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Chapter 8: Problem 10

Case studies showed that out of 10,351 convicts who escaped from U.S. prisons, only 7867 were recaptured (The Book of Odds, by Shook and Shook, Signet). (a) Let \(p\) represent the proportion of all escaped convicts who will eventually be recaptured. Find a point estimate for \(p\). (b) Find a \(99 \%\) confidence interval for \(p .\) Give a brief statement of the meaning of the confidence interval. (c) Is use of the normal approximation to the binomial justified in this problem? Explain.

Short Answer

Expert verified
(a) Point estimate: 0.7601. (b) 99% confidence interval: (0.7500, 0.7702). (c) Yes, the normal approximation is justified.

Step by step solution

01

Calculate the Point Estimate for p

The proportion of recaptured convicts, \(p\), can be estimated using the formula \(\hat{p} = \frac{x}{n}\), where \(x\) is the number of recaptured convicts, and \(n\) is the total number of escaped convicts. Substituting \(x = 7867\) and \(n = 10351\), we have:\[ \hat{p} = \frac{7867}{10351} \approx 0.7601 \]
02

Construct the 99% Confidence Interval for p

To find a \(99\%\) confidence interval for \(p\), use the formula:\[ \hat{p} \pm z_{\alpha/2} \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]For \(99\%\) confidence, \(z_{\alpha/2} \approx 2.576\). Substituting in the values:\[ 0.7601 \pm 2.576 \times \sqrt{\frac{0.7601 \times (1 - 0.7601)}{10351}} \]Calculate the margin of error:\[ 2.576 \times \sqrt{\frac{0.7601 \times 0.2399}{10351}} \approx 0.0101 \]The confidence interval is:\[ 0.7601 \pm 0.0101 = (0.7500, 0.7702) \]
03

Interpret the Confidence Interval

The \(99\%\) confidence interval indicates that we are \(99\%\) confident that the true proportion of all escaped convicts who are recaptured lies between \(75.00\%\) and \(77.02\%\).
04

Evaluate the Justification for the Normal Approximation

To use the normal approximation, both \(np\) and \(n(1-p)\) should be greater than 5. With \(\hat{p} = 0.7601\), calculate:\[ np = 10351 \times 0.7601 \approx 7867 > 5 \]\[ n(1-p) = 10351 \times 0.2399 \approx 2484 > 5 \]Both conditions are satisfied, so the normal approximation is justified.

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Key Concepts

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

Confidence Interval
A confidence interval gives us a range in which we believe a true population parameter lies, based on our sample data. It helps us understand the variability and reliability of our point estimate. In this exercise, we're constructing a 99% confidence interval for the proportion of escaped convicts who are recaptured.

The formula for a confidence interval involves the point estimate, a critical value from the standard normal distribution (often represented as a "z" value), and the standard error of the estimate. The confidence interval is expressed as:
  • Point Estimate 卤 (z-value 脳 Standard Error)
A 99% confidence interval indicates that if we were to take 100 different samples and compute a confidence interval for each sample, we would expect about 99 of the intervals to contain the true proportion.

In this case, using the sample data, the point estimate is approximately 76%, and the calculated 99% confidence interval is between 75% and 77.02%. This means we are 99% sure that the actual proportion of all escaped convicts who are recaptured lies within this range.
Point Estimate
The point estimate is a single value estimate of a population parameter. For proportions, the sample proportion is commonly used as a point estimate of the population proportion. It is calculated by dividing the number of successes by the total number of observations in the sample.

In the case of the escaped convicts:
  • The number of successes (recaptured convicts) is 7867.
  • The total number of observations (escaped convicts) is 10,351.
The point estimate, therefore, is calculated as:
yielding \[ \hat{p} = \frac{7867}{10351} \approx 0.7601 \]which means approximately 76% of the escaped convicts in the sample were recaptured.

This value serves as our best guess for the true proportion of recaptured convicts in the entire population.

It is important to remember that this is just an estimate, and it may not be perfectly accurate, which is why we use confidence intervals to give us a range of plausible values.
Normal Approximation
The normal approximation to the binomial distribution is an essential concept in statistics, especially when dealing with proportion problems. The binomial distribution can be approximated by a normal distribution under certain conditions.

For this approximation to be justified, two conditions must be met:
  • The sample size times the sample proportion, \(np\), must be greater than 5.
  • The sample size times one minus the sample proportion, \(n(1-p)\), must also be greater than 5.
In our exercise, these calculations were:
  • \( np = 10351 \times 0.7601 \approx 7867 \)
  • \( n(1-p) = 10351 \times 0.2399 \approx 2484 \)
Both of which clearly fulfill the condition of being greater than 5, thus validating the use of the normal approximation.

This means we could reliably use the normal distribution to make calculations about the confidence interval from our sample data.

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Most popular questions from this chapter

The following data represent crime rates per 1000 population for a random sample of 46 Denver neighborhoods (Reference: The Piton Foundation, Denver, Colorado). $$ \begin{array}{lrrrrrr} 63.2 & 36.3 & 26.2 & 53.2 & 65.3 & 32.0 & 65.0 \\ 66.3 & 68.9 & 35.2 & 25.1 & 32.5 & 54.0 & 42.4 \\ 77.5 & 123.2 & 66.3 & 92.7 & 56.9 & 77.1 & 27.5 \\ 69.2 & 73.8 & 71.5 & 58.5 & 67.2 & 78.6 & 33.2 \\ 74.9 & 45.1 & 132.1 & 104.7 & 63.2 & 59.6 & 75.7 \\ 39.2 & 69.9 & 87.5 & 56.0 & 154.2 & 85.5 & 77.5 \\ 84.7 & 24.2 & 37.5 & 41.1 & & & \end{array} $$ (a) Use a calculator with mean and sample standard deviation keys to verify that \(\bar{x} \approx 64.2\) and \(s \approx 27.9\) crimes per 1000 population. (b) Let us say the preceding data are representative of the population crime rates in Denver neighborhoods. Compute an \(80 \%\) confidence interval for \(\mu\), the population mean crime rate for all Denver neighborhoods. (c) Suppose you are advising the police department about police patrol assignments. One neighborhood has a crime rate of 57 crimes per 1000 population. Do you think that this rate is below the average population crime rate and that fewer patrols could safely be assigned to this neighborhood? Use the confidence interval to justify your answer. (d) Another neighborhood has a crime rate of 75 crimes per 1000 population. Does this crime rate seem to be higher than the population average? Would you recommend assigning more patrols to this neighborhood? Use the confidence interval to justify your answer. (e) Repeat parts (b), (c), and (d) for a \(95 \%\) confidence interval. (f) In previous problems, we assumed the \(x\) distribution was normal or approximately normal. Do we need to make such an assumption in this problem? Why or why not? Hint: See the central limit theorem in Section \(7.2\).

Sam computed a \(95 \%\) confidence interval for \(\mu\) from a specific random sample. His confidence interval was \(10.1<\mu<12.2 .\) He claims that the probability that \(\mu\) is in this interval is \(0.95\). What is wrong with his claim?

A random sample of medical files is used to estimate the proportion \(p\) of all people who have blood type \(B\). (a) If you have no preliminary estimate for \(p\), how many medical files should you include in a random sample in order to be \(85 \%\) sure that the point estimate \(\hat{p}\) will be within a distance of \(0.05\) from \(p ?\) (b) Answer part (a) if you use the preliminary estimate that about 8 out of 90 people have blood type B. (Reference: Manual of Laboratory and Diagnostic Tests, F. Fischbach.)

In a marketing survey, a random sample of 730 women shoppers revealed that 628 remained loyal to their favorite supermarket during the past year (i.e., did not switch stores). (Source: Trends in the United States: Consumer Attitudes and the Supermarket, The Research Department, Food Marketing Institute.) (a) Let \(p\) represent the proportion of all women shoppers who remain loyal to their favorite supermarket. Find a point estimate for \(p\). (b) Find a \(95 \%\) confidence interval for \(p\). Give a brief explanation of the meaning of the interval. (c) As a news writer, how would you report the survey results regarding the percentage of women supermarket shoppers who remained loyal to their favorite supermarket during the past year? What is the margin of error based on a \(95 \%\) confidence interval?

Consider a \(90 \%\) confidence interval for \(\mu\). Assume \(\sigma\) is not known. For which sample size, \(n=10\) or \(n=20\), is the confidence interval longer?
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