Question

Suppose a sample of 80 with a sample proportion of 0.58 is taken from a population. Which of the following is the approximate 95% confidence interval for the population parameter?

Answer 1

Answer:

The 95% confidence interval for the population parameter is (0.4718, 0.6882).

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 interval $1-\alpha$, 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:

$n = 80, p = 0.58$

95% confidence interval

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $z = 1.96$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.58 - 1.96\sqrt{\frac{0.58*0.42}{80}} = 0.4718$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.58 + 1.96\sqrt{\frac{0.58*0.42}{80}} = 0.6882$

The 95% confidence interval for the population parameter is (0.4718, 0.6882).

Answer 2

Answer:

Correct answer is 0.470 0.690.  

Step-by-step explanation:

Just got it right on apex.