Question

7. Of 101 randomly selected adults

Answer 1

Answer:

The 95% confidence interval for the true percentage of all adults over 30 who shop at that mall who admit to having lost their car at the mall is (25.37%, 43.93%).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, 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 $1 - \frac{\alpha}{2}$.

Of 101 randomly selected adults over 30 who frequent a very large mall, 35 admitted to having lost their car at the mall.

This means that $n = 101, \pi = \frac{35}{101} = 0.3465$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3465 - 1.96\sqrt{\frac{0.3465*0.6535}{101}} = 0.2537$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3465 + 1.96\sqrt{\frac{0.3465*0.6535}{101}} = 0.4393$

As percentages:

0.2537*100% = 25.37%

0.4393*100% = 43.93%

The 95% confidence interval for the true percentage of all adults over 30 who shop at that mall who admit to having lost their car at the mall is (25.37%, 43.93%).