Question

In a random sample of 200 people, 154 said that they watched educational television. Find the 90% confidence interval of the true proportion of people who watched educational television.

Answer 1

Answer:

The 90% confidence interval of the true proportion of people who watched educational television is (0.7117, 0.8283)

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:

$n = 200, \pi = \frac{154}{200} = 0.77$

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:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 - 1.96\sqrt{\frac{0.77*0.23}{200}} = 0.7117$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.77 + 1.96\sqrt{\frac{0.77*0.23}{200}} = 0.8283$

The 90% confidence interval of the true proportion of people who watched educational television is (0.7117, 0.8283)