A random sample found that 30% of 150 Americans were satisfied with the gun control laws in 2018. Compute a 98% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2018 Fill in the blanks appropriately. A 98% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2018 is ( , ) (round to 3 decimal places)

A 98% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2018 is (0.213, 0.387).

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 .

For this problem, we have that:

98% confidence level

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

The lower limit of this interval is:

The upper limit of this interval is:

A 98% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2018 is (0.213, 0.387).

example 2 – simplify: step 1: completely factor both the numerators and denominators of all fractions. step 2: change the division sign to a multiplication sign and flip (or reciprocate) the fraction after the division sign; essential you need to multiply by the reciprocal. step 3 : cancel or reduce the fractions.