Posted: August 27th, 2021
Applied Statistic Discussion 1-1
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Applied Statistic Discussion 1-1
Option 2
P1 and P2 are expected to be the two groups of students from the population as below;
Group 1 is the student’s names whose last name begins with A-K.
Group 2 is the student’s names whose last name begins with L-Z.
Let: be the size of group 1
: be the size of group 2
That is
Where: is the total number of students in the whole class.
From the above proportions, the variables can be expressed as follows;
According to Kosambi (2016), the probability of the whole population is equal to one and the proportion of n1 for group 1 (A-K) and owns an iPhone is= 0.08
Similarly, the proportion for group 2 (L-Z) and owns an iPhone is= 0.06
Therefore, the total probability of the two groups is
The probability of every student owning an iPhone in the whole population is, that is
Thus, the probability of the two groups is different from the whole population.
Consequently, P1 and P2 are not equal as the number of students whose last name begins with A-K is not the same as the number of students whose last name begins with (L-Z). Hence, the probability for all alphabet numbers is different, and the probability of group 1 is not equal to the probability of group 2. Thus
In conclusion, the student’s names whose last name begins with A-K and L-Z are not related to owning an iPhone, and the two variables are independent (Kaliyadan, & Kulkarni, 2019). The following assumptions are made to draw the above conclusion on the two variables;
Figure
1: Group one and two proportion
References
Kaliyadan, F., & Kulkarni, V. (2019). Types of variables, descriptive statistics, and sample size. Indian dermatology online journal, 10(1), 82.
Kosambi, D. D. (2016). Statistics in function space. In DD Kosambi (pp. 115-123). Springer, New Delhi.
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