Question

On the basis of sample of 25 students in a large campus laboratory, it appears that 8 students smoke. A 95% CI estimate of the proportion of smokers in the lab is: a. [0.137, 0.503] b. [0.037, 0.050] c. [1.137, 1.503]

Answer 1

Answer:

a. [0.137, 0.503]

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:

A sample of 25 students in the lab, of whom 8 smoke. So $n = 25, \pi = \frac{8}{25} = 0.32$

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:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.32 - 1.96\sqrt{\frac{0.32*0.68}{25}} = 0.137$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.32 + 1.96\sqrt{\frac{0.32*0.68}{25}}{119}} = 0.503$

The correct answer is:

a. [0.137, 0.503]