Question

The NWBC found that 42.1% of women-owned businesses provided retirement plans contributions.

Answer 1

Answer:

Sample size of 586 or higher.

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 margin of error is:

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

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 $Z = 1.96$.

What sample size could be 95% confident that the estimated (sample) proportion is within 4 percentage points of the true population proportion?

Sample size of at least n when $M = 0.04$

42.1% of women-owned businesses provided retirement plans contributions, which means that $p = 0.421$. So

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

$0.04 = 1.96\sqrt{\frac{0.421*0.579}{n}}$

$0.04\sqrt{n} = 0.9677$

$\sqrt{n} = \frac{0.9677}{0.04}$

$\sqrt{n} = 24.19$

$\sqrt{n}^{2} = (24.19)^(2)$

$n = 585.2$

We need a sample size of 586 or higher.