Respuesta :
Answer:
The 99% confidence interval for the true proportion of fish that are Black carp in Lake Ontario
(0.4828 , 0.7704)
Step-by-step explanation:
Step(i):-
Given that the fisherman responds that, of the 75 fish he has caught, 47 have been Black carp
Sample proportion = [tex]\frac{x}{n} = \frac{47}{75} = 0.6266[/tex]
We have to find the 99% confidence interval for the true proportion of fish that are Black carp in Lake Ontario
Step(ii):-
solution:-
Given that the level of significance = 0.01
Z-score = 2.576
The 99% confidence interval for the true proportion of fish that are Black carp in Lake Ontario
[tex](p^{-} - Z_{0.01} \sqrt{\frac{p(1-p)}{n} } , p^{-} +Z_{0.01} \sqrt{\frac{p(1-p)}{n} })[/tex]
[tex](0.6266 - 2.576 \sqrt{\frac{0.6266(1-0.6266)}{75} } ,0.6266 +2.576 \sqrt{\frac{0.6266(1-0.6266)}{75} })[/tex]
After calculation, we get
(0.6266 - 0.1438 , 0.6266 +0.1438)
(0.4828 , 0.7704)
Final answer:-
The 99% confidence interval for the true proportion of fish that are Black carp in Lake Ontario
(0.4828 , 0.7704)
Using the z-distribution, it is found that the lower bound of a 99% confidence interval for the true proportion of fish that are Black Carp in Lake Ontario is 0.4829.
What is a confidence interval of proportions?
A confidence interval of proportions is given by:
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which:
- [tex]\pi[/tex] is the sample proportion.
- z is the critical value.
- n is the sample size.
In this problem, for the parameters, we have that:
- 99% confidence level, hence[tex]\alpha = 0.99[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.99}{2} = 0.995[/tex], so [tex]z = 2.575[/tex].
- Of the 75 fish he has caught, 47 have been Black Carp, hence [tex]n = 75, \pi = \frac{47}{75} = 0.6267[/tex]
The lower bound of the interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6267 - 2.575\sqrt{\frac{0.6267(0.3733)}{75}} = 0.4829[/tex]
To learn more about the z-distribution, you can take a look at https://brainly.com/question/16236451
