Answer:
There is a relationship between the age of a young adult and the type of movie preferred.
Step-by-step explanation:
In this case we need to test whether there is a significance relationship between the age of a young adult and the type of movie preferred.
The hypothesis can be defined as:
<em>H₀</em>: Age of a young adult and movie preference are independent.
<em>Hₐ</em>: Age of a young adult and movie preference are not independent.
The test statistic is given as follows:
![\chi^{2}=\sum\limits^{n}_{i=1}{\frac{(O_{i}-E_{i})^{2}}{E_{i}}}](https://tex.z-dn.net/?f=%5Cchi%5E%7B2%7D%3D%5Csum%5Climits%5E%7Bn%7D_%7Bi%3D1%7D%7B%5Cfrac%7B%28O_%7Bi%7D-E_%7Bi%7D%29%5E%7B2%7D%7D%7BE_%7Bi%7D%7D%7D)
Consider the table below.
The formula to compute the expected frequencies is:
![E_{i}=\frac{\text{Row Total}\times \text{Column Total}}{\text{Total}}](https://tex.z-dn.net/?f=E_%7Bi%7D%3D%5Cfrac%7B%5Ctext%7BRow%20Total%7D%5Ctimes%20%5Ctext%7BColumn%20Total%7D%7D%7B%5Ctext%7BTotal%7D%7D)
The Chi-square test statistic value is:
![\chi^{2}=\sum\limits^{n}_{i=1}{\frac{(O_{i}-E_{i})^{2}}{E_{i}}}=2.6013](https://tex.z-dn.net/?f=%5Cchi%5E%7B2%7D%3D%5Csum%5Climits%5E%7Bn%7D_%7Bi%3D1%7D%7B%5Cfrac%7B%28O_%7Bi%7D-E_%7Bi%7D%29%5E%7B2%7D%7D%7BE_%7Bi%7D%7D%7D%3D2.6013)
The significance level of the test is, <em>α</em> = 0.05.
The degrees of freedom of the test is,
df = (r - 1)(c - 1)
= (3 - 1)(3 - 1)
= 2 × 2
= 4
Compute the <em>p</em>-value as follows:
<em>p</em>-value = 0.6266
*Use a Chi square table.
As the <em>p-</em>value is more than the significance level the null hypothesis was failed to be rejected.
Thus, concluding that there is a relationship between the age of a young adult (18 to 35 years old) and the type of movie preferred.