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The following results on summations will help us in calculating the sample variance \(s^{2}\). For any constant \(c\), a. \(\sum_{i=1}^{n} c=n c\). b. \(\sum_{i=1}^{n} c y_{i}=c \sum_{i=1}^{n} y_{i}\). c. \(\sum_{i=1}^{n}\left(x_{i}+y_{i}\right)=\sum_{i=1}^{n} x_{i}+\sum_{i=1}^{n} y_{i}\). Use (a), (b), and (c) to show that $$s^{2}=\frac{1}{n-1} \sum_{i=1}^{n}\left(y_{i}-\bar{y}\right)^{2}=\frac{1}{n-1}\left[\sum_{i=1}^{n} y_{i}^{2}-\frac{1}{n}\left(\sum_{i=1}^{n} y_{i}\right)^{2}\right].$$

Short Answer

Expert verified
Sample variance can be expressed as \(s^2 = \frac{1}{n-1}\left[\sum_{i=1}^{n} y_i^2 - \frac{1}{n}\left(\sum_{i=1}^{n} y_i\right)^2\right]\) using summation properties.

Step by step solution

01

Understand the Sample Variance Formula

The sample variance formula is defined as \(s^2 = \frac{1}{n-1} \sum_{i=1}^{n}(y_i - \bar{y})^2\), where \(\bar{y}\) is the sample mean.
02

Express \\(\bar{y}\\) in Terms of Summation

\(\bar{y} = \frac{1}{n} \sum_{i=1}^{n} y_i\) is the definition of the sample mean. We will substitute \(\bar{y}\) in the variance formula.
03

Substitute \((y_i - \bar{y})^2\)

Substitute \(\bar{y} = \frac{1}{n} \sum_{i=1}^{n} y_i\) into \(s^2\), resulting in \(s^2 = \frac{1}{n-1} \sum_{i=1}^{n}\left(y_i - \frac{1}{n} \sum_{i=1}^{n} y_i\right)^2\).
04

Expand \((y_i - \bar{y})^2\) using Algebra

Expand \(\left(y_i - \frac{1}{n} \sum_{i=1}^{n} y_i\right)^2\) into \(y_i^2 - 2y_i\frac{1}{n}\sum_{j=1}^{n} y_j + \left(\frac{1}{n}\sum_{j=1}^{n} y_j\right)^2\).
05

Apply Summation Properties

Using property (c): \(\sum_{i=1}^{n}(x_i + y_i) = \sum_{i=1}^{n} x_i + \sum_{i=1}^{n} y_i\), separate the summation into three parts: \(\sum_{i=1}^{n} y_i^2 - 2\sum_{i=1}^{n} y_i \cdot \frac{1}{n} \sum_{j=1}^{n} y_j + \sum_{i=1}^{n}\left(\frac{1}{n}\sum_{j=1}^{n} y_j\right)^2\).
06

Simplify the Middle Term

Recognize \(-2\sum_{i=1}^{n} y_i \cdot \frac{1}{n} \sum_{j=1}^{n} y_j\) uses property (b), resulting in \(-2 \cdot \frac{1}{n} \left(\sum_{i=1}^{n} y_i\right)^2\).
07

Simplify the Third Term

Using property (a): \(\sum_{i=1}^{n} c = nc\), the term \(\sum_{i=1}^{n}\left(\frac{1}{n}\sum_{j=1}^{n} y_j\right)^2 = \frac{1}{n}\left(\sum_{i=1}^{n} y_j\right)^2\).
08

Combine and Simplify the Expression

Combine the simplified terms into the equation: \[\frac{1}{n-1}\left[\sum_{i=1}^{n} y_i^2 - 2 \cdot \frac{1}{n}\left(\sum_{i=1}^{n} y_i\right)^2 + \frac{1}{n}\left(\sum_{i=1}^{n} y_i\right)^2\right]\].Simplifying further leads to:\[\frac{1}{n-1}\left[\sum_{i=1}^{n} y_i^2 - \frac{1}{n}\left(\sum_{i=1}^{n} y_i\right)^2\right]\].

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Key Concepts

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

Summation Properties
In mathematics, summation properties are essential tools that simplify complex calculations, especially when dealing with sequences and series. Understanding and applying these properties can make working with statistical formulas, such as the sample variance, more efficient and straightforward.

Let's unpack the three crucial summation properties used in the calculation of sample variance:

1. **Property (a): Constant Summation**
- For a summation involving a constant, the rule is \( \sum_{i=1}^{n} c = nc \).
- This means if you sum a constant \(c\) n times, the result is simply \( nc \), the constant multiplied by the number of terms.

2. **Property (b): Scalar Multiplication**
- This property states \( \sum_{i=1}^{n} c y_{i} = c \sum_{i=1}^{n} y_{i} \).
- It tells us that multiplying each term in a summation by a constant \(c\) can be factored out, making calculations simpler.

3. **Property (c): Summing Two Sequences**
- Expressed as \( \sum_{i=1}^{n}(x_{i} + y_{i}) = \sum_{i=1}^{n} x_{i} + \sum_{i=1}^{n} y_{i} \).
- This property helps separate the summation of sums into separate terms, allowing for easier manipulation and solving of equations.

By applying these properties, the expansion and simplification of mathematical expressions, like those in the sample variance formula, become more accessible. It allows for efficient transformation and simplification of terms, crucial for statistical analysis.
Algebraic Expansion
Algebraic expansion is a method used to simplify expressions that involve binomials and polynomials. In the context of the sample variance, algebraic expansion helps to decompose the expression \((y_i - \bar{y})^2\) into a more workable form.

Here's a step-by-step breakdown of the expansion process in this scenario:

- **Original Expression**
- The expression \((y_i - \bar{y})^2\) represents the squared deviation of each observation \(y_i\) from the sample mean \(\bar{y}\).

- **Expansion Steps**
- Expand the expression using the algebraic identity for squaring binomials: \((a - b)^2 = a^2 - 2ab + b^2\).
- Substitute \(y_i\) and \(\bar{y} = \frac{1}{n} \sum_{i=1}^{n} y_i\) to get: \[ y_i^2 - 2\left( y_i \frac{1}{n} \sum_{j=1}^{n} y_j \right) + \left( \frac{1}{n} \sum_{j=1}^{n} y_j \right)^2 \].

- **Purpose of Expansion**
- The expansion allows us to break down a complex quadratic term into manageable linear and constant components.
- It makes it easier to apply summation properties to further simplify and solve the variance formula.

Through algebraic expansion, we achieve a clearer structure that aids in applying further mathematical operations, ultimately simplifying the computational process of finding the variance.
Statistical Formulas
Statistical formulas are foundational tools in data analysis and interpretation, with the sample variance formula playing a key role in data variability assessments.

**Sample Variance Formula**
- The sample variance is a measure of how spread out the values in a data sample are. The formula used is:
\[ s^2 = \frac{1}{n-1} \sum_{i=1}^{n}(y_i - \bar{y})^2 \]

- **Components of the Formula**
- \(n\) is the total number of observations.
- \(y_i\) represents each observation in the sample.
- \(\bar{y}\) stands for the sample mean, calculated as \( \bar{y} = \frac{1}{n} \sum_{i=1}^{n} y_i \).
- The expression \((y_i - \bar{y})^2\) calculates the square of each observation's deviation from the mean.

- **Understanding Denominators**
- The "degrees of freedom" in the denominator, \(n-1\), adjusts for bias in the sample, as it compensates for using the sample mean as an estimator.

The formula can also be expressed through a different form by expanding it and simplifying the expressions:

- \[ s^2 = \frac{1}{n-1}\left[\sum_{i=1}^{n} y_i^2 - \frac{1}{n}\left(\sum_{i=1}^{n} y_i\right)^2\right] \]

This alternative form makes it easier to compute using the sum of individual squares and the square of the total sum. Understanding these aspects of statistical formulas enhances your ability to dissect data variability thoroughly.

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Most popular questions from this chapter

A pharmaceutical company wants to know whether an experimental drug has an effect on systolic blood pressure. Fifteen randomly selected subjects were given the drug and, after sufficient time for the drug to have an impact, their systolic blood pressures were recorded. The data appear below:$$\begin{array}{lllll} 172 & 140 & 123 & 130 & 115 \\ 148 & 108 & 129 & 137 & 161 \\ 123 & 152 & 133 & 128 & 142 \end{array}$$ a. Approximate the value of \(s\) using the range approximation. b. Calculate the values of \(\bar{y}\) and \(s\) for the 15 blood pressure readings. c. Use Tchebysheff's theorem (Exercise 1.32 ) to find values \(a\) and \(b\) such that at least \(75 \%\) of the blood pressure measurements lie between \(a\) and \(b\) d. Did Tchebysheff's theorem work? That is, use the data to find the actual percent of blood pressure readings that are between the values \(a\) and \(b\) you found in part (c). Is this actual percentage greater than \(75 \%\) ?

Compared to their stay-at-home peers, women employed outside the home have higher levels of high-density lipoproteins (HDL), the "good" cholesterol associated with lower risk for heart attacks. A study of cholesterol levels in 2000 women, aged \(25-64\), living in Augsburg, Germany, was conducted by Ursula Haertel, Ulrigh Keil, and colleagues \(^{\star}\) at the GSF-Medis Institut in Munich. Of these 2000 women, the \(48 \%\) who worked outside the home had HDL levels that were between 2.5 and 3.6 milligrams per deciliter (mg/dL) higher than the HDL levels of their stay-at-home counterparts. Suppose that the difference in HDL levels is normally distributed, with mean 0 (indicating no difference between the two groups of women) and standard deviation \(1.2 \mathrm{mg} / \mathrm{dL}\). If you were to select an employed woman and a stay-at-home counterpart at random, what is the probability that the difference in their HDL levels would be between 1.2 and \(2.4 ?\)

A set of 340 examination scores exhibiting a bell-shaped relative frequency distribution has a mean of \(\bar{y}=72\) and a standard deviation of \(s=8\). Approximately how many of the scores would you expect to fall in the interval from 64 to \(80 ?\) The interval from 56 to \(88 ?\)

According to the Environmental Protection Agency, chloroform, which in its gaseous form is suspected to be a cancer-causing agent, is present in small quantities in all the country's 240,000 public water sources. If the mean and standard deviation of the amounts of chloroform present in water sources are 34 and 53 micrograms per liter \((\mu g / \mathrm{L})\), respectively, explain why chloroform amounts do not have a normal distribution.

The manufacturer of a new food additive for beef cattle claims that \(80 \%\) of the animals fed a diet including this additive should have monthly weight gains in excess of 20 pounds. A large sample of measurements on weight gains for cattle fed this diet exhibits an approximately normal distribution with mean 22 pounds and standard deviation 2 pounds. Do you think the sample information contradicts the manufacturer's claim? (Calculate the probability of a weight gain exceeding 20 pounds.)

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