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Chapter 8: Problem 110

An insurance company selected a sample of 50 auto claims filed with it and investigated those claims carefully. The company found that \(12 \%\) of those claims were fraudulent. a. What is the point estimate of the percentage of all auto claims filed with this company that are fraudulent? b. Make a \(99 \%\) confidence interval for the percentage of all auto claims filed with this company that are fraudulent.

Short Answer

Expert verified
The point estimate of the proportion of fraudulent claims is 12%. The 99% confidence interval for this proportion is (0, 0.2471).

Step by step solution

01

Calculate Point Estimate

The point estimate is simply the observed proportion of fraudulent claims in the sample. Since there were 50 claims and 12% of them were found to be fraudulent, there are \(50 * 0.12 = 6\) fraudulent claims. The point estimate (p) for the proportion is therefore \(6/50 = 0.12\) or 12%.
02

Calculate Standard Error

The standard error of the proportion can be calculated using the formula \(\sqrt{ (p(1-p)) / n}\), where p is the point estimate and n is the sample size. Substituting in the numbers, we get \(\sqrt{ (0.12(1 - 0.12)) / 50 } = 0.0494\).
03

Determine the Z-Score for 99% Confidence Level

The Z-score associated with a 99% confidence level is typically 2.576 (which can be looked up in a standard statistical table).
04

Calculate the Confidence Interval

The 99% confidence interval can be calculated using the formula \(p 卤 Z*SE\), where p is the sample proportion, Z is the Z-score from step 3 and SE is the standard error from step 2. This gives \(0.12 卤 2.576*0.0494\), which simplifies to \((0.12 - 0.1271, 0.12 + 0.1271)\) or \((-0.0071, 0.2471)\). However, a proportion cannot be less than 0 or more than 1, so we adjust the interval to \((0, 0.2471)\).

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Most popular questions from this chapter

In a Time/Money Magazine poll of Americans of age 18 years and older, \(65 \%\) agreed with the statement, "We are less sure our children will achieve the American Dream" (Time, October 10,2011 ). Assume that this poll was based on a random sample of 1600 Americans. a. Construct a \(95 \%\) confidence interval for the proportion of all Americans of age 18 years and older who will agree with the aforementioned statement. b. Explain why we need to construct a confidence interval. Why can we not simply say that \(65 \%\) of all Americans of age 18 years and older agree with the aforementioned statement?

A gas station attendant would like to estimate \(p\), the proportion of all households that own more than two vehicles. To obtain an estimate, the attendant decides to ask the next 200 gasoline customers how many vehicles their households own. To obtain an estimate of \(p\), the attendant counts the number of customers who say there are more than two vehicles in their households and then divides this number by 200 . How would you critique this estimation procedure? Is there anything wrong with this procedure that would result in sampling and/or nonsampling errors? If so, can you suggest a procedure that would reduce this error?

Briefly explain the similarities and the differences between the standard normal distribution and the \(t\) distribution.

a. A sample of 100 observations taken from a population produced a sample mean equal to \(55.32\) and a standard deviation equal to \(8.4 .\) Make a \(90 \%\) confidence interval for \(\mu .\) b. Another sample of 100 observations taken from the same population produced a sample mean equal to \(57.40\) and a standard deviation equal to \(7.5 .\) Make a \(90 \%\) confidence interval for \(\mu .\) c. A third sample of 100 observations taken from the same population produced a sample mean equal to \(56.25\) and a standard deviation equal to \(7.9 .\) Make a \(90 \%\) confidence interval for \(\mu .\) d. The true population mean for this population is \(55.80 .\) Which of the confidence intervals constructed in parts a through c cover this population mean and which do not?

The following data give the number of pitches thrown by both teams in each of a random sample of 24 Major League Baseball games played between the beginning of the 2012 season and May 16, \(2012 .\) \(\begin{array}{llllllll}234 & 281 & 264 & 251 & 284 & 266 & 337 & 291 \\ 309 & 245 & 331 & 284 & 239 & 282 & 226 & 286 \\ 361 & 278 & 317 & 306 & 325 & 256 & 295 & 276\end{array}\) a. Create a histogram of these data using the class intervals 210 to less than 230,230 to less than 250,250 to less than 270 , and so on. Based on the histogram, does it seem reasonable to assume that these data are approximately normally distributed? b. Calculate the value of the point estimate of the corresponding population mean. c. Assuming that the distribution of total number of pitches thrown by both teams in Major League Baseball games is approximately normal, construct a \(99 \%\) confidence interval for the average number of pitches thrown by both teams in a Major League Baseball game.
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