Question

A U.S. Travel Data Center survey of 500 adults found that 195 said they would take more vacations this year than last year. Construct a 95% confidence interval for the true proportion of adults who said that they will travel more this year.

Answer 1

Using the z-distribution, it is found that the 95% confidence interval for the true proportion of adults who said that they will travel more this year is (0.347, 0.433).

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which z is the z-score that has a p-value of $\frac{1+\alpha}{2}$.

195 out of 500 people said they would take more vacations this year than last year, hence:

$n = 500, \pi = \frac{195}{500} = 0.39$

95% confidence level, hence$\alpha = 0.95$, z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$, so $z = 1.96$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.39 - 1.96\sqrt{\frac{0.39(0.61)}{500}} = 0.347$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.39 + 1.96\sqrt{\frac{0.39(0.61)}{500}} = 0.433$

The 95% confidence interval for the true proportion of adults who said that they will travel more this year is (0.347, 0.433).

A similar problem is given at brainly.com/question/15850972