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Margin of error and \(n\) The Gallup poll in Example 6 reported that during March \(2011,60 \%\) of Americans favored offshore drilling as a means of reducing U.S. dependence on foreign oil. The poll was based on the responses of \(n=1021\) individuals, and resulted in a margin of error of approximately \(3 \%\). Find the approximate margin of error had the poll been based on a sample of size (a) \(n=100,\) (b) \(n=400\), and (c) \(n=1600\). Explain how the margin of error changes as \(n\) increases.

Short Answer

Expert verified
(a) 9.6%, (b) 4.8%, (c) 2.4%; larger sample sizes decrease margin of error.

Step by step solution

01

Understanding Margin of Error Formula

The margin of error (ME) in a proportion is typically calculated using the formula: \[ ME = Z \times \sqrt{\frac{p(1-p)}{n}} \] where \(Z\) is the Z-score corresponding to the desired confidence level (typically 1.96 for 95%), \(p\) is the sample proportion, and \(n\) is the sample size.
02

Calculate for Initial Sample Size

Given: \( n = 1021 \), \( p = 0.60 \), and margin of error \( ME = 3\% = 0.03 \).Using the initial margin of error, we find the \(Z\)-score by rearranging the margin of error formula:\[ 0.03 = Z \times \sqrt{\frac{0.6 \times 0.4}{1021}} \]This calculation will help us maintain the same confidence level (approximate \(Z\)-score) for comparisons.
03

Determine Z-score

Calculate the standard error: \(SE = \sqrt{\frac{0.6 \times 0.4}{1021}}\).\(SE = 0.0154\). From \[ 0.03 = Z \times 0.0154 \],we solve for \(Z\):\(Z \approx \frac{0.03}{0.0154} \approx 1.948\). This is close to 1.96, consistent with a 95% confidence level.
04

Calculate Margin of Error for n=100

For \( n = 100 \):\[ ME = 1.96 \times \sqrt{\frac{0.6 \times 0.4}{100}} \] \[ ME = 1.96 \times \sqrt{0.0024} \] \[ ME \approx 1.96 \times 0.049 \approx 0.096 \].The new margin of error is approximately 9.6%.
05

Calculate Margin of Error for n=400

For \( n = 400 \):\[ ME = 1.96 \times \sqrt{\frac{0.6 \times 0.4}{400}} \] \[ ME = 1.96 \times \sqrt{0.0006} \] \[ ME \approx 1.96 \times 0.0245 \approx 0.048 \].The new margin of error is approximately 4.8%.
06

Calculate Margin of Error for n=1600

For \( n = 1600 \):\[ ME = 1.96 \times \sqrt{\frac{0.6 \times 0.4}{1600}} \] \[ ME = 1.96 \times \sqrt{0.00015} \] \[ ME \approx 1.96 \times 0.01225 \approx 0.024 \].The new margin of error is approximately 2.4%.
07

Analyze Margin of Error Behavior

As the sample size \(n\) increases, the margin of error decreases because the standard error \( \sqrt{\frac{p(1-p)}{n}} \) becomes smaller. This indicates greater precision in the sample estimates with larger sample sizes.

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Key Concepts

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

Sample Size
To understand the impact of sample size on the margin of error, think of it as the number of individuals or units used in a poll or study. This number plays a crucial role in determining the accuracy of the poll results. Larger sample sizes tend to yield more precise results.

- A **large sample size** reduces the margin of error, making estimates more reliable. - A **small sample size** can lead to a larger margin of error, leading to less certainty in the results.
For instance, in our example, using a sample of 1600 individuals instead of 1021 decreased the margin of error from 3% to 2.4%. Conversely, using a sample of just 100 increased the margin to 9.6%.

This concept explains why in national surveys or polls, organizations often try to obtain as large a sample size as feasible within their resources.
Confidence Level
The confidence level is the probability that the value of a parameter lies within a specified range of values, often expressed as a percentage. It reflects how sure you can be about your results.

- For most surveys, a 95% confidence level is standard. This means there's a 95% probability that the sample accurately reflects the population. - A higher confidence level provides more assurance, but it also increases the margin of error if the sample size stays the same.

This is important because it determines the Z-score used in margin of error calculations. In our example, with a Z-score of approximately 1.96, we approximated a 95% confidence level. Increasing the desired confidence level would require adjusting this Z-score, leading to an increase in the margin of error if the sample size doesn’t change.
Standard Error
The standard error is a measure that describes the variability of a sample estimate from the true population parameter. It indicates how much the sample proportion is expected to fluctuate.

- **Calculation**: It's computed using the formula: \[ SE = \sqrt{\frac{p(1-p)}{n}} \]where \( p \) is the sample proportion, and \( n \) is the sample size.
Like the sample size, reducing the standard error can drastically enhance the precision of the study's outcomes.

For our example, as the sample size increased from 100 to 1600, the standard error decreased markedly. This results in a smaller margin of error, which indicates how confident we are that the sample accurately reflects the entire population.
Z-score
The Z-score is a statistical measure that describes a value's relation to the mean of a group of values. It is used to find the confidence level in statistical studies.

- **Relation to Confidence Level**: The Z-score directly corresponds to your confidence level. For a 95% confidence level, the Z-score is typically around 1.96, aligning almost perfectly with our calculated value of approximately 1.948 from the sample proportion.- **Role in Margin of Error**: It serves as a multiplier in the margin of error formula:\[ ME = Z \times \text{Standard Error} \]In our exercise, this Z-score was crucial to determining the margin of error for different sample sizes.

By understanding the Z-score, we gain insight into interpreting the confidence of our results, allowing us to better measure the uncertainty inherent in survey results.

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