Question

7. Of 101 randomly selected adults

over 30 who frequent a very large
mall, 35 admitted to having lost their
car at the mall. Construct a 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.

Answer 1

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%).

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{(\pi(1-\pi))/(n)}`

In which

z is the z-score that has a p-value of
`1 - (\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 = (35)/(101) = 0.3465`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a p-value of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

The lower limit of this interval is:

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

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.3465 + 1.96\sqrt{(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%).