A travel agent wants to estimate the proportion of vacationers who plan to travel outside the United States in the next 12 months. A random sample of 130 vacationers revealed that 40 had plans for foreign travel in that time frame. Construct a 95% confidence interval estimate of the population proportion.

The 95% confidence interval estimate of the population proportion is (0.2284, 0.3870).

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:

A random sample of 130 vacationers revealed that 40 had plans for foreign travel in that time frame. This means that

95% 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 95% confidence interval estimate of the population proportion is (0.2284, 0.3870).

the range is 19

step-by-step explanation:

the range is biggest number minus the smallest, so in this case it would be

41 - 22 = 19

the answer is 45 i just did it on oddesy ware