A researcher wishes to estimate the proportion of left-handed people among a certain population. In a random sample of 820820 people from the population, 22.722.7% are left-handed. Find the margin of error for the 95% confidence interval for the population proportion. Round to four decimal places.

The margin of error for the 95% confidence interval for the population proportion is of 0.0287.

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 .

Find the margin of error for the 95% confidence interval for the population proportion.

We have that

So

The margin of error for the 95% confidence interval for the population proportion is of 0.0287.

its 8

step-by-step explanation:

2nd one in the first row is correct