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Chapter 8: Problem 36

According to statistics reported on \(\mathrm{CNBC}\), a surprising number of motor vehicles are not covered by insurance (CNBC, February 23,2006 ). Sample results, consistent with the CNBC report, showed 46 of 200 vehicles were not covered by insurance. a. What is the point estimate of the proportion of vehicles not covered by insurance? b. Develop a \(95 \%\) confidence interval for the population proportion.

Short Answer

Expert verified
The point estimate of the proportion of vehicles not covered by insurance is \(\hat{p} = 0.23\). The 95% confidence interval for the population proportion of vehicles not covered by insurance is (0.173, 0.287).

Step by step solution

01

a. Calculate the point estimate of the proportion of vehicles not covered by insurance.

To calculate the point estimate of the proportion of vehicles not covered by insurance, we divide the number of vehicles not covered by insurance by the total number of vehicles in the sample: \[ \hat{p} = \frac{\text{number of vehicles not covered}}{\text{total number of vehicles}} = \frac{46}{200} = 0.23\] So the point estimate is \(\hat{p} = 0.23\).
02

b. Develop a 95% confidence interval for the population proportion.

To develop a 95% confidence interval for the population proportion, we will use the formula for a confidence interval of a proportion: \[ CI = \hat{p} \pm z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\] We already have the point estimate \(\hat{p} = 0.23\) and the sample size n = 200. Now, we need to find the z-score corresponding to the 95% confidence level. This value can be found in a standard normal distribution table or using a calculator, and it is approximately 1.96. Now, plug these values into the formula: \[ CI = 0.23 \pm 1.96 \times \sqrt{\frac{0.23(1-0.23)}{200}}\] \[ CI = 0.23 \pm 1.96 \times \sqrt{\frac{0.1769}{200}}\] \[ CI = 0.23 \pm 1.96 \times 0.0291\] \[ CI = 0.23 \pm 0.057 \] So, the 95% confidence interval for the population proportion of vehicles not covered by insurance is (0.173, 0.287).

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Key Concepts

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

Understanding Point Estimate
A point estimate refers to the use of sample data to calculate a single value, or statistic, that serves as a best guess or approximation for an unknown population parameter. In simpler terms, it's like taking a small piece of the whole pie to understand what the entire pie tastes like.
For example, when we're interested in finding out the proportion of vehicles not covered by insurance, we use the data from our sample. In the original exercise, out of 200 vehicles, 46 aren't insured. To get the point estimate of the uninsured proportion, we divide 46 by 200.
  • This gives us a value of 0.23, which means we estimate that 23% of vehicles in the whole population would also be uninsured.
Point estimates are super handy because they provide a snapshot of what the larger population might look like based on just a small part of it.
Grasping Population Proportion
The population proportion is a specific statistical measure that indicates the fraction of a population that possesses a certain attribute. It is essentially what we're trying to estimate or understand when we talk about insured or uninsured vehicles.
In the context of our exercise, if every single vehicle in a vast country could be inspected, the population proportion would tell us exactly what percentage of vehicles are not covered by insurance.
  • However, because examining every vehicle is impractical, we rely on sample data and a point estimate as proxies.
  • These serve as our best guess for the true population proportion.
While point estimates give us a single figure to work with, remember that they are just estimates. They carry a bit of uncertainty because we are not examining the entire population.
Insight into Sample Statistics
Sample statistics play a crucial role in helping us make inferences about a larger population. They are the values or measures we derive from a sample, like sample mean, sample proportion (which we've just seen), standard deviation, etc.
When we aren't able to evaluate an entire population, we do use a sample, a smaller, manageable subset of the population, to perform our validations and analyses.
  • The proportion of uninsured vehicles in our sample (46 out of 200) is a key statistic as it helps us estimate the population proportion with a level of confidence.
  • Sample statistics like these allow us to employ various statistical methods, such as constructing confidence intervals, to measure the reliability of our estimates.
By understanding the sample statistics, we have a powerful toolkit that aids in scientific inquiry, allowing us to make educated guesses about the population's characteristics.

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Most popular questions from this chapter

Nielsen Media Research conducted a study of household television viewing times during the 8 P.M. to 11 P.M. time period. The data contained in the file named Nielsen are consistent with the findings reported (The World Almanac, 2003). Based upon past studies, the population standard deviation is assumed known with \(\sigma=3.5\) hours. Develop a \(95 \%\) confidence interval estimate of the mean television viewing time per week during the 8 P.M. to 11 P.M. time period.

According to Thomson Financial, through January \(25,2006,\) the majority of companies reporting profits had beaten estimates (Business Week, February 6, 2006). A sample of 162 companies showed that 104 beat estimates, 29 matched estimates, and 29 fell short. a. What is the point estimate of the proportion that fell short of estimates? b. Determine the margin of error and provide a \(95 \%\) confidence interval for the proportion that beat estimates. c. How large a sample is needed if the desired margin of error is \(.05 ?\)

How large a sample should be selected to provide a \(95 \%\) confidence interval with a margin of error of \(10 ?\) Assume that the population standard deviation is 40

Which would be hardest for you to give up: Your computer or your television? In a survey of 1677 U.S. Internet users, \(74 \%\) of the young tech elite (average age of 22 ) say their computer would be very hard to give up (PC Magazine, February 3, 2004). Only 48\% say their television would be very hard to give up. a. Develop a \(95 \%\) confidence interval for the proportion of the young tech elite that would find it very hard to give up their computer. b. Develop a \(99 \%\) confidence interval for the proportion of the young tech elite that would find it very hard to give up their television. c. In which case, part (a) or part (b), is the margin of error larger? Explain why.

A well-known bank credit card firm wishes to estimate the proportion of credit card holders who carry a nonzero balance at the end of the month and incur an interest charge. Assume that the desired margin of error is .03 at \(98 \%\) confidence. a. How large a sample should be selected if it is anticipated that roughly \(70 \%\) of the firm's card holders carry a nonzero balance at the end of the month? b. How large a sample should be selected if no planning value for the proportion could be specified?
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