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Chapter 7: Problem 24

A sample of 56 research cotton samples resulted in a sample average percentage elongation of \(8.17\) and a sample standard deviation of \(1.42\) ("An Apparent Relation Between the Spiral Angle \(\phi\), the Percent Elongation \(E_{1}\), and the Dimensions of the Cotton Fiber," Textile Research \(J ., 1978\) : 407-410). Calculate a \(95 \%\) large-sample CI for the true average percentage elongation \(\mu\). What assumptions are you making about the distribution of percentage elongation?

Short Answer

Expert verified
The 95% CI for the true average percentage elongation is (7.798, 8.542). Assuming normal distribution of elongation.

Step by step solution

01

Identify the Known Values

To calculate the confidence interval, we need to identify the given values:- Sample Mean (\(\bar{x}\)) = 8.17- Sample Standard Deviation (\(s\)) = 1.42- Sample Size (\(n\)) = 56- Confidence Level = 95%
02

Determine the Z-Score

A 95% confidence level corresponds to a Z-score of 1.96 (commonly used in standard normal distribution tables).
03

Calculate the Standard Error of the Mean (SEM)

The standard error of the mean can be calculated using the formula:\[SEM = \frac{s}{\sqrt{n}}\]Substituting the known values:\[SEM = \frac{1.42}{\sqrt{56}} \approx 0.1898\]
04

Calculate the Margin of Error (MOE)

The margin of error is calculated using the formula:\[MOE = Z \times SEM\]Substituting the values:\[MOE = 1.96 \times 0.1898 \approx 0.372\]
05

Determine the Confidence Interval

The confidence interval (CI) is calculated using the formula:\[\text{CI} = \bar{x} \pm \text{MOE}\]Substitute the sample mean and margin of error:\[\text{CI} = 8.17 \pm 0.372 = (7.798, 8.542)\]
06

Assumptions

We assume that the sample observations are independent and identically distributed (i.i.d) and that the sample size is large enough for the Central Limit Theorem to apply, so the distribution of the sample mean can be approximated as normal.

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Key Concepts

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

Central Limit Theorem
The Central Limit Theorem (CLT) is a key concept in statistics that explains why many distributions approximate a normal distribution when the sample size is large enough. According to the CLT, regardless of the population distribution's shape, the distribution of the sample mean will tend to be normally distributed as the sample size increases.
The CLT allows us to make inferences about population parameters when we're working with sample data.
  • The theorem holds for sample sizes of 30 or greater, which is often considered adequate for the approximation.
  • In the given problem, our sample size is 56, which is comfortably large enough for the CLT to apply.
  • This makes it valid to use the normal distribution to calculate the confidence interval for the sample mean.
Even if the initial data is skewed or non-normal, the mean of the sample will likely follow a normal distribution. This is vital in forming and calculating confidence intervals, as a normal distribution is assumed.
Sample Mean
The sample mean, denoted as \(\bar{x}\), is a critical statistic for summarizing and understanding a data set. It is calculated by summing all sample observations and dividing by the sample size \(n\). In this exercise, the sample mean of the percentage elongation is 8.17.
  • The sample mean provides a measure of the central tendency of the data, serving as a point estimate for the population mean \(\mu\).
  • In confidence interval calculations, the sample mean is used as the midpoint around which the interval is constructed.
  • The reliability of the sample mean improves with larger sample sizes, which is supported by the CLT.
Understanding the sample mean helps in representing the typical value in your data, making it integral in both descriptive statistics and inferential statistical techniques like confidence intervals.
Standard Deviation
Standard deviation, represented as \(s\), reflects the variability or dispersion of a data set around the mean. In this problem, the standard deviation of the sample is 1.42. It indicates how much the sample observations deviate on average from the sample mean.
  • The formula for standard deviation in a sample is \(s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}\).
  • Standard deviation is crucial for understanding the spread of the data and is essential when calculating the standard error.
  • A higher standard deviation indicates a wider spread, while a lower standard deviation suggests data that closely clusters around the mean.
Recognizing the standard deviation's role in statistical calculations underscores its importance in assessing data reliability and variability, contributing to the precision of confidence interval estimations.
Z-Score
The Z-score is a statistical metric that quantifies a data point's deviation from the mean in terms of standard deviations. It is often used in the context of the normal distribution. For the exercise, we used a Z-score of 1.96, which corresponds to a 95% confidence level.
  • A Z-score gives insight into how extreme a data point is compared to the rest of the data.
  • When constructing confidence intervals, the Z-score determines how much to extend from the mean to create the interval.
  • The Z-score selection depends on the confidence level desired; for example, 1.96 for 95% and 2.58 for 99% are common.
The Z-score is essential for transforming data into a standard normal context, allowing us to make probabilistic statements about intervals and predictions based on different confidence levels.

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Most popular questions from this chapter

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The article "Measuring and Understanding the Aging of Kraft Insulating Paper in Power Transformers" (IEEE Electrical Insul. Mag., 1996: 28-34) contained the following observations on degree of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range: \(\begin{array}{lllllll}418 & 421 & 421 & 422 & 425 & 427 & 431 \\ 434 & 437 & 439 & 446 & 447 & 448 & 453 \\ 454 & 463 & 465 & & & & \end{array}\) a. Construct a boxplot of the data and comment on any interesting features. b. Is it plausible that the given sample observations were selected from a normal distribution? c. Calculate a two-sided \(95 \%\) confidence interval for true average degree of polymerization (as did the authors of the article). Does the interval suggest that 440 is a plausible value for true average degree of polymerization? What about 450 ?
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