Question

36 out of 100 randomly selected taxpayers knew about tax incentives for installing energy-saving furnaces. Find a 90% confidence interval for the population proportion who knew about the incentives. (Round to two places after the decimal point.)

Answer 1

Answer:

The 90% confidence interval for the population proportion who knew about the incentives is (0.28, 0.44).

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:

$n = 100, \pi = \frac{36}{60} = 0.6$

90% confidence level

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

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.36 - 1.645\sqrt{\frac{0.36*0.64}{100}} = 0.28$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.36 + 1.645\sqrt{\frac{0.36*0.64}{100}} = 0.44$

The 90% confidence interval for the population proportion who knew about the incentives is (0.28, 0.44).