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A state legislator wishes to survey residents of her district to see what proportion of the electorate is aware of her position on using state funds to pay for abortions. a. What sample size is necessary if the \(95 \%\) CI for \(p\) is to have a width of at most \(.10\) irrespective of \(p\) ? b. If the legislator has strong reason to believe that at least \(2 / 3\) of the electorate know of her position, how large a sample size would you recommend?

Short Answer

Expert verified
The sample size needed is 97 if no information on \(p\) is known. If \(p\) is estimated as \(\frac{2}{3}\), the sample size is 84.

Step by step solution

01

Determine Required Sample Size Formula

The formula for the sample size required for a confidence interval of width \(w\) at a given confidence level is: \[ n = \left( \frac{Z_{\alpha/2}}{w} \right)^2 p(1-p) \] where \(Z_{\alpha/2}\) is the critical value from the normal distribution for \(95\%\), \(w\) is the desired width, and \(p\) is the estimated proportion.
02

Use Maximum Width Formula for Unknown p

When \(p\) is unknown, the worst-case scenario for the largest sample size is used by setting \(p = 0.5\), since this maximizes the product \(p(1-p)\). The formula becomes: \[ n = \left( \frac{1.96}{0.10} \right)^2 \times 0.25 \]
03

Calculate Maximum Sample Size

Plug the values into the equation to get \( n = \left( 19.6 \right)^2 \times 0.25 = 96.04 \). Since sample size needs to be a whole number, round up to 97.
04

Calculate Sample Size with Estimated p

If \(2/3\) of the electorate is aware, this implies \(p = \frac{2}{3} = 0.67\). Use the formula: \[ n = \left( \frac{1.96}{0.10} \right)^2 \times 0.67 \times (1-0.67) \]
05

Compute Optimal Sample Size

Calculate the values: \[ n = 19.6^2 \times 0.67 \times 0.33 = 83.48 \]. Round up to the nearest whole number, giving 84.

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Key Concepts

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

Understanding Confidence Intervals
A confidence interval is a range of values that estimate a particular parameter of a population. It provides an idea of how accurate our sample statistic is in estimating the population parameter. In the context of survey sampling, confidence intervals are crucial as they tell us the degree to which we can trust our sample results to reflect the broader population.
Confidence intervals are typically associated with a confidence level, like 95%, which tells us the probability that the interval contains the true parameter. A 95% confidence level means that if we repeated our sampling over and over, 95% of these intervals would contain the true proportion of the population.
To calculate a confidence interval, especially for proportions, we usually start by determining the sample mean or proportion and then use statistical formulas involving the standard deviation to find the interval. This interval helps us understand the potential margin of error in our survey results.
Proportion Estimation in Surveys
Proportion estimation is a technique used to determine the relative frequency of a particular characteristic within a population. For example, estimating what proportion of a district's voters are aware of a politician's stance on an issue. Proportion estimation is especially useful when dealing with categorical data, like survey responses that yield 'yes' or 'no'.
To estimate proportions accurately, a sample size must be carefully considered. A small sample might give an inaccurate picture, while a larger sample would yield a more reliable estimate of the true population proportion. When planning surveys, it's essential to estimate your proportion accurately. This is often expressed as a fraction (like \( p = \frac{2}{3} \)) representing the part of the population with the characteristic of interest.
All About Survey Sampling
Survey sampling is the process of selecting a group of individuals from a larger population to study. The aim is to make inferences about the entire population from the sample. Correct sampling techniques are crucial as they ensure that the survey results are representative and not biased.
There are several types of sampling methods, including simple random sampling, stratified sampling, and cluster sampling. Each method has its own pros and cons and is chosen based on the research objectives.
In survey sampling, the size of the sample is crucial. A larger sample size generally leads to more accurate survey results. However, the sample size decision also depends on the desired confidence interval width and the estimated proportion of the population with the desired characteristic.
Normal Distribution and Its Importance
The normal distribution, often referred to as the bell curve, is a vital concept in statistics and is used extensively in many areas, including survey sampling. It is a continuous probability distribution characterized by its symmetrical bell shape and is defined by its mean and standard deviation.
Most real-world phenomena, like heights, test scores, and measurement errors, tend to follow a normal distribution. This property makes the normal distribution incredibly valuable for statistical analyses, including confidence intervals. The central limit theorem further adds to its importance by stating that the means of samples from a population will be normally distributed, regardless of the original distribution, if the sample size is large enough.
The concept of the normal distribution is integral when determining the critical value for calculating confidence intervals. For a 95% confidence interval, the critical value from the normal distribution is often 1.96, which is used alongside the sample size and proportion to estimate the confidence interval width.

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Most popular questions from this chapter

The article "Measuring and Understanding the Aging of Kraft Insulating Paper in Power Transformers" (IEEE Electrical Insul. Mago, 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 ?

Determine the \(t\) critical value for a two-sided confidence interval in each of the following situations: a. Confidence level \(=95 \%\), df \(=10\) b. Confidence level \(=95 \%\), df \(=15\) c. Confidence level \(=99 \%\), df \(=15\) d. Confidence level \(=99 \%, n=5\) e. Confidence level \(=98 \%\), df \(=24\) f. Confidence level \(=99 \%, n=38\)

A sample of 56 research cotton samples resulted in a sample average percentage elongation of \(8.17\) and a sample standard deviation of \(1.42(*\) An Apparent Relation Between the Spiral Angle \(\phi\), the Percent Elongation \(E_{1}\), and the Dimensions of the Cotton Fiber," Textile Research \(J_{.}, 1978: 407-410\) ). Calculate a \(95 \%\) large-sample CI for the true average percentage elongation \(\mu\). What assumptions are you making about the distribution of percentage elongation?

Consider a normal population distribution with the value of \(\sigma\) known. a. What is the confidence level for the interval \(\bar{x} \pm\) \(2.81 \sigma / \sqrt{n}\) ? b. What is the confidence level for the interval \(\bar{x} \pm\) \(1.44 \sigma / \sqrt{n} ?\) c. What value of \(z_{\alpha / 2}\) in the CI formula (7.5) results in a confidence level of \(99.7 \%\) ? d. Answer the question posed in part (c) for a confidence level of \(75 \%\).

The article "Gas Cooking, Kitchen Ventilation, and Exposure to Combustion Products" (Indoor Air, 2006: 65-73) reported that for a sample of 50 kitchens with gas cooking appliances monitored during a oneweek period, the sample mean \(\mathrm{CO}_{2}\) level (ppm) was \(654.16\), and the sample standard deviation was 164.43. a. Calculate and interpret a \(95 \%\) (two-sided) confidence interval for true average \(\mathrm{CO}_{2}\) level in the population of all homes from which the sample was selected. b. Suppose the investigators had made a rough guess of 175 for the value of \(s\) before collecting data. What sample size would be necessary to obtain an interval width of \(50 \mathrm{ppm}\) for a confidence level of \(95 \%\) ?

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