Respuesta :
Answer: A. (0.432, 0.629)
Step-by-step explanation:
Let [tex]\hat{p}[/tex] be the sample proportion.
As peer given, we have
n= 98
[tex]\hat{p}=\dfrac{52}{98}=\dfrac{26}{49}[/tex]
Critical value for 95% confidence interval : [tex]z_{\alpha/2}=1.960[/tex]
Formula for confidence interval :-
[tex]\dfrac{26}{49}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
i.e. [tex]\dfrac{26}{49}\pm (1.96)\sqrt{\dfrac{\dfrac{26}{49}(1-\dfrac{26}{49})}{98}}[/tex]
[tex]\approx\dfrac{26}{49}\pm 0.0988092369474= (\dfrac{26}{49}+0.0988092369474,\ 0.531+0.0988092369474)\\\\=(0.431803007951,\ 0.629421481845)\\\\\approx(0.432,\ 0.629)[/tex]
Hence, the correct 95% confidence interval : A. (0.432, 0.629)
Using the z-distribution, as we are working with a proportion, it is found that the 95% confidence interval for the proportion of all Americans who have been active in a veteran's organization is:
A. (0.432, 0.629).
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, the parameters are:
[tex]n = 98, \pi = \frac{52}{98} = 0.531, z = 1.96[/tex]
Hence:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.531 - 1.96\sqrt{\frac{0.531(0.469)}{98}} = 0.432[/tex]
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.531 + 1.96\sqrt{\frac{0.531(0.469)}{98}} = 0.629[/tex]
Hence option A is correct.
More can be learned about the z-distribution at https://brainly.com/question/25730047
