Question

Last week, 108 cars received parking violations in the main university parking lot. Of these, 27 had unpaid parking tickets from a previous violation. Assuming that last week was a random sample of all parking violators, find the 95 percent confidence interval for the percentage of parking violations that have prior unpaid parking tickets.

Answer 1

Answer:

95 percent confidence interval

[0.208,0.29166]

Step-by-step explanation:

First we identificate the sample proportion p = 27 /108, and we find the 95 percent confidence interval for P(population proportion) in this context the parking violations that have prior unpaid parking tickets.

Then α = 0.05 and 1 - α/2 = 0.975

Thus $Z_{0.975} = 1.96$

The 95 percent confidence interval is

[ p ± 1.96 $\sqrt{\frac{p(1-p)}{n} }$ ]

[0.25  ± 1.96 $\sqrt{\frac{0.25*0.75}{108} }$ ]

[0.208,0.29166]