The American Heart Association is about to conduct an anti-smoking campaign and wants to know the fraction of Americans over 44 who smoke. Step 2 of 2 : Suppose a sample of 632 Americans over 44 is drawn. Of these people, 462 don't smoke. Using the data, construct the 90% confidence interval for the population proportion of Americans over 44 who smoke. Round your answers to three decimal places.

The 90% confidence interval for the population proportion of Americans over 44 who smoke is (0.24, 0.298)

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:

632 American, 462 don't smoke, 632 - 462 = 170 smoke.

Proportion who smoke, so

90% 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:

The 90% confidence interval for the population proportion of Americans over 44 who smoke is (0.24, 0.298)

his chances of getting a coin of higher value is 4/19

step-by-step explanation: