/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 Does the \(x\) distribution need... [FREE SOLUTION] | 91Ó°ÊÓ

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Does the \(x\) distribution need to be normal in order to use the chi-square distribution to test the variance? Is it acceptable to use the chi-square distribution to test the variance if the \(x\) distribution is simply mound- shaped and more or less symmetric?

Short Answer

Expert verified
The distribution must be normal to use the chi-square test for variance; simply being mound-shaped and symmetric is not enough.

Step by step solution

01

Understanding Chi-Square Test Requirements

The chi-square test for variance requires the sample data to come from a population that follows a normal distribution. This is because the chi-square distribution is only applicable under the assumption that the underlying data's distribution is normal.
02

Analyzing Mound-Shaped and Symmetric Distributions

A distribution that is mound-shaped and symmetric might resemble a normal distribution but is not guaranteed to be normal. Such distributions may not meet the specific requirements needed to apply the chi-square distribution in testing variances.
03

Conclusion on Applicability

Since the chi-square test relies on the assumption of normality, using it on distributions that are merely mound-shaped and symmetric, but not confirmed to be normal, is not appropriate. The test might give inaccurate results if the data do not have the required normal distribution.

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Key Concepts

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

Test for Variance
The test for variance often employs the chi-square distribution to determine if there is a statistically significant difference between the variance of a sample and the variance of a known or assumed population. This is particularly useful in fields like quality control and experimental research where maintaining consistency is crucial.

A hypothesis regarding variance can be tested using the chi-square statistic, which follows the formula:- \[ \chi^2 = \frac{(n-1)s^2}{\sigma^2} \] where:- \(n\) is the sample size,- \(s^2\) is the sample variance,- \(\sigma^2\) is the population variance.So, if you want to test whether the spread of a dataset matches an expected range, the chi-square test for variance is the way to go. Be vigilant, however, as this test assumes you have a normally distributed dataset.
Normal Distribution Assumption
The normal distribution assumption is a vital part of many statistical tests, including the chi-square test for variance. This assumption requires that the data used come from a population with a normal distribution. If this assumption is not met, the results of the statistical test may not be valid.

A normal distribution is characterized by several key features:- Symmetry around the mean: The left and right sides of the graph are mirror images.- Bell-shaped curve: Most of the data points lie close to the mean.- Mean, median, and mode are all equal.- Determined by two parameters: mean (\(\mu\)) and standard deviation (\(\sigma\)).In the context of the chi-square test, assuming a normal distribution makes the test applicable and powerful. If a distribution is only mound-shaped and symmetric but not strictly normal, applying the chi-square test can lead to errors. Instead, always verify that your data follows a normal distribution, or consider other statistical tests if this assumption cannot be met.
Statistical Testing Assumptions
Every statistical test comes with its own set of assumptions that must be met for its results to be considered valid. Understanding these assumptions is essential in data analysis. For the chi-square test for variance, let’s review these assumptions:
  • Normal Distribution: The primary assumption is that the population from which the sample is drawn must follow a normal distribution. Without this, the chi-square test might mislead.
  • Independence: Observations must be independent of each other, meaning that the value of one observation should not influence another.
  • Random Sampling: Samples should be randomly selected to ensure each member of the population has an equal chance of selection.
Not abiding by these statistical testing assumptions could invalidate your results, leading to wrong inferences. It's essential to verify that your dataset aligns with these conditions before carrying out any statistical test, particularly when involving the chi-square distribution.

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Most popular questions from this chapter

The following table shows the Myers-Briggs personality preference and area of study for a random sample of 519 college students (Applications of the Myers- Briggs Type Indicator in Higher Education, edited by Provost and Anchors). In the table, IN refers to introvert, intuitive; EN refers to extrovert, intuitive; IS refers to introvert, sensing; and ES refers to extrovert, sensing. $$\begin{array}{l|c|c|c|c} \hline \begin{array}{l} \text { Myers-Briggs } \\ \text { Preference } \end{array} & \begin{array}{c} \text { Arts \& } \\ \text { Science } \end{array} & \text { Business } & \begin{array}{c} \text { Allied } \\ \text { Health } \end{array} & \text { Row Total } \\ \hline \text { IN } & 64 & 15 & 17 & 96 \\ \hline \text { EN } & 82 & 42 & 30 & 154 \\ \hline \text { IS } & 68 & 35 & 12 & 115 \\ \hline \text { ES } & 75 & 42 & 37 & 154 \\ \hline \text { Column Total } & 289 & 134 & 96 & 519 \\ \hline \end{array}$$ Use a chi-square test to determine if Myers-Briggs preference type is independent of area of study at the 0.05 level of significance.

The following table shows ceremonial ranking and type of pottery sherd for a random sample of 434 sherds at a location in the Sand Canyon Archaeological Project, Colorado (The Architecture of Social Integration in Prehistoric Pueblos, edited by Lipe and Hegmon). $$\begin{array}{lcccc} \hline \begin{array}{l} \text { Ceremonial } \\ \text { Ranking } \end{array} & \begin{array}{c} \text { Cooking Jar } \\ \text { Sherds } \end{array} & \begin{array}{c} \text { Decorated Jar Sherds } \\ \text { (Noncooking) } \end{array} & \text { Row Total } \\ \hline \mathrm{A} & 86 & 49 & 135 \\ \hline \mathrm{B} & 92 & 53 & 145 \\ \hline \mathrm{C} & 79 & 75 & 154 \\ \hline \text { Column Total } & 257 & 177 & 434 \\ \hline \end{array}$$ Use a chi-square test to determine if ceremonial ranking and pottery type are independent at the 0.05 level of significance.

The following problem is based on information taken from Academe, Bulletin of the American Association of University Professors. Let \(x\) represent the average annual salary of college and university professors (in thousands of dollars) in the United States. For all colleges and universities in the United States, the population variance of \(x\) is approximately \(\sigma^{2}=47.1 .\) However, a random sample of 15 colleges and universities in Kansas showed that \(x\) has a sample variance \(s^{2}=83.2 .\) Use a \(5 \%\) level of significance to test the claim that the variance for colleges and universities in Kansas is greater than \(47.1 .\) Find a \(95 \%\) confidence interval for the population variance.

For chi-square distributions, as the number of degrees of freedom increases, does any skewness increase or decrease? Do chi-square distributions become more symmetric (and normal) as the number of degrees of freedom becomes larger and larger?

Vegetation Wild irises are beautiful flowers found throughout the United States, Canada, and northern Europe. This problem concerns the length of the sepal (leaf-like part covering the flower) of different species of wild iris. Data are based on information taken from an article by R. A. Fisher in Annals of Eugenics (Vol. 7, part 2, pp. 179-188). Measurements of sepal length in centimeters from andom samples of Iris setosa (I), Iris versicolor (II), and Iris virginica (III) are as follows:$$\begin{array}{ccc}\mathrm{I} & \mathrm{II} & \mathrm{III} \\\5.4 & 5.5 & 6.3 \\\4.9 & 6.5 & 5.8 \\\5.0 & 6.3 & 4.9 \\\5.4 & 4.9 & 7.2 \\\4.4 & 5.2 & 5.7 \\\5.8 & 6.7 & 6.4 \\\5.7 & 5.5 & \\\& 6.1 &\end{array}$$,Shall we reject or not reject the claim that there are no differences among the population means of sepal length for the different species of iris? Use a \(5 \%\) level of significance.

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