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:

Consider the table below.
The formula to compute the expected frequencies is:

The Chi-square test statistic value is:

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.