Question

A random sample found that 30% of 150 Americans were satisfied with the gun control laws in 2018. Compute a 98% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2018 Fill in the blanks appropriately. A 98% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2018 is ( , ) (round to 3 decimal places)

Answer 1

Answer:

A 98% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2018 is (0.213, 0.387).

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:

$n = 150, \pi = 0.3$

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:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3 - 2.327\sqrt{\frac{0.3*0.7}{150}} = 0.213$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3 + 2.327\sqrt{\frac{0.3*0.7}{150}} = 0.387$

A 98% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2018 is (0.213, 0.387).