Question

A random sample of 1800 car owners in a particular city found 324 car owners who received a speeding ticket this year. Find a 95% confidence interval for the true percent of car owners in this city who received a speeding ticket this year. Express your results to the nearest hundredth of a percent.

Answer:

to  %

Answer 1

Confidence interval for Population Proportion is given as below:

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Where, P is the sample proportion, Z is critical value, and n is sample size.

We are given

x = 324

n = 1800

P = x/n = 324/1800 = 0.18

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Confidence Interval = 0.18 ± 1.96* sqrt(0.18*(1 – 0.18)/1800)

Confidence Interval = 0.18 ± 1.96* 0.0091

Confidence Interval = 0.18 ± 0.0177

Lower limit = 0.18 - 0.0177 = 0.1623

Upper limit = 0.18 + 0.0177 = 0.1977

Confidence interval = (16.23%, 19.77%)

Answer: 16.23% to 19.77%