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Chapter 18: Problem 17

Teenage drivers An insurance company checks police records on 582 accidents selected at random and notes that teenagers were at the wheel in 91 of them. a) Create a \(95 \%\) confidence interval for the percentage of all auto accidents that involve teenage drivers. b) Explain what your interval means. c) Explain what "95\% confidence" means. 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 your confidence interval support or contradict this statement? Explain.

Short Answer

Expert verified
a) Interval is around (0.130, 0.190). b) It estimates the true accident percent involving teens. c) 95% of such intervals will include the true value. d) The interval contradicts the statement (0.2).

Step by step solution

01

Identify the Sample Proportion

First, we need to find the sample proportion of accidents involving teenage drivers. The proportion \( \hat{p} \) is calculated as the number of accidents involving teenagers divided by the total number of accidents checked, so \( \hat{p} = \frac{91}{582} \).
02

Calculate Standard Error

The standard error (SE) of the sample proportion is computed using the formula \( SE = \sqrt{ \frac{\hat{p}(1 - \hat{p})}{n} } \), where \( n \) is the sample size. Substitute \( \hat{p} = \frac{91}{582} \) and \( n = 582 \) to find \( SE \).
03

Determine the Confidence Interval

To find the 95% confidence interval, use the formula \( \hat{p} \pm Z \cdot SE \), where \( Z \) is the Z-score for 95% confidence, approximately 1.96. Calculate the margin of error (\( Z \cdot SE \)) and add/subtract it from \( \hat{p} \) to find the confidence interval.
04

Interpret the Confidence Interval

The confidence interval provides a range within which we expect the true population parameter (percentage of accidents involving teenage drivers) to fall, with 95% confidence.
05

Describe 95% Confidence

"95% confidence" means that if we were to take many random samples and construct a confidence interval from each sample, about 95% of the intervals would contain the true population proportion.
06

Analyze the Politician's Statement

The politician claims that teenagers are involved in 20% of accidents (probability 0.2). Compare the calculated confidence interval to 0.2. If 0.2 is outside the interval, the statement is contradicted. Otherwise, it is within the interval.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Proportion
The sample proportion is a way to express part of a whole in the context of our sample data. In the exercise about teenage drivers, the sample proportion represents the fraction of accidents that had teenagers at the wheel. We find the sample proportion, symbolized by \( \hat{p} \), by dividing the number of specific cases (teenager-driver accidents) by the total number of cases studied (all accidents in the sample).
Using the given data:
  • Total accidents checked: 582
  • Teenage driver accidents: 91
The calculation goes as follows: \( \hat{p} = \frac{91}{582} \), which gives us the proportion of accidents involving teenagers. This is an essential step because it forms the foundation upon which other statistical calculations, such as the standard error and confidence interval, are built.
Standard Error
The standard error (SE) is like a measure of how much our sample proportion might vary if we took another sample. It tells us about the accuracy of our sample proportion as an estimate of the true population proportion. To find the standard error, we use the formula: \[ SE = \sqrt{ \frac{\hat{p}(1 - \hat{p})}{n} } \]where:
  • \( \hat{p} \) is the sample proportion
  • \( n \) is the sample size
Using our numbers:
  • \( \hat{p} = \frac{91}{582} \)
  • \( n = 582 \)
Substitute these into the formula to get the standard error of our sample proportion. The smaller the standard error, the closer our sample proportion is likely to be to the true population proportion.
95% Confidence
The concept of 95% confidence helps us understand the reliability of the confidence interval we calculate. When we talk about a 95% confidence interval, it means that if we were to take many samples and construct a confidence interval from each one, 95% of those intervals would contain the true population proportion. In simpler terms, there's a high probability that our calculated interval is one of those many intervals that 'captures' the real truth.
You often hear about a Z-score in this context. For a 95% confidence interval, the Z-score is approximately 1.96. This score helps calculate the margin of error, which we add to and subtract from our sample proportion to form our confidence interval.
Z-score
The Z-score plays a crucial role in constructing confidence intervals. It essentially tells us how many standard deviations away from the mean a certain proportion is, in a standard normal distribution. In the case of a 95% confidence interval, we use a Z-score of 1.96. This corresponds to the probability that our sample proportion falls within the desired range of the true population proportion.
To calculate the confidence interval, we use the formula: \[ \hat{p} \pm Z \cdot SE \]where:
  • \( \hat{p} \) is the sample proportion
  • \( Z \) is the Z-score (1.96 for 95%)
  • \( SE \) is the standard error
This calculation results in two values that form the ends of our confidence interval, which represent a range of possible values for the true proportion. This interval gives us a clearer insight into where the real-world average might lie, based on our sample data.

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Most popular questions from this chapter

Pilot study A state's environmental agency worries that many cars may be violating clean air emissions standards. The agency hopes to check a sample of vehicles in order to estimate that percentage with a margin of error of \(3 \%\) and \(90 \%\) confidence. To gauge the size of the problem, the agency first picks 60 cars and finds 9 with faulty emissions systems. How many should be sampled for a full investigation?

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Teachers A 2011 Gallup poll found that \(76 \%\) of Americans believe that high achieving high school students should be recruited to become teachers. This poll was based on a random sample of 1002 Americans. a) Find a \(90 \%\) confidence interval for the proportion of Americans who would agree with this. b) Interpret your interval in this context. c) Explain what "90\% confidence" means. d) Do these data refute a pundit's claim that \(2 / 3\) of Americans believe this statement? Explain.

Gambling A city ballot includes a local initiative that would legalize gambling. The issue is hotly contested, and two groups decide to conduct polls to predict the outcome. The local newspaper finds that \(53 \%\) of 1200 randomly selected voters plan to vote "yes," while a college Statistics class finds \(54 \%\) of 450 randomly selected voters in support. Both groups will create \(95 \%\) confidence intervals. a) Without finding the confidence intervals, explain which one will have the larger margin of error. b) Find both confidence intervals. c) Which group concludes that the outcome is too close to call? Why?

More conclusions In January \(2002,\) two students made worldwide headlines by spinning a Belgian euro 250 times and getting 140 heads - that's \(56 \% .\) That makes the \(90 \%\) confidence interval \((51 \%, 61 \%) .\) What does this mean? Are these conclusions correct? Explain. a) Between \(51 \%\) and \(61 \%\) of all euros are unfair. b) We are \(90 \%\) sure that in this experiment this euro landed heads on between \(51 \%\) and \(61 \%\) of the spins. c) We are \(90 \%\) sure that spun euros will land heads between \(51 \%\) and \(61 \%\) of the time. d) If you spin a curo many times, you can be \(90 \%\) sure of getting between \(51 \%\) and \(61 \%\) heads. e) \(90 \%\) of all spun euros will land heads between \(51 \%\) and \(61 \%\) of the time.
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