Question

From a large number of actuarial exam scores, a random sample of scores is selected, and it is found that of these are passing scores. Based on this sample, find a confidence interval for the proportion of all scores that are passing. Then find the lower limit and upper limit of the confidence interval. Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places.

Answer 1

Supposing 60 out of 100 scores are passing scores, the 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).

• The lower limit is 0.5.
• The upper limit is 0.7.

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $\alpha$, we have the following confidence interval of proportions.

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

In which

z is the z-score that has a p-value of $\frac{1+\alpha}{2}$.

60 out of 100 scores are passing scores, hence $n = 100, \pi = \frac{60}{100} = 0.6$

95% confidence level

So $\alpha = 0.95$, z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$, so $z = 1.96$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6 - 1.96\sqrt{\frac{0.6(0.4)}{100}} = 0.5$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6 + 1.96\sqrt{\frac{0.6(0.4)}{100}} = 0.7$

The 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).

• The lower limit is 0.5.
• The upper limit is 0.7.

A similar problem is given at brainly.com/question/16807970