A private opinion poll is conducted for a politician to determine what proportion of the population favors adding more national parks. How large a sample is needed in order to be 90% confident that the sample proportion will not differ from the true proportion by more than 5%

The size of the sample that is needed in order to be 90% confident that the sample proportion will not differ from the true proportion by more than 5% is; n = 271

We are given;

Confidence level of 90%

Margin of error; E = 5% = 0.05

We are not given the population proportion but we will assume that it is 0.5

Formula for margin of error here is;

E = z(√(p(1 - p)/n)

where z is the critical value at the confidence level

n is the sample size

Now, z at CL of 90% is 1.645

Making n the subject gives;

n = z²(p(1 - p)/E²

Plugging in the relevant values gives;

n = 1.645²(0.5(1 - 0.5))/0.05²

n ≈ 271

Read more about confidence intervals with proportion at; link

We need a sample size of at least 271

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:

90% confidence level

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

How large a sample is needed in order to be 90% confident that the sample proportion will not differ from the true proportion by more than 5%

We need a sample size of at least n, in which n is found when

We do not know the true proportion, so we use , which is when we are going to need to use the largest sample size.

Rounding up

We need a sample size of at least 271