Answer:
The calculated value Z = 0.063< 1.96 at 5% level of significance
Null hypothesis is accepted
There is no significant difference between age and watching television in teenagers
Explanation:
Step(i):-
Given random sample size 'n' = 905
Given data 198 say they watch 3 or more hours of television per day.
Given random first sample size
n₁ = 905
First sample proportion
![p_(1) = (x_(1) )/(n_(1) ) = (198)/(905) = 0.2187](https://img.qammunity.org/2021/formulas/mathematics/college/29lilmv9afvf4hmp0jhh57kw58evvhniwj.png)
Given random second sample size
n₂ = 503
second sample proportion
![p_(2) = (x_(2) )/(n_(1) ) = (97)/(503) = 0.1928](https://img.qammunity.org/2021/formulas/mathematics/college/qu4czjnj8gn7yofp9qn8rcmxl2bejg5rky.png)
Step(ii):-
Null Hypothesis : H₀
There is no significant difference between age and watching television in teenagers
Alternative Hypothesis :H₁
There is significant difference between age and watching television in teenagers
Step(iii):-
Test statistic
![Z = \frac{p_(1)-p_(2) }{\sqrt{PQ((1)/(n_(1) )+(1)/(n_(2) ) }) }](https://img.qammunity.org/2021/formulas/mathematics/college/3o85mxr7k0p829g6tfuhxqeekrpqy6bxdd.png)
Where
![P = (n_(1)p_(1) + n_(2) p_(2) )/(n_(1)+n_(2) )](https://img.qammunity.org/2021/formulas/mathematics/college/puhux93o3a9rh67fytwx1lb24tx7v2q04d.png)
P = 0.2094
Q = 1 - 0.2094 = 0.7906
![Z = \frac{0.2187-0.1928}{\sqrt{0.2094 X0.7906((1)/(905) +(1)/(503) } )}](https://img.qammunity.org/2021/formulas/mathematics/college/z1cc12cpdqoaozd41rvbhdqt8vybmmmxbf.png)
on calculation , we get
Z = 0.063
The critical value Z₀.₀₅ = 1.96
The calculated value Z = 0.063< 1.96 at 5% level of significance
Conclusion:-
Null hypothesis is accepted
There is no significant difference between age and watching television in teenagers