Question

A board-game club meets each week at the local library. People bring different board games to play, like checkers and chess. Out of the 50 people that meet each week, 67% prefer to play checkers over chess. Construct an 83% confidence interval for the population mean of people that prefer to play checkers over chess.

Answer 1

Using the z-distribution, as we are working with a proportion, it is found that the 83% confidence interval is (0.5789, 0.7611).

What is a confidence interval of proportions?

A confidence interval of proportions is given by:

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which:

• $\pi$ is the sample proportion.

• z is the critical value.

• n is the sample size.

In this problem, we have an 83% confidence level, hence$\alpha = 0.93$, z is the value of Z that has a p-value of $\frac{1+0.83}{2} = 0.915$, so the critical value is z = 1.37.

The other parameters are $n = 50, \pi = 0.67$.

Then, the bounds of the interval are given as follows:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.67 - 1.37\sqrt{\frac{0.67(0.33)}{50}} = 0.5789$

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.67 + 1.37\sqrt{\frac{0.67(0.33)}{50}} = 0.7611$

The 83% confidence interval is (0.5789, 0.7611).

More can be learned about the z-distribution at brainly.com/question/25890103