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Chapter 7: Problem 23

The Pew Forum on Religion and Public Life reported on Dec. 9,2009 , that in a survey of 2003 American adults, \(25 \%\) said they believed in astrology. a. Calculate and interpret a confidence interval at the \(99 \%\) confidence level for the proportion of all adult Americans who believe in astrology. b. What sample size would be required for the width of a \(99 \%\) CI to be at most \(.05\) irrespective of the value of \(\hat{p}\) ?

Short Answer

Expert verified
a) CI is (0.2251, 0.2749); b) Sample size needed is 6648.

Step by step solution

01

Identify Given Information

We have a sample proportion \( \hat{p} = 0.25 \) (since 25% of the sample believes in astrology). The sample size \( n = 2003 \). The task is to find a 99% confidence interval for the true population proportion \( p \).
02

Determine Z-Score for 99% Confidence Level

For a 99% confidence interval, the corresponding Z-score \( Z \) is approximately 2.576 (from the standard normal distribution table).
03

Calculate the Standard Error

The standard error (SE) of \( \hat{p} \) is calculated using the formula: \[ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \] where \( \hat{p} = 0.25 \) and \( n = 2003 \). So, \[ SE = \sqrt{\frac{0.25 \times 0.75}{2003}} \approx 0.00967 \]
04

Compute the Confidence Interval

The confidence interval is given by \( \hat{p} \pm Z \times SE \). This is \[ 0.25 \pm 2.576 \times 0.00967 \approx 0.25 \pm 0.02489 \]. Thus, the confidence interval is approximately \( (0.2251, 0.2749) \).
05

Interpret the Confidence Interval

We are 99% confident that the true proportion of all American adults who believe in astrology lies between 22.51% and 27.49%.
06

Calculate Required Sample Size for Given Margin of Error

The formula for sample size \( n \) when the margin of error (MOE) is at most \(.05\) is: \[ n = \left(\frac{Z^2 \cdot 0.25}{MOE^2}\right) \] For a 99% confidence interval, \( Z = 2.576 \) and \( 0.25 \) is used as an estimate for \( \hat{p}(1-\hat{p}) \) since \( \hat{p} \approx 0.5 \) provides the maximum variance. Calculate as\[ n = \left(\frac{2.576^2 \cdot 0.25}{0.05^2}\right) \approx 6648 \].
07

Interpretation of Sample Size Calculation

To achieve a 99% confidence interval with a width no more than 0.05, irrespective of \( \hat{p} \), at least 6648 people should be surveyed.

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Key Concepts

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

Sample Proportion
The sample proportion is an estimate of the proportion of a certain trait or characteristic within a population, based on a sample. In statistics, it's represented by the symbol \( \hat{p} \). To find the sample proportion, you divide the number of individuals in the sample displaying the characteristic by the total sample size.

In this problem, we have a sample of 2003 American adults, out of which 25% believe in astrology. This gives us a sample proportion \( \hat{p} = 0.25 \).
  • Importance: Provides a basis for inferring the population proportion.
  • Limitation: Subject to sample variability鈥攊t might not always accurately reflect the whole population.
Z-Score
The Z-score is a statistical measurement describing a value's relation to the mean of a group of values. It indicates how many standard deviations an element is from the mean. In the context of confidence intervals, the Z-score helps to determine how far, within a certain confidence level, we should extend from the sample estimate to capture the population parameter.

For a 99% confidence interval, the Z-score is typically 2.576. This means we would expect 99% of sampling instances to have a sample proportion within about 2.576 standard deviations of the population proportion.
  • Significance: Provides the critical value for constructing confidence intervals.
  • Context: Dependent on the desired confidence level, with common levels being 90%, 95%, and 99%.
Standard Error
In statistics, standard error (SE) represents the standard deviation of the sampling distribution of a statistic, most commonly the mean. It gives us an idea of how much variability we can expect in our sample estimate from one sample to another.

The standard error for a sample proportion is calculated using the formula: \[ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \] where \( \hat{p} \) is the sample proportion, and \( n \) is the sample size.
In the given exercise, \( SE = \sqrt{\frac{0.25 \times 0.75}{2003}} \approx 0.00967 \).
  • Purpose: Indicates how accurately the sample proportion estimates the population proportion.
  • Variability: Smaller SE suggests a more precise estimate.
Population Proportion
The population proportion refers to the proportion of individuals in a population that possess a specific characteristic. It is denoted by \( p \), and unlike the sample proportion, it's a fixed value, although often unknown. We use sample data to make inferences about the population proportion.

In this scenario, we aim to estimate the proportion of all American adults who believe in astrology. We use the sample proportion to estimate \( p \) while acknowledging that there is some degree of uncertainty represented by the confidence interval.
  • Estimation: Typically estimated using the sample proportion \( \hat{p} \).
  • Uncertainty: Accuracy of estimation improves as sample size increases.
Margin of Error
The margin of error (MOE) represents the range of values below and above the sample statistic in a confidence interval. It tells us how precise our estimate is from the true population value.

The formula to calculate the margin of error in the context of a confidence interval is:
\[ MOE = Z \times SE \] where \( Z \) is the Z-score for the desired confidence level, and \( SE \) is the standard error of the sample proportion. In our exercise, the margin of error is \( 2.576 \times 0.00967 = 0.02489 \).
  • Interpretation: With a 99% confidence level, the true population proportion is expected to lie within the bounds set by the margin of error around the sample proportion.
  • Influence: A larger sample size typically results in a smaller margin of error, all else being equal.

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Most popular questions from this chapter

An April 2009 survey of 2253 American adults conducted by the Pew Research Center's Internet \& American Life Project revealed that 1262 of the respondents had at some point used wireless means for online access. a. Calculate and interpret a \(95 \% \mathrm{CI}\) for the proportion of all American adults who at the time of the survey had used wireless means for online access. b. What sample size is required if the desired width of the \(95 \% \mathrm{CI}\) is to be at most .04, irrespective of the sample results? c. Does the upper limit of the interval in (a) specify a \(95 \%\) upper confidence bound for the proportion being estimated? Explain. 52\. High concentration of the toxic element arsenic is all too common in groundwater. The article "Evaluation of Treatment Systems for the Removal of Arsenic from Groundwater" (Practice Periodical of Hazardous, Toxic, and Radioactive Waste Mgmt., 2005: 152-157) reported that for a sample of \(n=5\) water specimens selected for treatment by coagulation, the sample mean arsenic concentration was \(24.3 \mu \mathrm{g} / \mathrm{L}\), and the sample standard deviation was 4.1. The authors of the cited article used \(t\)-based methods to analyze their data, so hopefully had reason to believe that the distribution of arsenic concentration was normal. a. Calculate and interpret a \(95 \%\) CI for true average arsenic concentration in all such water specimens. b. Calculate a \(90 \%\) upper confidence bound for the standard deviation of the arsenic concentration distribution. c. Predict the arsenic concentration for a single water specimen in a way that conveys information about precision and reliability.

It is important that face masks used by firefighters be able to withstand high temperatures because firefighters commonly work in temperatures of \(200-500^{\circ} \mathrm{F}\). In a test of one type of mask, 11 of 55 masks had lenses pop out at \(250^{\circ}\). Construct a \(90 \% \mathrm{Cl}\) for the true proportion of masks of this type whose lenses would pop out at \(250^{\circ}\).

A random sample of 110 lightning flashes in a certain region resulted in a sample average radar echo duration of \(.81 \mathrm{sec}\) and a sample standard deviation of \(34 \mathrm{sec}\) ("Lightning Strikes to an Airplane in a Thunderstorm," \(J .\) of Aircraft, 1984: 607-611). Calculate a \(99 \%\) (two- sided) confidence interval for the true average echo duration \(\mu\), and interpret the resulting interval.

The alternating current (AC) breakdown voltage of an insulating liquid indicates its dielectric strength. The article "Testing Practices for the AC Breakdown Voltage Testing of Insulation Liquids" (IEEE Electrical Insulation Magazine, 1995: 21-26) gave the accompanying sample observations on breakdown voltage (kV) of a particular circuit under certain conditions. \(\begin{array}{llllllllllllllll}62 & 50 & 53 & 57 & 41 & 53 & 55 & 61 & 59 & 64 & 50 & 53 & 64 & 62 & 50 & 68 \\ 54 & 55 & 57 & 50 & 55 & 50 & 56 & 55 & 46 & 55 & 53 & 54 & 52 & 47 & 47 & 55 \\ 57 & 48 & 63 & 57 & 57 & 55 & 53 & 59 & 53 & 52 & 50 & 55 & 60 & 50 & 56 & 58\end{array}\) a. Construct a boxplot of the data and comment on interesting features. b. Calculate and interpret a \(95 \%\) CI for true average breakdown voltage \(\mu\). Does it appear that \(\mu\) has been precisely estimated? Explain. c. Suppose the investigator believes that virtually all values of breakdown voltage are between 40 and 70 . What sample size would be appropriate for the \(95 \% \mathrm{Cl}\) to have a width of \(2 \mathrm{kV}\) (so that \(\mu\) is estimated to within \(1 \mathrm{kV}\) with \(95 \%\) confidence)?

The article "Measuring and Understanding the Aging of Kraft Insulating Paper in Power Transformers" (IEEE Electrical Insul. Mag., 1996: 28-34) contained the following observations on degree of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range: \(\begin{array}{lllllll}418 & 421 & 421 & 422 & 425 & 427 & 431 \\ 434 & 437 & 439 & 446 & 447 & 448 & 453 \\ 454 & 463 & 465 & & & & \end{array}\) a. Construct a boxplot of the data and comment on any interesting features. b. Is it plausible that the given sample observations were selected from a normal distribution? c. Calculate a two-sided \(95 \%\) confidence interval for true average degree of polymerization (as did the authors of the article). Does the interval suggest that 440 is a plausible value for true average degree of polymerization? What about 450 ?
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