Question

A random sample of 1700 car owners in a particular city found 799 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 1

The 95% confidence interval for the true percent of car owners in this city who received a speeding ticket this year is 44.63% to 49.37%

The formula for calculating the confidence interval is expressed as:

CI = p±z√p(1-p)/n

Given the following parameters

n = 1700

p = 799/1700

p = 0.47

z-score at 95% interval = 1.96

Substitute the given values

CI = 0.47 ±1.96√0.47(1-0.47)/1700

CI =0.47 ±1.96(0.0121)

CI = 0.47±0.0237

CI = 0.47+0.0237, 0.47-0.0237

CI = (0.4463, 0.4937)

Hence the 95% confidence interval for the true percent of car owners in this city who received a speeding ticket this year is 44.63% to 49.37%

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