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Suppose \(x\) has a mound-shaped distribution with \(\sigma=3\). (a) Find the minimal sample size required so that for a \(95 \%\) confidence interval, the maximal margin of error is \(E=0.4\). (b) Check Requirements Based on this sample size, can we assume that the \(\bar{x}\) distribution is approximately normal? Explain.

Short Answer

Expert verified
(a) The sample size required is 217. (b) Yes, because the sample size is large enough (\(n = 217\)).

Step by step solution

01

Understand the Problem

We need to find the minimum sample size, denoted as \(n\), required for a 95% confidence interval where the margin of error, \(E\), is 0.4. We have the population standard deviation, \(\sigma = 3\).
02

Formula for Margin of Error

The formula for the margin of error \(E\) in a confidence interval is \(E = z \cdot \frac{\sigma}{\sqrt{n}}\), where \(z\) is the z-score corresponding to the desired confidence level (95% in this case).
03

Find the Z-Score for 95% Confidence

For a 95% confidence level, the z-score, \(z\), can be found from statistical tables or a calculator, and it is approximately \(z = 1.96\).
04

Set up the Inequality for Sample Size

Using the margin of error formula, we set up the inequality as \(0.4 = 1.96 \cdot \frac{3}{\sqrt{n}}\). We need to solve for \(n\).
05

Solve for Sample Size

Rearrange the inequality to solve for \(n\):\[0.4 = 1.96 \cdot \frac{3}{\sqrt{n}}\]\[\frac{0.4}{1.96} = \frac{3}{\sqrt{n}}\]\[\sqrt{n} = \frac{3 \cdot 1.96}{0.4}\]\[\sqrt{n} = \frac{5.88}{0.4}\]\[\sqrt{n} = 14.7\]Square both sides to find \(n\):\[n = (14.7)^2 \approx 216.09\]Since \(n\) must be an integer, round up to get \(n = 217\).
06

Check Normality Requirement

For the distribution of \(\bar{x}\) to be approximately normal, according to the Central Limit Theorem, a sample size \(n \geq 30\) is generally sufficient if the population is mound-shaped. Here, \(n = 217\), which is well above 30, so the requirement is satisfied.

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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 an important concept when dealing with confidence intervals in statistics. This value represents how much we expect our sample statistics, like the sample mean, to differ from the true population parameter.

In a 95% confidence interval, the margin of error tells us the range within which the true population mean is likely to lie. We derive it using the formula:
  • \(E = z \cdot \frac{\sigma}{\sqrt{n}}\)
Here, \(E\) is the margin of error, \(z\) is the z-score related to our confidence level, \(\sigma\) is the population standard deviation, and \(n\) is the sample size.

The goal is to make \(E\) as small as practical, ensuring greater precision. In our example from the original exercise, given a target margin of error of 0.4 and a 95% confidence level, we used the z-score of 1.96 (common for 95% confidence) to calculate the necessary sample size. Lower confidence levels or larger margins of error would decrease the z-score or increase \(E\), altering required sample sizes.
Sample Size Calculation
Calculating the right sample size is key to ensuring reliable results in statistical studies. Larger samples provide more accurate estimates of the population, but they also require more resources. So, knowing how to find the right balance is crucial.

To calculate sample size particularly for a confidence interval with a predetermined margin of error, we rearrange the margin of error formula as follows:
  • \[\sqrt{n} = \frac{z \cdot \sigma}{E}\]
  • Square the result: \(n = \left(\frac{z \cdot \sigma}{E}\right)^2\)
This ensures our sample adequately represents the population given our confidence level and desired precision.

In the original problem, solving provides \(n \approx 216.09\). Since sample size must be a whole number, round up to the next whole number, 217. By adjusting this formula, one can predict how changes in z-score, \(\sigma\), or margin of error impact \(n\). It’s a useful equation for planning data collection!
Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental principle in statistics that greatly simplifies analysis. It states that given a sufficiently large sample size, the sampling distribution of the sample mean will be approximately normally distributed, even if the original population distribution is not perfect.

As a rule of thumb, a sample size of 30 or greater is often enough; however, more may be needed depending on the shape and spread of the original population distribution.
  • This theorem is central to the standard practice because it provides a basis for making inferences about populations from samples.
For our example from the exercise, since we calculated a sample size of 217, this amount surpasses the threshold requirement for CLT application even more confident statements about the population mean.

Thus, the Central Limit Theorem reassures us that, as n increases, the sampling distribution becomes more normal, facilitating the use of traditional statistical models and formulas to predict population characteristics accurately.

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Most popular questions from this chapter

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