STATISTICAL ANALYSIS

Assignment



Assignment

Answer 1

(a), (b), (c)

Statistics

Incubation period

N

Valid

7

Missing

7

Mean

10.2143

Median

10.5000

Std. Deviation

2.71442

(d)

Statistics

Incubation period

N

Valid

7

Missing

7

Mean

9.5286

Median

10.5000

Std. Deviation

4.11368

From the above table, it can be said that the mean value of incubation period has declined from 10.2 to 9.52. However, it is observed that the median value is same in both the cases that is median is not affected by the change of value of incubation period from 6.3 to 1.5. Moreover, it is also observed that the standard deviation of the incubation period is increased that is 2.71 to 4.11 which reflects that there is possibility of deviation from first exposure to HIV infection to AIDS diagnosis.

(e)

Statistics

Incubation period

N

Valid

14

Missing

0

Mean

9.8500

Median

10.0000

Std. Deviation

3.05885

It is assumed that the due to the increase in the sample size of incubation period for the first exposure to HIV infection to AIDS diagnosis, the mean value of the incubation period will increase which will affect the standard deviation of the incubation period that is it will decrease.

Thus, after increasing the sample size of incubation period for the first exposure to HIV infection to AIDS diagnosis, it is found that the mean value of incubation period decreased and standard deviation of the incubation period increased. This change in the mean and standard deviation of the incubation period might be taken place due to the reason that the sample data was randomly distributed.

Answer 2

The new sample standard deviation would tend to be smaller than the first and approximately about one- third the size.

Answer 3

(a)

Data:

X

=

< 60

Mean of X

=

77

SD

=

11.6

P (x)

=

X - µ

s

P (x)

=

60

-

77

11.6

P (x)

=

- 1.47

(b)

Data:

X

=

> 90

Mean of X

=

77

SD

=

11.6

P (x)

=

X - µ

s

P (x)

=

90

-

77

11.6

P (x)

=

1.12

P (x)

=

1 - 1.12

P (x)

=

- 0.12

(c)

Data:

X

=

between 60 and 90

Mean of X

=

77

SD

=

11.6

P (x)

=

X - µ

s

P (x)

=

60

-

77

11.6

P (x)

=

-1.47

P (y)

=

X - µ

s

P (y)

=

90

-

77

11.6

P (y)

=

1.12

P (y)

=

1 - 1.12

P (y)

=

-0.12

P (x < Z < y) = P (Z < x) - P (Z < y)

P (x < Z < y) = P (Z < 60 x) - P (Z < 90 y)

P (x < Z < y) = (- 1.47) - (- 0.12)

P (x < Z < y) = 1.59

Answer 4

(a)

Data:

n

=

200

C I

=

95 %

s

=

25

Mean of X

140

Mean of X

-

1.96

s

<

µ

<

Mean of X

+

1.96

s

Square root of n

Square root of n

140

-

1.96

25

<

µ

<

140

+

1.96

25

14.14

14.14

140

-

1.96

1.77

<

µ

<

140

+

1.96

1.77

140

-

3.46

<

µ

<

140

+

3.46

136.54

<

µ

<

143.46

(b)

Data:

n

=

100

C I

=

95 %

s

=

25

Mean of X

140

Mean of X

-

1.96

s

<

µ

<

Mean of X

+

1.96

s

Square root of n

Square root of n

140

-

1.96

25

<

µ

<

140

+

1.96

25

10.00

10.00

140

-

1.96

2.50

<

µ

<

140

+

1.96

2.50

140

-

4.90

<

µ

<

140

+

4.90

135.10

<

µ

<

144.90

From the above calculation of the interval, it can be said that the interval has become wider; the reason of this is that as the sample size decreases the interval becomes wide because increasing the sample size from the population gives more accurate results or the interval.

Answer 5

(a)

In the study, determining whether exposure to organochlorine DDT used as an insecticide for many years, is associated with breast cancer in women, quantitative study design is used in which non-probability sampling is used. The non-probability sampling comprises a broad category of approaches that are similar primarily in that subjective criteria are used in case selection rather than randomization (Joubert, & Ehrlich, Katzenellenbogen & Karim, 2007). The method in question provides no assurance that each item has a chance of being included ...