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Chapter 7: Problem 37

Let \(Y_{1}, Y_{2}, \ldots, Y_{5}\) be a random sample of size 5 from a normal population with mean 0 and variance 1 and let \(\bar{Y}=(1 / 5) \sum_{i=1}^{5} Y_{i} .\) Let \(Y_{6}\) be another independent observation from the same population. What is the distribution of a. \(W=\sum_{i=1}^{5} Y_{i}^{2} ?\) Why? b. \(U=\sum_{i=1}^{5}\left(Y_{i}-\bar{Y}\right)^{2} ?\) Why? c. \(\sum_{i=1}^{5}\left(Y_{i}-\bar{Y}\right)^{2}+Y_{6}^{2} ?\) Why?

Short Answer

Expert verified
a. \(\chi^2(5)\); b. \(\chi^2(4)\); c. \(\chi^2(5)\).

Step by step solution

01

Understanding the Distribution of W

The variable \(W = \sum_{i=1}^{5} Y_i^2\) is the sum of the squares of five independent standard normal variables \(Y_i\). Each \(Y_i\) is \(N(0,1)\), so \(Y_i^2\) follows a \(\chi^2\) distribution with 1 degree of freedom, \(\chi^2(1)\). Since there are five such variables, \(W\) follows a \(\chi^2(5)\) distribution. This is because the sum of \(n\) squared independent standard normal variables is \(\chi^2(n)\).
02

Understanding the Distribution of U

The variable \(U = \sum_{i=1}^{5}(Y_i - \bar{Y})^2\) represents the sum of squared deviations of \(Y_i\) from their sample mean \(\bar{Y}\). For a sample from a normal distribution, \((n-1)S^2\) follows a \(\chi^2(n-1)\) distribution, where \(S^2\) is the sample variance. Thus, given 5 observations, \(U\) follows a \(\chi^2(4)\) distribution, since it's based on 4 degrees of freedom (5 samples minus 1).
03

Understanding the Distribution of the Combined Sum

The variable \(\sum_{i=1}^{5}(Y_i - \bar{Y})^2 + Y_6^2\) combines the distribution of \(U\) and an independent normal variable \(Y_6^2\). We previously determined that \(U\) is \(\chi^2(4)\) and \(Y_6^2\) is \(\chi^2(1)\). Because \(U\) and \(Y_6^2\) are independent, the sum of these two variables, \(\sum_{i=1}^{5}(Y_i - \bar{Y})^2 + Y_6^2\), follows a \(\chi^2(5)\) distribution by the property of chi-squared distributions.

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Key Concepts

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

Random Sample
A random sample is a set of observations drawn from a population where each individual observation has an equal chance of being selected. Imagine you have a big jar filled with marbles, and you close your eyes and pick out five marbles without looking. Each pick has the same chance of being any marble in the jar.

In the context of statistics, this concept ensures that the sample you collect is representative of the entire population. This means it is not biased by any particular selection method. The goal is to capture the natural variation present in the population.
  • Ensures each sample is likely similar to every other part of the population.
  • Helps in making inferences and generalizations about the population.
This randomness also means the mathematical tools and tests, like those involving normal distributions or chi-squared distributions, can be applied reliably.
Normal Distribution
The normal distribution is a bell-shaped curve that describes how the values of a random variable are distributed. Many phenomena in the natural world exhibit this pattern, making it a cornerstone of statistics. Imagine plotting the heights of adults in a country; you'd notice most heights cluster around an average, with fewer individuals at the extremes.

Mathematically, a normal distribution is characterized by two parameters: the mean and the variance. In problems involving standard normal variables, the mean is typically 0, and the variance is 1.
  • Mean (average) indicates where the center of the data lies.
  • Variance measures the spread or variability of the data.
This distribution is crucial for inferential statistics because many statistical tests assume data are normally distributed, especially in large samples due to the Central Limit Theorem.
Degrees of Freedom
Degrees of freedom (df) are a concept that refers to the number of values in a calculation that are free to vary. Consider it as the number of independent ways by which a dynamic system can move. It's an important concept in statistics that affects various calculations, including those of variances and statistical tests.

In practice, the degrees of freedom often arise in contexts involving estimating population parameters, where the constraint is that the sum of the deviations from a mean must equal zero.
  • Degrees of freedom adjust as you fit data to models.
  • For a sample size of 5, calculated deviations from the sample mean leave 4 degrees of freedom (n - 1).
Understanding degrees of freedom helps in determining the adequacy of various statistical models, providing insight into the stability and variability of the estimators.
Sum of Squares
The sum of squares is a way of measuring the variance in a dataset, highlighting the deviations of each point from the average. Picture a classroom full of students recording their heights. The sum of squares helps in figuring out how much variety there is in those heights.

The calculation involves subtracting the mean from each data point, squaring the result, and summing these squares. This approach is foundational in statistical analyses, such as variance and standard deviation calculations.
  • Helps in estimating the variability within a sample.
  • Forms the basis of several important statistical measures.
When the sum of squares is used in the context of chi-squared testing, it further allows statisticians to understand how observed data compares to the expected model. This process creates a clear picture of the spread and dispersion in the variable being studied.

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Most popular questions from this chapter

The quality of computer disks is measured by the number of missing pulses. Brand X is such that \(80 \%\) of the disks have no missing pulses. If 100 disks of brand \(X\) are inspected, what is the probability that 15 or more contain missing pulses?

The fracture strength of tempered glass averages 14 (measured in thousands of pounds per square inch) and has standard deviation 2. a. What is the probability that the average fracture strength of 100 randomly selected pieces of this glass exceeds \(14.5 ?\) b. Find an interval that includes, with probability \(0.95,\) the average fracture strength of 100 randomly selected pieces of this glass.

a. Find \(t_{05}\) for a \(t\) -distributed random variable with 5 df. b. Refer to part (a). What is \(P\left(T^{2}>t_{.05}^{2}\right) ?\) c. Find \(F_{10}\) for an \(F\) -distributed random variable with 1 numerator degree of freedom and 5 denominator degrees of freedom. d. Compare the value of \(F_{10}\) found in part \((c)\) with the value of \(t_{.05}^{2}\) from parts \((a)\) and \((b)\). e. In Exercise 7.33 , you will show that if \(T\) has a \(t\) distribution with \(\nu\) df, then \(U=T^{2}\) has an \(F\) distribution with 1 numerator degree of freedom and \(\nu\) denominator degrees of freedom. How does this explain the relationship between the values of \(F_{.10}\) (1 num. df, 5 denom df) and \(t_{05}^{2}(5 \mathrm{df})\) that you observed in part \((d)?\)

In this section, we provided the rule of thumb that the normal approximation to the binomial distribution is adequate if \(p \pm 3 \sqrt{p q / n}\) lies in the interval \((0,1)-\) that is, if $$0 < p-3 \sqrt{p q / n} \quad \text { and } \quad p+3 \sqrt{p q / n} < 1$$ a. Show that $$p+3 \sqrt{p q / n} < 1 \quad \text { if and only if } \quad n > 9(p / q)$$ b. Show that $$0 < p-3 \sqrt{p q / n} \quad \text { if and only if } \quad n > 9(q / p)$$ c. Combine the results from parts \((\underline{a})\) and \((\underline{b})\) to obtain that the normal approximation to the binomial is adequate if $$n > 9\left(\frac{p}{q}\right) \quad \text { and } \quad n > 9\left(\frac{q}{p}\right)$$ or, equivalently \(n>9\left(\frac{\text { larger of } p \text { and } q}{\text { smaller of } p \text { and } q}\right)\)

Many bulk products-such as iron ore, coal, and raw sugar- -are sampled for quality by a method that requires many small samples to be taken periodically as the material is moving along a conveyor belt. The small samples are then combined and mixed to form one composite sample. Let \(Y_{i}\) denote the volume of the \(i\) th small sample from a particular lot and suppose that \(Y_{1}, Y_{2}, \ldots, Y_{n}\) constitute a random sample, with each \(Y_{i}\) value having mean \(\mu\) (in cubic inches) and variance \(\sigma^{2}\). The average volume \(\mu\) of the samples can be set by adjusting the size of the sampling device. Suppose that the variance \(\sigma^{2}\) of the volumes of the samples is known to be approximately 4. The total volume of the composite sample must exceed 200 cubic inches with probability approximately .95 when \(n=50\) small samples are selected. Determine a setting for \(\mu\) that will allow the sampling requirements to be satisfied.
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