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Consider two data sets with equal sample standard deviations. The first data set has 20 data values that are not all equal, and the second has 50 data values that are not all equal. For which data set is the difference between \(s\) and \(\sigma\) greater? Explain. Hint: Consider the relationship \(\sigma=s \sqrt{(n-1) / n}\).

Short Answer

Expert verified
The first data set (20 values) has a greater difference between \( s \) and \( \sigma \).

Step by step solution

01

Understand the formulas

The sample standard deviation \( s \) is related to the population standard deviation \( \sigma \) through the formula \( \sigma = s \sqrt{\frac{n-1}{n}} \), where \( n \) is the sample size. This formula accounts for Bessel's correction, which adjusts the bias in the estimation of the standard deviation from a sample.
02

Set up the expressions for each data set

For the first data set with 20 values (\( n_1 = 20 \)), the formula becomes \( \sigma_1 = s \sqrt{\frac{19}{20}} \). For the second data set with 50 values (\( n_2 = 50 \)), it becomes \( \sigma_2 = s \sqrt{\frac{49}{50}} \).
03

Analyze the difference between \( s \) and \( \sigma \)

The difference between the sample and population standard deviations, \( s - \sigma \), depends on the term \( \sqrt{\frac{n-1}{n}} \). For each data set: \( s - \sigma_1 = s \left( 1 - \sqrt{\frac{19}{20}} \right) \) and \( s - \sigma_2 = s \left( 1 - \sqrt{\frac{49}{50}} \right) \). Since the square root value closer to 1 reduces this difference, a smaller \( n \) results in a larger difference.
04

Compare the two terms

Calculate each term: \( 1 - \sqrt{\frac{19}{20}} \) and \( 1 - \sqrt{\frac{49}{50}} \). We know \( \frac{19}{20} < \frac{49}{50} \), hence \( \sqrt{\frac{19}{20}} < \sqrt{\frac{49}{50}} \). Therefore, \( 1 - \sqrt{\frac{19}{20}} \) is greater than \( 1 - \sqrt{\frac{49}{50}} \).
05

Conclusion

The first data set with 20 data values will have a greater difference between \( s \) and \( \sigma \) compared to the second data set with 50 values because reducing \( n \) means \( 1 - \sqrt{\frac{n-1}{n}} \) increases.

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Key Concepts

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

Sample Standard Deviation
The sample standard deviation, often denoted as \( s \), is a measure of how much individual data points in a specific sample deviate from the mean of that sample. It is an important statistic used to estimate the variance within a sample and serves as an unbiased estimator of the population standard deviation when adjusted properly.

Calculating the sample standard deviation involves the following steps:
  • Compute the mean (average) of the sample data.
  • Subtract the mean from each data point to find deviation scores.
  • Square each deviation score.
  • Find the average of these squared deviations (this is known as the sample variance).
  • Take the square root of the sample variance to get the sample standard deviation.
Notice that when calculating the sample variance, we divide by \( n-1 \) instead of \( n \), where \( n \) is the number of data points in the sample. This specific adjustment is crucial and is known as Bessel's correction.

The reason for this adjustment is to correct the bias that would occur if we treated our sample as if it were the entire population. Using \( n-1 \) instead of \( n \) makes our sample variance a more accurate estimate of the population variance, even though it slightly inflates our variance estimate.
Population Standard Deviation
The population standard deviation, represented by \( \sigma \), is a parameter that describes the spread of data points in an entire population. Unlike the sample standard deviation, the population standard deviation is calculated using all members of a population, so there is no need for Bessel's correction.

The calculation for the population standard deviation follows these steps:
  • Determine the mean of the entire population.
  • Subtract the population mean from each data value to find deviation scores.
  • Square these deviation scores.
  • Average the squared deviations to obtain the population variance.
  • Finally, take the square root of the population variance to find the population standard deviation.
Since everyone in the population is observed, we divide by \( n \), which is the population size. The population standard deviation is a fixed value because it measures a defined group completely.

In the context of inferential statistics, when we don’t have access to the full population data, we use the sample standard deviation as an estimate for the population standard deviation.
Bessel's Correction
Bessel's correction is a small yet crucial adjustment applied when calculating the sample variance and subsequently the sample standard deviation. This adjustment corrects for the bias that occurs when estimating the population standard deviation from a sample.

Here’s how it works:
  • When calculating the sample variance, instead of dividing the sum of squared deviations by the sample size \( n \), we divide by \( n-1 \).
  • This adjustment increases the calculated variance slightly, making it a better estimate of the true variance.
  • As a result, when you compute the sample standard deviation, it's also slightly larger, better mirroring the population standard deviation.
Bessel's correction is necessary because a sample is an approximation of the population. Without this correction, the calculated sample variance (and thus the sample standard deviation) would systematically underestimate the actual population variance.

Given the formula \( \sigma = s \sqrt{\frac{n-1}{n}} \), it is evident that as the sample size \( n \) increases, the factor \( \sqrt{\frac{n-1}{n}} \) approaches 1, reducing the difference between \( s \) and \( \sigma \). For small samples, the correction plays a significant role, contributing to a larger discrepancy between \( s \) and \( \sigma \).

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Most popular questions from this chapter

What is the age distribution of adult shoplifters \((21\) years of age or older) in supermarkets? The following is based on information taken from the National Retail Federation. A random sample of 895 incidents of shoplifting gave the following age distribution: \begin{tabular}{l|ccc} \hline Age range (years) & \(21-30\) & \(31-40\) & 41 and over \\ \hline Number of shoplifters & 260 & 348 & 287 \\ \hline \end{tabular} Estimate the mean age, sample variance, and sample standard deviation for the shoplifters. For the class 41 and over, use \(45.5\) as the class midpoint.

If you like mathematical puzzles or love algebra, try this! Otherwise, just trust that the computational formula for the sum of squares is correct. We have a sample of \(x\) values. The sample size is \(n\). Fill in the details for the following steps. $$ \begin{aligned} \Sigma(x-\bar{x})^{2} &=\Sigma x^{2}-2 \bar{x} \sum x+n \bar{x}^{2} \\ &=\Sigma x^{2}-2 n \bar{x}^{2}+n \bar{x}^{2} \\ &=\Sigma x^{2}-\frac{(\Sigma x)^{2}}{n} \end{aligned} $$

Given the sample data \(\begin{array}{llllll}x: & 23 & 17 & 15 & 30 & 25\end{array}\) (a) Find the range. (b) Verify that \(\Sigma x=110\) and \(\Sigma x^{2}=2568\). (c) Use the results of part (b) and appropriate computation formulas to compute the sample variance \(s^{2}\) and sample standard deviation \(s\). (d) Use the defining formulas to compute the sample variance \(s^{2}\) and sample standard deviation \(s\). (e) Suppose the given data comprise the entire population of all \(x\) values. Compute the population variance \(\sigma^{2}\) and population standard deviation \(\sigma\).

Consider the numbers \(\begin{array}{lllll}2 & 3 & 4 & 5 & 5\end{array}\) (a) Compute the mode, median, and mean. (b) If the numbers represent codes for the colors of T-shirts ordered from a catalog, which average(s) would make sense? (c) If the numbers represent one-way mileages for trails to different lakes, which average(s) would make sense? (d) Suppose the numbers represent survey responses from 1 to 5, with \(1=\) disagree strongly, \(2=\) disagree, \(3=\) agree, \(4=\) agree strongly, and \(5=\) agree very strongly. Which averages make sense?

What was the age distribution of prehistoric Native Americans? Extensive anthropologic studies in the southwestern United States gave the following information about a prehistoric extended family group of 80 members on what is now the Navajo Reservation in northwestern New Mexico (Source: Based on information taken from Prehistory in the Navajo Reservation District, by F. W. Eddy, Museum of New Mexico Press). \begin{tabular}{l|cccc} \hline Age range (years) & \(1-10^{*}\) & \(11-20\) & \(21-30\) & 31 and over \\ \hline Number of individuals & 34 & 18 & 17 & 11 \\ \hline \end{tabular} "Includes infants. For this community, estimate the mean age expressed in years, the sample variance, and the sample standard deviation. For the class 31 and over, use \(35.5\) as the class midpoint.

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