Respuesta :
Using the z-distribution, as we are working with a proportion, the 90% confidence level estimate for the true proportion of employees in the district who have gotten a flu shot is (0.2355, 0.4784).
What is a confidence interval of proportions?
As long as there are 10 successes and 10 failures in the sample, that is, [tex]n\pi \geq 10[/tex] and [tex]n(1 - \pi) \geq 10[/tex], 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 = 42, \pi = \frac{15}{42} = 0.3571, z = 1.645[/tex]
We have the [tex]n\pi = 15, n(1 - \pi) = 27[/tex], hence you can find the interval.
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3571 - 1.645\sqrt{\frac{0.3571(0.6429)}{42}} = 0.2355[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3571 + 1.645\sqrt{\frac{0.3571(0.6429)}{42}} = 0.4787[/tex]
The 90% confidence level estimate for the true proportion of employees in the district who have gotten a flu shot is (0.2355, 0.4784).
More can be learned about the z-distribution at https://brainly.com/question/25890103
