Answer :
a) The 90% confidence interval for this statistic for males is of: (0.1497, 0.2103).
b) The 90% confidence level for this statistic is of: (0.3661, 0.4339).
What is a confidence interval of proportions?
A confidence interval of proportions has the bounds given by the rule presented as follows:
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which the variables used to calculated these bounds are listed as follows:
- [tex]\pi[/tex] is the sample proportion, which is also the estimate of the parameter.
- z is the critical value.
- n is the sample size.
The confidence level is of 90%, hence the critical value z is the value of Z that has a p-value of [tex]\frac{1+0.9}{2} = 0.95[/tex], so the critical value is z = 1.645.
For males, the parameters are given as follows:
[tex]\pi = 0.18, n = 435[/tex]
Hence the lower bound of the interval is given as follows:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.18 - 1.645\sqrt{\frac{0.18(0.82)}{435}} = 0.1497[/tex]
The upper bound of the interval is of:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.18 + 1.645\sqrt{\frac{0.18(0.82)}{435}} = 0.2103[/tex]
For the statistic, the parameters are given as follows:
[tex]\pi = 0.4, n = 566[/tex]
Hence the lower bound of the interval is given as follows:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4 - 1.645\sqrt{\frac{0.4(0.6)}{566}} = 0.3661[/tex]
The upper bound of the interval is of:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4 + 1.645\sqrt{\frac{0.4(0.6)}{566}} = 0.4339[/tex]
More can be learned about the z-distribution at https://brainly.com/question/25890103
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