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Farm size An estimate is needed of the mean acreage of farms in Ontario, Canada. A \(95 \%\) confidence interval should have a margin of error of 25 acres. A study 10 years ago in this province had a sample standard deviation of 200 acres for farm size. a. About how large a sample of farms is needed? b. A sample is selected of the size found in part a. However, the sample has a standard deviation of 300 acres, rather than 200 . What is the margin of error for a \(95 \%\) confidence interval for the mean acreage of farms?

Short Answer

Expert verified
a. A sample of 246 farms is needed. b. The margin of error is approximately 37.5 acres.

Step by step solution

01

Understand the Problem

We need to determine the sample size required for a confidence interval with a specific margin of error. We have a desired margin of error \(ME = 25\) acres and a past study standard deviation \(s = 200\) acres. The confidence level is \(95\%\), corresponding to a Z-score of approximately 1.96 for a normal distribution.
02

Calculate Sample Size (Part a)

To find the necessary sample size \(n\), use the formula for the margin of error: \(ME = \frac{Z \cdot s}{\sqrt{n}}\). First, rearrange to find \(n\): \(n = \left(\frac{Z \cdot s}{ME}\right)^2\). Substitute the values: \(n = \left(\frac{1.96 \times 200}{25}\right)^2\). Calculate this to get \(n = \left(15.68\right)^2 = 245.8624\). Since sample size must be a whole number, round up to \(n = 246\).
03

Re-evaluate margin of error with new standard deviation (Part b)

Now, consider the sample standard deviation is actually \(s = 300\) acres instead of 200. Use the same margin of error formula: \(ME = \frac{Z \cdot s}{\sqrt{n}}\). Substitute the values: \(ME = \frac{1.96 \times 300}{\sqrt{246}}\). Calculate the root of 246, which is approximately 15.68. Compute the margin of error: \(ME = \frac{588}{15.68} \approx 37.5\) acres.

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Key Concepts

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

Sample Size Calculation
Determining the right sample size is crucial for estimating the mean of a population with a given confidence level and margin of error. Here, we're seeking to find out how many farms need to be sampled to estimate the average farm size in Ontario, Canada, within a 25-acre margin of error. The 95% confidence level means we're 95% confident that the true mean lies within this range. To calculate sample size, we use the formula: \[ n = \left(\frac{Z \cdot s}{ME}\right)^2 \]where:
  • n is the sample size.
  • Z is the Z-score corresponding to the confidence level (1.96 for 95%).
  • s is the sample standard deviation.
  • ME stands for Margin of Error.
Plugging in our values: using 1.96 for the Z-score and 200 for the standard deviation, we calculate the sample size as \[ n = \left(\frac{1.96 \times 200}{25}\right)^2 = 246 \] (rounded up to the nearest whole number). Choosing an appropriate sample size ensures the sample accurately reflects the population, allowing us to make reliable inferences.
Margin of Error
The margin of error gives us a window in which we can expect the true mean of our population to fall. It's a critical component for ensuring the accuracy and reliability of our confidence intervals. In this case, we initially target a margin of error of 25 acres. However, our margin of error can change if the sample standard deviation varies from what we expected. The original margin of error was calculated using a standard deviation of 200 acres. If the true sample standard deviation turns out to be 300 acres, as experienced in the revision of our problem, the margin of error must be recalculated using the same formula: \[ ME = \frac{Z \cdot s}{\sqrt{n}} \]With a sample size of 246, Z-score of 1.96, and standard deviation of 300, the new margin of error becomes:\[ ME = \frac{1.96 \times 300}{\sqrt{246}} \approx 37.5 \text{ acres} \]This recalculated margin of error shows an increase, indicating a wider confidence interval. It's critical for decision-makers to be aware of such adjustments, as they affect the degree of confidence we can place in the interval.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In the context of our farm size problem, it reflects how much individual farm sizes deviate from the average farm size. A larger standard deviation indicates more variability in farm sizes across the sample population. In the original calculation for determining sample size and margin of error, we used a standard deviation of 200 acres, based on past data. When the actual sample showed a standard deviation of 300 acres, it signaled greater variability in farm sizes than the previous study suggested. This affects our confidence interval as more variability necessitates a larger margin of error to maintain a given confidence level. To illustrate, if you have a narrow standard deviation, your data points are clustered closely around the mean, indicating consistency. Conversely, a wide standard deviation suggests that your data points are spread out, reflecting greater diversity in your sample observations. Understanding the standard deviation helps in adapting the sample size and accommodating the margin of error to provide more robust confidence intervals, ultimately leading to better decision-making based on statistical analyses.

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Most popular questions from this chapter

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