Answer:
His 95% confidence interval is (0.065, 0.155).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
For this problem, we have that:
[tex]n = 186, \pi = 0.11[/tex]
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.11 - 1.96\sqrt{\frac{0.11*0.89}{186}} = 0.065[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.11 + 1.96\sqrt{\frac{0.11*0.89}{186}} = 0.155[/tex]
His 95% confidence interval is (0.065, 0.155).
Based on the standard deviation, the sample proportion and the population, the 95% confidence interval is  (0.1035, 0.1165).
A 95% confidence interval means a Z score of 1.96.
The 95% interval lower limit is:
= Sample proportion - Z score x Standard deviation / √population size
= 0.11 - 1.96 x 0.045 / √186
= 0.1035
Upper limit is:
=  0.11 + 1.96 x 0.045 / √186
= 0.1165
95% Confidence interval is (0.1035, 0.1165).
Find out more on Confidence intervals at https://brainly.com/question/26658887.
