Question

It is important that face masks used by firefighters be able to withstand high temperatures because firefighters commonly work in temperatures of 200-500 degrees. In a test of one type of mask, 24 of 55 were found to have their lenses pop out at 325 degrees. Construct and interpret a 93% confidence interval for the true proportion of masks of this type whose lenses would pop out at 325 degrees.

Answer 1

The 93% confidence interval for the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574). This means that we are 93% sure that the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574).

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{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

For this problem, we have that:

`n = 55, \pi = (24)/(55) = 0.4364`

93% confidence level

So
`\alpha = 0.07`, z is the value of Z that has a pvalue of
`1 - (0.07)/(2) = 0.965`, so
`Z = 1.81`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.4364 - 1.81\sqrt{(0.4364*0.5636)/(55)} = 0.3154`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.4364 + 1.81\sqrt{(0.4364*0.5636)/(55)} = 0.5574`

The 93% confidence interval for the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574). This means that we are 93% sure that the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574).