Question

A newsgroup is interested in constructing a 90% confidence interval for the proportion of all Americans who are in favor of a new Green initiative. Of the 533 randomly selected Americans surveyed, 351 were in favor of the initiative. Round answers to 4 decimal places where possible.

Answer 1

Answer:

The 90% confidence interval for the proportion of all Americans who are in favor of a new Green initiative is (0.6247, 0.6923).

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 $1 - \frac{\alpha}{2}$.

Of the 533 randomly selected Americans surveyed, 351 were in favor of the initiative.

This means that $n = 533, \pi = \frac{351}{533} = 0.6585$

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:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6585 - 1.645\sqrt{\frac{0.6585*0.3415}{533}} = 0.6247$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6585 + 1.645\sqrt{\frac{0.6585*0.3415}{533}} = 0.6923$

The 90% confidence interval for the proportion of all Americans who are in favor of a new Green initiative is (0.6247, 0.6923).