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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

A poll for the presidential campaign sampled 491 potential voters in June. A primary purpose of the poll was to obtain an estimate of the proportion of potential voters who favored each candidate. Assume a planning value of \(p^{*}=.50\) and a \(95 \%\) confidence level. a. For \(p^{*}=.50,\) what was the planned margin of error for the June poll? b. Closer to the November election, better precision and smaller margins of error are desired. Assume the following margins of error are requested for surveys to be conducted during the presidential campaign. Compute the recommended sample size for each survey.

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

If a perfectly competitive firm sells 100 units of output at a market price of \(\$ 100\) per unit, its marginal revenue per unit is a. \(\$ 1\) b. \(\$ 100\) c. more than \(\$ 1,\) but less than \(\$ 100\). d. less than \(\$ 100\).

A sample survey of 54 discount brokers showed that the mean price charged for a trade of 100 shares at \(\$ 50\) per share was \(\$ 33.77\) (AAII Journal, February 2006 ). The survey is conducted annually. With the historical data available, assume a known population standard deviation of \(\$ 15\) a. Using the sample data, what is the margin of error associated with a \(95 \%\) confidence interval? b. Develop a \(95 \%\) confidence interval for the mean price charged by discount brokers for a trade of 100 shares at \(\$ 50\) per share.

The following sample data are from a normal population: 10,8,12,15,13,11,6,5 a. What is the point estimate of the population mean? b. What is the point estimate of the population standard deviation? c. With \(95 \%\) confidence, what is the margin of error for the estimation of the population mean? d. What is the \(95 \%\) confidence interval for the population mean?
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