Question

An insurance company checks police records on 568 accidents selected at random and notes that teenagers were at the wheel in 92 of them. Complete parts ​a) through ​d).

​a) Construct the​ 95% confidence interval for the percentage of all auto accidents that involve teenage drivers.

​95% CI = ( ___________ %, _________ % )

​(Round to one decimal place as​ needed.)

​b) Explain what your interval means.

1. We are​ 95% confident that the true percentage of accidents involving teenagers falls inside the confidence interval limits.
2. We are​ 95% confident that a randomly sampled accident would involve a teenager a percent of the time that falls inside the confidence interval limits.
3. We are​ 95% confident that the percent of accidents involving teenagers is 16.2​%.

​c) Explain what​ "95% confidence" means.

About​ 95% of random samples of size 568 will produce (1) __________ that​ contain(s) the

(2) _________   of accidents involving teenagers.

​d) A politician urging tighter restrictions on​ drivers' licenses issued to teens​ says, "In one of every five auto​ accidents, a teenager is behind the​ wheel." Does the confidence interval support or contradict this​ statement?

1. The confidence interval supports the assertion of the politician. The figure quoted by the politician is inside the interval.
2. The confidence interval contradicts the assertion of the politician. The figure quoted by the politician is inside the interval.
3. The confidence interval contradicts the assertion of the politician. The figure quoted by the politician is outside the interval.
4. The confidence interval supports the assertion of the politician. The figure quoted by the politician is outside the interval.

(1)

• confidence intervals
• a true proportion
• sample proportion

(2)

• true proportion
• sample proportion
• confidence interval

Answer 1

Part a

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 n = 568, x = 92

P = x/n = 92/568 = 0.161971831

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

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

Confidence Interval = 0.161971831 ± 1.96* sqrt(0.161971831*(1 – 0.161971831)/568)

Confidence Interval = 0.161971831 ± 1.96* 0.0155

Confidence Interval = 0.161971831 ± 0.0303

Lower limit = 0.161971831 ± 0.0303 = 0.1317

Upper limit = 0.161971831 ± 0.0303 = 0.1923

95% CI = (13.2%, 19.2%)

Part b

a. We are​ 95% confident that the true percentage of accidents involving teenagers falls inside the confidence interval limits.

Part c

About 95% of random samples of size 568 will produce confidence interval that contains the true proportion of accidents involving teenagers.

Part d

We are given n = 5, x = 1, so P = 1/5 = 0.20 = 20%

The value 20% is not included in the given confidence interval (13.2%, 19.2%), so we reject the claim.

This confidence interval contradicts this statement.

Answer: C

The confidence interval contradicts the assertion of the politician. The figure quoted by the politician is outside the interval.