Answer:
The 93% confidence interval of the true proportion of people who watched educational TV
(0.73716 , 0.80684)
Step-by-step explanation:
Step(i):-
Given sample size 'n' = 145
A random sample of 145 people, 112 said that they watched educational TV.
Sample proportion
[tex]p^{-} = \frac{112}{145} = 0.772[/tex]
Step(ii):-
The 93% confidence interval of the true proportion of people who watched educational TV
[tex](p- Z_{0.07}\sqrt{\frac{pq}{n} } , p + Z_{0.07} \sqrt{\frac{pq}{n} } )[/tex]
[tex](0.772- 1.427\sqrt{\frac{0.772X0.228}{145} } , 0.772 + 1.427 \sqrt{\frac{0.772X0.228}{145} } )[/tex]
( 0.772 - 0.03484 , 0.772 + 0.03484)
(0.73716 , 0.80684)
Final answer:-
The 93% confidence interval of the true proportion of people who watched educational TV
(0.73716 , 0.80684)
