Question

There is no prior information about the proportion of Americans who support Medicare-for-all in 2019. If we want to estimate 95% confidence interval for the true proportion of Americans who support Medicare-for-all in 2019 with a 0.175 margin of error, how many randomly selected Americans must be surveyed? Answer: (Round up your answer to nearest whole number, do not include any decimals)

Answer 1

Answer:

32 randomly selected Americans must be surveyed

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 $Z = 1.96$.

If we want to estimate 95% confidence interval for the true proportion of Americans who support Medicare-for-all in 2019 with a 0.175 margin of error, how many randomly selected Americans must be surveyed?

n randomly selected Americans must be surveyed. n is found when M = 0.175.

We have no prior estimate, so we work with the worst case scenario, which is $\pi = 0.5$.

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

$0.175 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.175\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.175}$

$(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.175})^{2}$

$n = 31.36$

Rounding up to the nearest whole number

32 randomly selected Americans must be surveyed