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Explain why quadrupling the sample size causes the margin of error to be cut in half.

Short Answer

Expert verified
Quadrupling the sample size results in the margin of error being halved because the margin of error is inversely proportional to the square root of the sample size.

Step by step solution

01

Understand the Margin of Error Formula

The margin of error (ME) in statistics is defined as \[ ME = z \frac{\sigma}{\sqrt{n}} \] where \( z \) is the z-score corresponding to the desired confidence level, \( \sigma \) is the population standard deviation, and \( n \) is the sample size.
02

Analyze the Effect of Quadrupling the Sample Size

To understand the effect of increasing the sample size, substitute \( n \) with \( 4n \) in the formula: \[ ME = z \frac{\sigma}{\sqrt{4n}} = z \frac{\sigma}{2\sqrt{n}} \].
03

Simplify the New Margin of Error

Note that \[ \frac{\sigma}{2\sqrt{n}} = \frac{1}{2} \cdot \frac{\sigma}{\sqrt{n}} \]. Therefore, the new margin of error is: \[ ME_{new} = \frac{1}{2} ME_{original} \].
04

Conclude the Relationship

From the simplified margin of error formula, it is evident that quadrupling the sample size \(n\) results in a new margin of error that is half the original margin of error.

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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 is a crucial concept in statistics, often used to describe the uncertainty in an estimate. It reflects the amount by which the sample estimate may differ from the true population value.
Mathematically, the margin of error (ME) is given by the formula:
\[ ME = z \frac{\sigma}{\sqrt{n}} \] Here,
  • z represents the z-score at a given confidence level.
  • \(\sigma\) is the population standard deviation.
  • \(n\) is the sample size.
The formula shows that the margin of error is directly proportional to the population standard deviation and inversely proportional to the square root of the sample size. This means larger sample sizes generally lead to a smaller margin of error, indicating more precise estimates.
Sample Size
Sample size (\(n\)) plays a vital role in determining the precision of a statistical estimate. It refers to the number of observations or data points collected in a sample.
In the formula for margin of error:

\[ ME = z \frac{\sigma}{\sqrt{n}} \]
You can see that the sample size appears in the denominator under a square root. As the sample size increases, the denominator of the fraction becomes larger, which makes the margin of error smaller. This means that larger samples provide more accurate estimates of the population parameter.

When discussing the effect of quadrupling the sample size,
If we increase \(n\) to \(4n\), the new margin of error becomes:
\[ ME new = z \frac{\sigma}{\sqrt{4n}} = z \frac{\sigma}{2\sqrt{n}} \]
Thus, the new margin of error is half of the original, highlighting how sample size directly affects the accuracy of statistical results.
Confidence Level
Confidence level is an essential element in interpreting statistical results. It represents the probability that the true population parameter lies within the calculated confidence interval.
Common confidence levels are 90%, 95%, and 99%. A higher confidence level means a larger z-score, which increases the margin of error, leading to wider confidence intervals. For example:
  • At a 95% confidence level, the z-score is approximately 1.96.
  • At a 99% confidence level, the z-score is approximately 2.576.

Thus, a higher confidence level provides greater assurance that the true parameter is within the interval but at the cost of precision, i.e., a larger margin of error. The balance between confidence and precision is key in statistical analysis.
Population Standard Deviation
The population standard deviation (σ) measures the spread or variability of a population distribution.
In the margin of error formula:
\[ ME = z \frac{\sigma}{\sqrt{n}} \]
\(\sigma\) is in the numerator, signifying that higher variability within the population results in a larger margin of error. This underscores the challenge of estimating parameters in populations with high variability.

Accurately estimating σ from sample data can be difficult, especially with small samples. However, in smaller samples, the same principle applies: higher population variability leads to less precise estimates. Recognizing the role of standard deviation in margin of error calculations helps understand the reliability of statistical results.

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Most popular questions from this chapter

Clayton Kershaw of the Los Angeles Dodgers is one of the premier pitchers in baseball. His most popular pitch is a four-seam fastball. The data in the next column represent the pitch speed (in miles per hour) for a random sample of 18 of his four-seam fastball pitches. $$ \begin{array}{llllll} \hline 93.63 & 93.83 & 94.18 & 94.71 & 95.52 & 95.07 \\ \hline 95.12 & 95.35 & 94.15 & 94.62 & 96.08 & 93.86 \\ \hline 94.75 & 94.70 & 95.28 & 95.49 & 95.77 & 93.34 \\ \hline \end{array} $$ (a) Is "pitch speed" a quantitative or qualitative variable? Why is it important to know this when determining the type of confidence interval you may construct? (b) Draw a normal probability plot to verify that "pitch speed" could come from a population that is normally distributed. (c) Draw a boxplot to verify the data set has no outliers. (d) Are the requirements for constructing a confidence interval for the mean pitch speed of a Clayton Kershaw four-seam fastball satisfied? (e) Construct and interpret a \(95 \%\) confidence interval for the mean pitch speed of a Clayton Kershaw four-seam fastball. (f) Do you believe that a \(95 \%\) confidence interval for the mean pitch speed of all major league pitchers' four-seam fastbal would be narrower or wider? Why?

Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 20 students, she finds 2 who eat cauliflower. Obtain and interpret a \(95 \%\) confidence interval for the proportion of students who eat cauliflower on Jane's campus using Agresti and Coull's method.

True or False: A \(95 \%\) confidence interval for a population proportion with lower bound 0.45 and upper bound 0.51 means there is a \(95 \%\) probability the population proportion is between 0.45 and 0.51

(a) Find the \(t\) -value such that the area in the right tail is 0.10 with 25 degrees of freedom. (b) Find the \(t\) -value such that the area in the right tail is 0.05 with 30 degrees of freedom. (c) Find the \(t\) -value such that the area left of the \(t\) -value is 0.01 with 18 degrees of freedom. [Hint: Use symmetry.] (d) Find the critical \(t\) -value that corresponds to \(90 \%\) confidence. Assume 20 degrees of freedom.

In a survey conducted by the marketing agency 11 mark, 241 of 1000 adults 19 years of age or older confessed to bringing and using their cell phone every trip to the bathroom (confessions included texting and answering phone calls). (a) What is the sample in this study? What is the population of interest? (b) What is the variable of interest in this study? Is it qualitative or quantitative? (c) Based on the results of this survey, obtain a point estimate for the proportion of adults 19 years of age or older who bring their cell phone every trip to the bathroom. (d) Explain why the point estimate found in part (c) is a statistic. Explain why it is a random variable. What is the source of variability in the random variable? (e) Construct and interpret a \(95 \%\) confidence interval for the population proportion of adults 19 years of age or older who bring their cell phone every trip to the bathroom. (f) What ensures that the results of this study are representative of all adults 19 years of age or older?

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