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The article "Viewers Speak Out Against Reality TV" (Associated Press, September 12,2005 ) included the following statement: "Few people believe there's much reality in reality TV: a total of 82 percent said the shows are either 'totally made up' or 'mostly distorted'." This statement was based on a survey of 1002 randomly selected adults. Compute and interpret a bound on the error of estimation for the reported percentage.

Short Answer

Expert verified
To calculate the bound of the error of estimation for the reported percentage in a survey, use the formula for standard error of a proportion, substituting the given proportion and sample size. The resultant standard error represents the bound on the error of estimation. Calculate this to obtain a numerical value.

Step by step solution

01

Recall the formula for standard error of a proportion

The standard error of a proportion is defined by the formula \(SE = \sqrt{p(1-p)/n}\), where \(p\) is the proportion (in this case, 0.82) and \(n\) is the sample size (in this case, 1002).
02

Insert the values into the formula and calculate

Insert the given values into the standard error formula: \(SE = \sqrt{0.82(1-0.82)/1002} \). Carry out the calculation to find the standard error which can give a rough approximation of the bound on the error.
03

Interpretation

The standard error calculated in the previous step is a measure of the accuracy of our sample proportion in estimating the true population proportion. The smaller the standard error, the more accurate the estimate. It represents the standard deviation of the sampling distribution of the sample proportion. This calculated standard error then serves as an estimate of the bound for the error of estimation for the reported percentage.

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Key Concepts

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

Standard Error
When we talk about standard error, we refer to a statistic that quantifies the precision with which a sample estimate approximates the true population parameter. It's important because it gives us insight into the reliability of the estimate obtained from our sample data. In the context of proportions, the formula becomes \[\begin{equation}SE = \sqrt{\frac{p(1-p)}{n}}\end{equation}\]where p is the sample proportion and n is the sample size. If you were dealing with the average of some measurements, you would use a different version of the formula, one involving the standard deviation and the square root of the sample size.
  • The smaller the standard error, the closer your sample proportion is likely to be to the true population proportion.
  • A larger sample size will generally lead to a smaller standard error, making your estimate more reliable.
  • The standard error can be influenced by the variability of the population—the more varied the population, the larger the standard error might be, even with a large sample size.
Understanding the standard error is crucial for interpreting survey results and can help you judge the margin of error for the estimates provided.
Sample Proportion
The sample proportion is a statistic that represents the fraction of the sample with a certain characteristic. In our exercise, it relates to the proportion of adults who think that reality TV shows are either 'totally made up' or 'mostly distorted'. The given proportion (p) is 0.82 or 82%, meaning that out of the randomly selected adults, 82% shared this opinion about reality TV.

Why is this important?

  • Sample proportions are used as estimates for the true population proportion.
  • The accuracy of the sample proportion as an estimator depends on the sample size and the variability within the population.
  • With a larger sample size, the sample proportion typically becomes a better estimator of the population proportion.
This specific measure is vital to obtaining an accurate picture of public opinion. It's what pollsters use to gauge the sentiments of the broader population based on a smaller group of individuals. That said, it's not without its limitations; sampling error and the potential bias in how the sample was collected could affect its accuracy.
Sampling Distribution
Lastly, the sampling distribution is a theoretical distribution that represents how the sample statistic would behave if we were to take many samples from the same population.

Key aspects:

  • It's the probability distribution of a given statistic based on a random sample.
  • The sampling distribution of the sample proportion can tell us how the proportion will vary from sample to sample.
  • The shape of this distribution is often approximately normal when the sample size is large, thanks to the Central Limit Theorem.
For the exercise in question, this concept allows us to understand the variability and therefore the reliability of the sample proportion of 82%. The standard error of the sample proportion, which we calculated, can be viewed as the standard deviation of its sampling distribution. If we assume a normal distribution, we can use this to create confidence intervals around the sample proportion to make educated guesses about the true population proportion. This ties back into the concept of the error of estimation, giving us insight into how much the sample proportion may differ from the actual population proportion.

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Most popular questions from this chapter

The interval from \(-2.33\) to \(1.75\) captures an area of \(.95\) under the \(z\) curve. This implies that another largesample \(95 \%\) confidence interval for \(\mu\) has lower limit \(\bar{x}-2.33 \frac{\sigma}{\sqrt{n}}\) and upper limit \(\bar{x}+1.75 \frac{\sigma}{\sqrt{n}} .\) Would you recommend using this \(95 \%\) interval over the \(95 \%\) interval \(\bar{x} \pm 1.96 \frac{\sigma}{\sqrt{n}}\) discussed in the text? Explain. (Hint: Look at the width of each interval.)

The Center for Urban Transportation Research released a report stating that the average commuting distance in the United States is \(10.9 \mathrm{mi}\) (USA Today, August 13 , 1991). Suppose that this average is actually the mean of a random sample of 300 commuters and that the sample standard deviation is \(6.2 \mathrm{mi}\). Estimate the true mean commuting distance using a \(99 \%\) confidence interval.

The article "Most Canadians Plan to Buy Treats, Many Will Buy Pumpkins, Decorations and/or Costumes" (Ipsos-Reid, October 24, 2005) summarized results from a survey of 1000 randomly selected Canadian residents. Each individual in the sample was asked how much he or she anticipated spending on Halloween during 2005 . The resulting sample mean and standard deviation were \(\$ 46.65\) and \(\$ 83.70\) respectively. a. Explain how it could be possible for the standard deviation of the anticipated Halloween expense to be larger than the mean anticipated expense. b. Is it reasonable to think that the distribution of the variable anticipated Halloween expense is approximately normal? Explain why or why not. c. Is it appropriate to use the \(t\) confidence interval to estimate the mean anticipated Halloween expense for Canadian residents? Explain why or why not. d. If appropriate, construct and interpret a \(99 \%\) confidence interval for the mean anticipated Halloween expense for Canadian residents.

A random sample of \(n=12\) four-year-old red pine trees was selected, and the diameter (in inches) of each tree's main stem was measured. The resulting observations are as follows: \(\begin{array}{llllllll}11.3 & 10.7 & 12.4 & 15.2 & 10.1 & 12.1 & 16.2 & 10.5\end{array}\) \(\begin{array}{llll}11.4 & 11.0 & 10.7 & 12.0\end{array}\) a. Compute a point estimate of \(\sigma\), the population standard deviation of main stem diameter. What statistic did you use to obtain your estimate? b. Making no assumptions about the shape of the population distribution of diameters, give a point estimate for the population median diameter. What statistic did you use to obtain the estimate? c. Suppose that the population distribution of diameter is symmetric but with heavier tails than the normal distribution. Give a point estimate of the population mean diameter based on a statistic that gives some protection against the presence of outliers in the sample. What statistic did you use? d. Suppose that the diameter distribution is normal. Then the 90 th percentile of the diameter distribution is \(\mu\) t \(1.28 \sigma\) (so \(90 \%\) of all trees have diameters less than this value). Compute a point estimate for this percentile. (Hint: First compute an estimate of \(\mu\) in this case; then use it along with your estimate of \(\sigma\) from Part (a).)

The following data on gross efficiency (ratio of work accomplished per minute to calorie expenditure per minute) for trained endurance cyclists were given in the article "Cycling Efficiency Is Related to the Percentage of Type I Muscle Fibers" (Medicine and Science in Sports and Exercise [1992]: \(782-88\) ): \(\begin{array}{llllllll}18.3 & 18.9 & 19.0 & 20.9 & 21.4 & 20.5 & 20.1 & 20.1\end{array}\) \(\begin{array}{llllllll}20.8 & 20.5 & 19.9 & 20.5 & 20.6 & 22.1 & 21.9 & 21.2\end{array}\) \(\begin{array}{lll}20.5 & 22.6 & 22.6\end{array}\) a. Assuming that the distribution of gross energy in the population of all endurance cyclists is normal, give a point estimate of \(\mu\), the population mean gross efficiency. b. Making no assumptions about the shape of the population distribution, estimate the proportion of all such cyclists whose gross efficiency is at most 20 .

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