##### Data Analysis and Decision Modelling Oz Assignments

Delivery in day(s): 4

**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

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

Number of Older Brother(s)

Number of Younger Brother(s)

Number of Older Sister(s)

Number of Younger Sister(s)

IQ Score

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

| |||||||||||||

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 |

**C. Graph showing the mean IQ Score by gender**

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.

1. Population mean of the number of siblings () is 1.8

2. Population standard Deviation of the number of siblings) is 0.78

Since,Population Standard Deviation of the number of standard deviation of siblings) 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)**

Therefore z-computed is 0.573

**To**make decision critical of z at, is needed,

**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

**1. Number of new male and female participants.**

The proportion of males is 44% of 25,=

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

Also, the proportion of females is 56% of 25

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

**2. New mean IQ Score of the five new participants**

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

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

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

Mean IQ score the new recruit will be

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

**New standard deviation of IQ Score for new sample, given****and**

Using the two given parameters, variance can determined

Therefore,

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

Hence the standard deviation ofIQ Score for the new sample is 13.467

**A. Given that a person has a-score of 0.5**

1. Computation of his/her raw IQ Score

According toBlack (2009), -Score is given by the formula

Hence, the raw IQ Score is 107.5

**2. Relative position for IQ Score in the population**

This will be given by

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

**B. Given that the sample size is 36, compute:**

**1. The standard error of the mean**

It was given by

Hence, the standard error of the mean is 2.5

2. **Analyze and determine the probability that the mean of IQ Score, from a random sample with****N****= 36, is > 102.5**

Required probability is

Let, denote the participants IQ Score from the population sample. Assuming normal distribution forwith mean 100 and standard deviation 15 then central limit theorem, mean of the sample given by, will follow a normal distribution, with mean of 100 and standard deviation of(Hassett & Stewart, 2006).

From these,will be computed using z-score as follow

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

**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.**

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.

Therefore, for, the number of people to be sampled from the population will be 195.

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

2. Goos, P., & Meintrup, D. (2016). *Statistics with JMP: Hypothesis Tests, ANOVA and Regression*. John Wiley & Sons.

3. Hassett, M. J., & Stewart, D. (2006). *Probability for risk management*. Actex Publications. Ruppert, D. (2014). *Statistics and finance: an introduction*. Springer.