Question

Construct a confidence interval of the population proportion at the given level of confidence. x =860​, n =1100​, 96​% confidence.
a) lower bound of the confidence interval is: ________

b) upper bound of the confidence interval is:________

Answer 1

Answer:

a) 0.7562

b) 0.8074

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

$n = 1100, \pi = \frac{860}{1100} = 0.7818$

96% confidence level

So $\alpha = 0.04$, z is the value of Z that has a pvalue of $1 - \frac{0.04}{2} = 0.98$, so $Z = 2.054$.  

a) The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.7818 - 2.054\sqrt{\frac{0.7818*0.2182}{1100}} = 0.7562$

B) The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.7818 + 2.054\sqrt{\frac{0.7818*0.2182}{1100}} = 0.8074$