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Chapter 9: Problem 42

A researcher wishes to estimate the mean number of miles on 3-year-old Chevy Cavaliers. (a) How many cars should be in a sample to estimate the mean number of miles within 2000 miles with 98\% confidence, assuming that \(\sigma=16,100 ?\) (b) How many cars should be in a sample to estimate the mean number of miles within 1000 miles with 98\% confidence, assuming that \(\sigma=16,100 ?\) (c) What effect does doubling the required accuracy have on the sample size? Why is this the expected result?

Short Answer

Expert verified
The sample sizes are (a) 353 cars, (b) 1410 cars. Doubling accuracy quadruples sample size because sample size varies with the square of the margin of error.

Step by step solution

01

- Understand the Formula for Sample Size

The formula for determining the sample size needed to estimate a population mean within a given margin of error (E) at a certain confidence level is: \[ n = \left( \frac{Z \sigma}{E} \right)^2\] Here, \( n \) is the sample size,\( Z \) is the Z-value corresponding to the confidence level,\( \sigma \) is the standard deviation, and \( E \) is the margin of error.
02

- Identify Given Values for Part (a)

For part (a): Confidence level = 98%, therefore \(Z\) = 2.33 (from Z-tables),\( \sigma = 16,100 \) miles,\( E = 2000 \) miles.
03

- Calculate Sample Size for Part (a)

Substitute the given values into the formula: \[ n = \left( \frac{2.33 \times 16100}{2000} \right)^2 = \left( \frac{37513}{2000} \right)^2 = 352.45\] Since the sample size must be a whole number, round up to get \( n = 353 \).
04

- Identify Given Values for Part (b)

For part (b): Confidence level = 98%, therefore \(Z\) = 2.33 (from Z-tables),\( \sigma = 16,100 \) miles,\( E = 1000 \) miles.
05

- Calculate Sample Size for Part (b)

Substitute the given values into the formula: \[ n = \left( \frac{2.33 \times 16100}{1000} \right)^2 = \left( \frac{37513}{1000} \right)^2 = 1409.8\] Since the sample size must be a whole number, round up to get \( n = 1410 \).
06

- Analyze the Effect of Doubling Accuracy

Doubling the required accuracy means halving the margin of error: If \( E \) is halved, from 2000 miles to 1000 miles, the sample size is multiplied by 4. This is because the sample size formula is inversely proportional to the square of the margin of error: \(n \propto \frac{1}{E^2}\).

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Key Concepts

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

Margin of Error
The margin of error (E) is a crucial component in sample size determination. It represents the maximum expected difference between the true population parameter and the sample estimate. A smaller margin of error requires a larger sample size to ensure statistical confidence. In the exercise:
  • For part (a), E is set to 2000 miles.
  • For part (b), it is reduced to 1000 miles.
Notice how reducing the margin of error increases the required sample size significantly. This is because smaller margins of error demand more precise measurements.
Confidence Level
The confidence level is the probability that the confidence interval contains the true population mean. It is usually expressed as a percentage. Higher confidence levels require larger sample sizes:
  • In this exercise, the confidence level is 98%, corresponding to a Z-value of 2.33.
This level of confidence indicates that if we were to take multiple samples, 98% of them would contain the true population mean. Higher confidence levels provide more reliable results, but they also increase the sample size needed.
Standard Deviation
The standard deviation (\( \sigma \)) measures the amount of variation or dispersion in a set of values. It affects how spreads out the data points are and, in turn, influences the required sample size. In the exercise:
  • The standard deviation is given as 16,100 miles for both parts (a) and (b).
A higher standard deviation typically requires a larger sample size to capture the same level of confidence and precision.
The standard deviation indicates the extent of deviation from the mean, which directly affects the margin of error and sample size.
Z-Value
The Z-value, derived from the Z-table, corresponds to the chosen confidence level in standard normal distribution. It helps determine the number of standard deviations a data point is from the mean. In the exercise:
  • For a 98% confidence level, the Z-value is 2.33.
The Z-value helps translate the confidence level into a numerical factor that influences the sample size. Higher Z-values mean higher confidence levels and, therefore, larger sample sizes.
To find the Z-value for any confidence level, use the Z-table or statistical tools. In formulas, the Z-value is combined with the standard deviation and margin of error to compute the sample size.

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Most popular questions from this chapter

For a simple random sample of forty 2005 Honda Accords (4 cylinder, 2.4 liter, 5 -speed automatic), the mean gas mileage was 23 miles per gallon with a standard deviation of 1.5 miles per gallon. Construct a \(95 \%\) confidence interval for the mean gas mileage of similar 2005 Honda Accords. Interpret the interval.

The following data represent the concentration of organic carbon \((\mathrm{mg} / \mathrm{L})\) collected from organic soil. (TABLE CAN'T COPY) Construct a \(99 \%\) confidence interval for the mean concentration of dissolved organic carbon collected from organic soil. Interpret the interval. (Note: \(\bar{x}=15.92 \mathrm{mg} / \mathrm{L}\) and \(s=7.38 \mathrm{mg} / \mathrm{L} .)\)

State the circumstances under which we construct a \(t\) interval. What are the circumstances under which we construct a Z-interval?

Explain why you should be wary of surveys that do not report a margin of error.

Suppose the arrival of cars at Burger King's drive-through follows a Poisson process with \(\mu=4\) cars every 10 minutes. (a) Simulate obtaining 30 samples of size \(n=40\) from this population. (b) Construct \(90 \%\) confidence intervals for each of the 30 samples. [Note: \(\sigma=\sqrt{\mu}\) in a Poisson process.] (c) How many of the intervals do you expect to include the population mean? How many actually contain the population mean?
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