on average, about *3 percentage points than in the eastern
states.

b.Based on the shapes of the distributions for population
growth in eastern and western states, how would you expect the means to compare
to the medians? Explain.

Because both distributions are , you would expect the means to be the medians.

c.How would
you expect the mean percentage change in the western states to change if you
removed Nevada from the analysis? Explain.

You should expect the mean to
if you remove Nevada from the analysis
because the mean .

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Answer *1:the absolute tolerance is +/-0.1
Answer *2:
the absolute tolerance is
+/-0.2
Answer *3:the absolute
tolerance is +/-0.2


HS311: Epidemiology and Biostatistics I –
1503A / Unit 7 Assignment

*Unit 2, Topic 09, Exercise 9-11

Baby Weights

Three-month-old Benjamin
Chance weighed 14.8 pounds. The national average for the weight of a 3-month-old baby is 12.5 pounds, with a standard deviation of 1.5 pounds.

Round your answers to two decimal
places.

a.Determine the z-score for Benjamin’s weight at age 3 months, and write a sentence interpreting the
score.

z-scoreequals*1

Benjamin was *2 standard deviations the average weight.

b.
For a 6-month-old, the national average weight is 17.25 pounds, with a standard deviation of 2.0 pounds. Determine what Benjamin’s weight would have been at the
age of 6 months, in order for him to have the same z-score at 6 months that he had at 3 months.

*3

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Answer *1:the absolute tolerance is +/-0.01
Answer *2:
the absolute tolerance is
+/-0.01
Answer *3:the absolute
tolerance is +/-0.02

Unit 2, Topic 10, Exercise 10-15

Diabetes Diagnoses

The National Health and Nutrition Examination Survey (NHANES) is a
large-scale study conducted annually by the National Center for Health
Statistics. It involves over 10,000 Americans, randomly selected according to a
multistage sampling plan. All sampled subjects are asked to complete a survey
and take a physical examination. One of the questions asked in the 2003–2004
NHANES survey pertained only to subjects who had been diagnosed with diabetes.
Subjects were asked to indicate the age at which they were first diagnosed with
diabetes by a health professional. Responses for the 548 subjects with diabetes
are summarized in the following frequency table:

Age

Tally

Age

Tally

Age

Tally

Age

Tally

Age

Tally

1

1

18

4

37

7

54

8

71

4

2

3

19

3

38

10

55

22

72

4

3

7

20

1

39

9

56

13

73

9

4

5

21

3

40

15

57

8

74

4

5

5

24

2

41

9

58

11

75

8

6

2

25

1

42

8

59

16

76

7

7

4

26

4

43

9

60

19

77

5

8

5

27

4

44

5

61

5

78

6

9

1

28

2

45

12

62

16

79

3

10

5

29

1

46

8

63

9

80

2

11

1

30

6

47

7

64

10

81

3

12

2

31

5

48

11

65

10

82

2

13

1

32

3

49

12

66

9

83

1

14

0

33

8

50

25

67

6

84

2

15

2

34

5

51

9

68

12

85

1

16

3

35

15

52

9

69

7

87

1

17

1

36

4

53

7

70

13

88

1

a. Use this frequency table to determine the five-number summary of
these ages.

Min =

QL =

Median =

QU =

Max =

b. Check for outliers, according to the 1.5 × IQR criterion.

There are low outliers. There are high outliers.

c. Produce a well-labeled, modified boxplot of the distribution of
ages.

Explanation & Answer

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