Question

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. n=83, x=45, 98 percent
Could you please explain the steps and how to get to an answer? Thank you!

Answer 1

Answer:

The 98% confidence interval for the population proportion p is (0.4149, 0.6695).

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:

$n = 83, \pi = \frac{45}{83} = 0.5422$

98% confidence level

So $\alpha = 0.02$, z is the value of Z that has a pvalue of $1 - \frac{0.02}{2} = 0.99$, so $Z = 2.327$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5422 - 2.327\sqrt{\frac{0.5422*0.4578}{83}} = 0.4149$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5422 + 2.327\sqrt{\frac{0.5422*0.4578}{83}} = 0.6695$

The 98% confidence interval for the population proportion p is (0.4149, 0.6695).