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? (Round up your answer to nearest whole number, do not include any decimals)

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 , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

The margin of error is:

95% confidence level

So , z is the value of Z that has a pvalue of , so .

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 .

Rounding up to the nearest whole number

32 randomly selected Americans must be surveyed

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