Question

In a random sample of 810 women employees, it is found that 81 would prefer working for a female boss. The width of the 95% confidence interval for the proportion of women who prefer a female boss is
A) ± .0288
B) ± .0105
C) ± .0196
D) ± .0207

Answer 1

Answer:

D) ± .0207

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 $1 - \frac{\alpha}{2}$.

The width of the interval is given by the following formula:

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

In a random sample of 810 women employees, it is found that 81 would prefer working for a female boss.

This means that $n = 810, p = \frac{81}{810} = 0.1$

95% confidence level

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 width of the 95% confidence interval for the proportion of women who prefer a female boss is

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

$W = 1.94\sqrt{\frac{0.1*0.9}{810}} = 0.0207$

So the correct answer is:

D) ± .0207