Question

Construct the 90% confidence interval for the proportion of students at the college who have completed their required English 101 course. Enter your answers as decimals (not percents) accurate to three decimal places. The Confidence Interval is

Answer 1

Answer:

The confidence interval has an lower limit of $\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = \pi - 1.645\sqrt{\frac{\pi(1-\pi)}{n}}$ and an upper limit of $\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = \pi + 1.645\sqrt{\frac{\pi(1-\pi)}{n}}$, in which $\pi$ is the sample proportion and $n$ is the size of the sample.

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 z-score that has a p-value of $1 - \frac{\alpha}{2}$.

90% confidence level

So $\alpha = 0.1$, z is the value of Z that has a p-value of $1 - \frac{0.1}{2} = 0.95$, so $Z = 1.645$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = \pi - 1.645\sqrt{\frac{\pi(1-\pi)}{n}}$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = \pi + 1.645\sqrt{\frac{\pi(1-\pi)}{n}}$

The confidence interval has an lower limit of $\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = \pi - 1.645\sqrt{\frac{\pi(1-\pi)}{n}}$ and an upper limit of $\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = \pi + 1.645\sqrt{\frac{\pi(1-\pi)}{n}}$, in which $\pi$ is the sample proportion and $n$ is the size of the sample.