Data Analysis Assignment on Birth Order and Intelligence Study

This document presents a comprehensive data analysis assignment solution, focusing on statistical analysis of data related to birth order and IQ scores. The assignment begins by identifying the appropriate scales of measurement for various variables, including participant ID, gender, number of siblings, and IQ scores. Descriptive statistics, such as mean, median, standard deviation, and variance, are calculated and compared for male and female participants. The analysis includes the creation and interpretation of a box plot to visualize the distribution of IQ scores by gender. The solution further explores hypothesis testing using z-scores to determine if the population mean of the number of siblings is greater than a specified value. The document then addresses scenarios involving new participants, calculating the number of new male and female recruits, the new mean IQ score, and the new standard deviation. Finally, the assignment delves into probability calculations, determining raw IQ scores based on z-scores, assessing the relative position of IQ scores within the population, and calculating the standard error of the mean. The solution also computes probabilities related to sample means and determines the required sample size to achieve a specific probability level. The analysis utilizes statistical formulas and concepts, providing detailed interpretations and conclusions based on the provided data.

Data Analysis
Data analysis Paper
Student’s Name
Institution Affiliation

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

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

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.

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

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

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

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

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 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

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

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.