# Data Analysis Paper: Descriptive Statistics, Hypothesis Testing, and Probability

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This paper discusses the appropriate scale of measurement for variables, descriptive statistics for male and female, a hypothesis test at a 5% significance level, and probability calculations. It includes graphs, tables, and computations using SPSS software.

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Data Analysis

Data analysis Paper

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Data analysis Paper

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Data Analysis

Figure 1: Data View and Variable View

Question One

a. Identifying the appropriated scale of measurement:

Below is a table showing the variable plus their associated scale of measurement

Variable Appropriate scale of measurement

i. Participant ID Nominal

ii. Gender Nominal

iii. Number of Older brother(s) Ratio

iv. Number of Younger Sister(s) Ratio

v. IQ Score Interval

Figure 1: Data View and Variable View

Question One

a. Identifying the appropriated scale of measurement:

Below is a table showing the variable plus their associated scale of measurement

Variable Appropriate scale of measurement

i. Participant ID Nominal

ii. Gender Nominal

iii. Number of Older brother(s) Ratio

iv. Number of Younger Sister(s) Ratio

v. IQ Score Interval

Data Analysis

b. Descriptive Statistics for male and female. Variables to be considered:

i. Number of Older Brother(s)

ii. Number of Younger Brother(s)

iii. Number of Older Sister(s)

iv. Number of Younger Sister(s)

v. IQ Score

The statistics were computed using SPSS software. The table below is a summary of the results

obtained.

Report

Gender Number

Of Older

Brothers

Number of

Younger

Brother

Number of

Older Sisters

Number of

Younger

Sisters

Total

Number

Siblings

IQ Score

Female

Mean .40 .70 .30 .50 1.90 109.90

N 10 10 10 10 10 10

Std. Deviation .699 .675 .675 .527 1.197 14.715

Median .00 1.00 .00 .50 2.00 109.50

Range 2 2 2 1 4 44

Variance .489 .456 .456 .278 1.433 216.544

Male

Mean .40 .60 .30 .60 1.90 109.60

N 10 10 10 10 10 10

Std. Deviation .699 .699 .675 .843 .994 14.501

Median .00 .50 .00 .00 2.00 112.50

Range 2 2 2 2 3 39

Variance .489 .489 .456 .711 .989 210.267

Total

Mean .40 .65 .30 .55 1.90 109.75

N 20 20 20 20 20 20

Std. Deviation .681 .671 .657 .686 1.071 14.220

Median .00 1.00 .00 .00 2.00 112.50

Range 2 2 2 2 4 44

Variance .463 .450 .432 .471 1.147 202.197

b. Descriptive Statistics for male and female. Variables to be considered:

i. Number of Older Brother(s)

ii. Number of Younger Brother(s)

iii. Number of Older Sister(s)

iv. Number of Younger Sister(s)

v. IQ Score

The statistics were computed using SPSS software. The table below is a summary of the results

obtained.

Report

Gender Number

Of Older

Brothers

Number of

Younger

Brother

Number of

Older Sisters

Number of

Younger

Sisters

Total

Number

Siblings

IQ Score

Female

Mean .40 .70 .30 .50 1.90 109.90

N 10 10 10 10 10 10

Std. Deviation .699 .675 .675 .527 1.197 14.715

Median .00 1.00 .00 .50 2.00 109.50

Range 2 2 2 1 4 44

Variance .489 .456 .456 .278 1.433 216.544

Male

Mean .40 .60 .30 .60 1.90 109.60

N 10 10 10 10 10 10

Std. Deviation .699 .699 .675 .843 .994 14.501

Median .00 .50 .00 .00 2.00 112.50

Range 2 2 2 2 3 39

Variance .489 .489 .456 .711 .989 210.267

Total

Mean .40 .65 .30 .55 1.90 109.75

N 20 20 20 20 20 20

Std. Deviation .681 .671 .657 .686 1.071 14.220

Median .00 1.00 .00 .00 2.00 112.50

Range 2 2 2 2 4 44

Variance .463 .450 .432 .471 1.147 202.197

Data Analysis

c. Graph showing the mean IQ Score by gender

Box plot compares the median, and quartiles (first and third). From the boxplot, it’s clear that

median of IQ Score of males is higher than that of females. Also, the interquartile range of the IQ

Score of males is bigger due to the elongated box than that of Female. This implies that there’s

great dispersion of IQ Score in males than in females.

c. Graph showing the mean IQ Score by gender

Box plot compares the median, and quartiles (first and third). From the boxplot, it’s clear that

median of IQ Score of males is higher than that of females. Also, the interquartile range of the IQ

Score of males is bigger due to the elongated box than that of Female. This implies that there’s

great dispersion of IQ Score in males than in females.

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Data Analysis

Question Two

Data:

Population mean of the number of siblings ( μ) is 1.8

Population standard Deviation of the number of siblings σ ) is 0.78

a. Hypothesis test at 5% significance level

H0 : μ=1.8

H0 : μ>1.8

Since, Population Standard Deviation of the number of standard deviation of siblingsσ ) is

known the hypothesis test will be based on z-score (Goos& Meintrup, 2016). . The table below

shows he results of computation of the sample means of the total number of siblings that each

participant had.

From the table in part b of question one, the sample mean of the number of siblings is 1.90.

Computation of z-score (z)

z x−μ

σ

√ n

= 1.90−1.8

0.78

√ 20

¿ 0.573

Therefore z-computed is 0.573

To make decision critical of z at α=0.05, is needed, zα =0.05 ¿−tailed ¿=1.645

Interpretation

Since, z-computed, 0.573, is less than critical z, 1.645, null hypothes will not be rejected (Rupert,

2014). This implies that the population mean is not greater than 1.8

Question Two

Data:

Population mean of the number of siblings ( μ) is 1.8

Population standard Deviation of the number of siblings σ ) is 0.78

a. Hypothesis test at 5% significance level

H0 : μ=1.8

H0 : μ>1.8

Since, Population Standard Deviation of the number of standard deviation of siblingsσ ) is

known the hypothesis test will be based on z-score (Goos& Meintrup, 2016). . The table below

shows he results of computation of the sample means of the total number of siblings that each

participant had.

From the table in part b of question one, the sample mean of the number of siblings is 1.90.

Computation of z-score (z)

z x−μ

σ

√ n

= 1.90−1.8

0.78

√ 20

¿ 0.573

Therefore z-computed is 0.573

To make decision critical of z at α=0.05, is needed, zα =0.05 ¿−tailed ¿=1.645

Interpretation

Since, z-computed, 0.573, is less than critical z, 1.645, null hypothes will not be rejected (Rupert,

2014). This implies that the population mean is not greater than 1.8

Data Analysis

b. Given that five new participant were recruited into the study. The proportion of

males and females in the sample 44% and 55% respectively.

The new mean IQ Score is 108.8

i. Number of new male and female participants.

Total number of Parcipants=Intialnumber +new recruited number

¿ 20+5=25

The proportion of males is 44% of 25,= 44

100∗25=11 Males

Since the initial number of male was 10, then the number of newly recruited male is

11−10=1 male

Also, the proportion of females is 56% of 25

56∗25

100 =14 Females

Therefore, since the initial number of females was10, then, the newly recruited females is

14−10=4 Females

ii. New mean IQ Score of the five new participants

First, the percentage increase of the number of participants will be computed

Number of Recruits

Intial Number of Participants∗100 %= 5

20∗100 %=25 %

This implies that after the new recruitment, the mean of IQ Score of participants will be 1.25

times the initial mean of IQ Score number of participants

1.25∗109.75=137.875

b. Given that five new participant were recruited into the study. The proportion of

males and females in the sample 44% and 55% respectively.

The new mean IQ Score is 108.8

i. Number of new male and female participants.

Total number of Parcipants=Intialnumber +new recruited number

¿ 20+5=25

The proportion of males is 44% of 25,= 44

100∗25=11 Males

Since the initial number of male was 10, then the number of newly recruited male is

11−10=1 male

Also, the proportion of females is 56% of 25

56∗25

100 =14 Females

Therefore, since the initial number of females was10, then, the newly recruited females is

14−10=4 Females

ii. New mean IQ Score of the five new participants

First, the percentage increase of the number of participants will be computed

Number of Recruits

Intial Number of Participants∗100 %= 5

20∗100 %=25 %

This implies that after the new recruitment, the mean of IQ Score of participants will be 1.25

times the initial mean of IQ Score number of participants

1.25∗109.75=137.875

Data Analysis

To computed mean IQ Score of the five participants, initial total of IQ score for the first 20

participants and total IQ score for the new number of participants is required

Initial IQ Score Total=Sample ¿ IQ score=20∗109.75=2195

new IQ ScoreTotal=25∗137.875=3446.875

Mean IQ score the new recruit will be

New IQ ScoreTotal−Initial IQ Score Total

5

¿ 3446.875−2195

5 =250.37 5

Therefore, the mean IQ score of the five new participants is approximately, 250

iii. New standard deviation of IQ Score for new sample, given ∑ x2=30050 5 and

(∑ X)2

=7403814

Using the two given parameters, variance can determined

s2=n¿ ¿

¿ n ¿ ¿

¿ 7512625−7403814

600

¿ 108811

600 =181.35 2

Therefore, s2=181.352

To computed mean IQ Score of the five participants, initial total of IQ score for the first 20

participants and total IQ score for the new number of participants is required

Initial IQ Score Total=Sample ¿ IQ score=20∗109.75=2195

new IQ ScoreTotal=25∗137.875=3446.875

Mean IQ score the new recruit will be

New IQ ScoreTotal−Initial IQ Score Total

5

¿ 3446.875−2195

5 =250.37 5

Therefore, the mean IQ score of the five new participants is approximately, 250

iii. New standard deviation of IQ Score for new sample, given ∑ x2=30050 5 and

(∑ X)2

=7403814

Using the two given parameters, variance can determined

s2=n¿ ¿

¿ n ¿ ¿

¿ 7512625−7403814

600

¿ 108811

600 =181.35 2

Therefore, s2=181.352

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Data Analysis

To obtain the standard deviation (s ¿, square root of the variance will be determined

s= √s2

¿ √ 181.352=13.46 7

Hence the standard deviation of IQ Score for the new sample is 13.467

Question Three

To obtain the standard deviation (s ¿, square root of the variance will be determined

s= √s2

¿ √ 181.352=13.46 7

Hence the standard deviation of IQ Score for the new sample is 13.467

Question Three

Data Analysis

Data: Population mean of IQ Score is 100 and Standard deviation is 15

a. Given that a person has a z-score of 0.5

i. Computation of his/her raw IQ Score

According to Black (2009), z-Score is given by the formula

z= x−μ

σ

0.5= x−100

15

7.5= x−10 0

x=107. 5

Hence, the raw IQ Score is 107.5

ii. Relative position for IQ Score in the population

This will be given by

100 ±15 ( z )=100 ± 15 ( 0.5 )

¿ 100 ±7. 5

This gives the range of the his/her IQ score in the population as (92.5 , 107.5)

b. Given that the sample size is 36, compute:

i. Standard error of the mean

It given by

Standard error ( mean ) = Standard deviation

√ Sample ¿ n ¿ ¿= 15

√ 36 =2. 5

Data: Population mean of IQ Score is 100 and Standard deviation is 15

a. Given that a person has a z-score of 0.5

i. Computation of his/her raw IQ Score

According to Black (2009), z-Score is given by the formula

z= x−μ

σ

0.5= x−100

15

7.5= x−10 0

x=107. 5

Hence, the raw IQ Score is 107.5

ii. Relative position for IQ Score in the population

This will be given by

100 ±15 ( z )=100 ± 15 ( 0.5 )

¿ 100 ±7. 5

This gives the range of the his/her IQ score in the population as (92.5 , 107.5)

b. Given that the sample size is 36, compute:

i. Standard error of the mean

It given by

Standard error ( mean ) = Standard deviation

√ Sample ¿ n ¿ ¿= 15

√ 36 =2. 5

Data Analysis

Hence, the standard error of the mean is 2.5

ii. Analyze and determine the probability that the mean of IQ Score, from a

random sample with N = 36, is > 102.5

Required probability is

P( X> 102.5)

Let Xi =x1 , x2 , … … … x36, denote the participants IQ Score from the population sample.

Assuming normal distribution for Xi with mean 100 and standard deviation 15 then central limit

theorem, mean of the sample given by X = 1

25 ( x1+ x2 +… . , … .+x36 ), will follow a normal

distribution, with mean of 100 and standard deviation of σ

√n (Hassett & Stewart, 2006).

σ

√n = 15

√36 =2.5

From these, P( X>102.5) will be computed using z-score as follow

P ( X >102.5 ) =P (( z = X−100

2.5 ) > ( z= 102.5−100

2.5 ))

¿ P ( ( z= X−100

2.5 ) > ( z< 1 ) )

¿ 1−P ( z <1 )=1−0.8413

¿ 0.158 7

Hence, the probability that mean IQ Score is greater than 102.5 is 0.1587

Hence, the standard error of the mean is 2.5

ii. Analyze and determine the probability that the mean of IQ Score, from a

random sample with N = 36, is > 102.5

Required probability is

P( X> 102.5)

Let Xi =x1 , x2 , … … … x36, denote the participants IQ Score from the population sample.

Assuming normal distribution for Xi with mean 100 and standard deviation 15 then central limit

theorem, mean of the sample given by X = 1

25 ( x1+ x2 +… . , … .+x36 ), will follow a normal

distribution, with mean of 100 and standard deviation of σ

√n (Hassett & Stewart, 2006).

σ

√n = 15

√36 =2.5

From these, P( X>102.5) will be computed using z-score as follow

P ( X >102.5 ) =P (( z = X−100

2.5 ) > ( z= 102.5−100

2.5 ))

¿ P ( ( z= X−100

2.5 ) > ( z< 1 ) )

¿ 1−P ( z <1 )=1−0.8413

¿ 0.158 7

Hence, the probability that mean IQ Score is greater than 102.5 is 0.1587

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Data Analysis

c. If we want to the probability of the sample mean > 102.5 to be 1%, how many

people should we randomly sample from the population? Apply the necessary

formula to obtain the answer.

P ( X >102.5 ) =0.0 1

First, find the z-score whose probability is 0.01, from tables, which -2.325. Here the standardized

formula for finding z-score will play a major role.

z= X −μ

σ

√ n

−2.325=102.5−100

15

√ n

( 34.875 )2= ( 2.5 )2 n

n= ( 34.875 )2

( 2.5 ) 2 =194.6025 ≈ 19 5

Therefore, forP ( X >102.5 ) =0.01, the number of people to be sampled from the population will

be 195.

References

c. If we want to the probability of the sample mean > 102.5 to be 1%, how many

people should we randomly sample from the population? Apply the necessary

formula to obtain the answer.

P ( X >102.5 ) =0.0 1

First, find the z-score whose probability is 0.01, from tables, which -2.325. Here the standardized

formula for finding z-score will play a major role.

z= X −μ

σ

√ n

−2.325=102.5−100

15

√ n

( 34.875 )2= ( 2.5 )2 n

n= ( 34.875 )2

( 2.5 ) 2 =194.6025 ≈ 19 5

Therefore, forP ( X >102.5 ) =0.01, the number of people to be sampled from the population will

be 195.

References

Data Analysis

Black, K. (2009). Business statistics: Contemporary decision making. John Wiley & Sons.

Goos, P., & Meintrup, D. (2016). Statistics with JMP: Hypothesis Tests, ANOVA and

Regression. John Wiley & Sons.

Hassett, M. J., & Stewart, D. (2006). Probability for risk management. Actex Publications.

Ruppert, D. (2014). Statistics and finance: an introduction. Springer.

Black, K. (2009). Business statistics: Contemporary decision making. John Wiley & Sons.

Goos, P., & Meintrup, D. (2016). Statistics with JMP: Hypothesis Tests, ANOVA and

Regression. John Wiley & Sons.

Hassett, M. J., & Stewart, D. (2006). Probability for risk management. Actex Publications.

Ruppert, D. (2014). Statistics and finance: an introduction. Springer.

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