Answer :
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 [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
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 [tex]n = 101, \pi = \frac{35}{101} = 0.3465[/tex]
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3465 - 1.96\sqrt{\frac{0.3465*0.6535}{101}} = 0.2537[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3465 + 1.96\sqrt{\frac{0.3465*0.6535}{101}} = 0.4393[/tex]
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%).