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Find the tabled value for \(a \chi^{2}\) variable based on \(n-1\) degrees of freedom with an area of a to its right. \(n=41, a=.025\)

Short Answer

Expert verified
Answer: The tabled value for a chi-squared variable with 41 observations and an area of 0.025 to its right is approximately 58.12.

Step by step solution

01

Find the degrees of freedom

The degrees of freedom for a \(\chi^2\) distribution is given by \(n - 1\). In this case, we have \(n = 41\) observations. So, the degrees of freedom is $$ df = n - 1 = 41 - 1 = 40 $$
02

Look up the \(\chi^2\) value in the table

With the degrees of freedom (\(df = 40\)) and the area to the right (\(a = 0.025\)), we can now consult a \(\chi^2\) distribution table to find our desired value. After looking at the table, we can see that for \(df = 40\) and \(a = 0.025\), the tabled value for the \(\chi^2\) variable is approximately: $$ \chi^2 = 58.12 $$ So, the tabled value for a \(\chi^2\) variable with \(40\) degrees of freedom and an area of \(0.025\) to its right is approximately \(58.12\).

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Key Concepts

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

Degrees of Freedom
In statistics, when we talk about 'degrees of freedom' (often abbreviated as df), we're referring to the number of independent values or quantities which can be assigned to a statistical distribution. To understand this, envision that you're trying to solve a puzzle with 41 pieces, but one of these pieces is the key piece that once placed, determines how the others fit around it. You can think of this key piece as being the 'restriction' on your puzzle. In the case of the \(\chi^2\) distribution, we calculate the degrees of freedom by taking the total number of observations and subtracting one. This subtraction accounts for the overall sample mean, which is a calculated value and not an independent observation. So, for our example of 41 observations, the degrees of freedom would be:\[df = n - 1 = 41 - 1 = 40\]
Understanding degrees of freedom is crucial because it helps us choose the correct statistical tables to determine probabilities or critical values, especially when working with sample data to estimate population parameters.
Statistical Tables
Statistical tables are indispensable tools for statisticians. They come in very handy when performing calculations that would otherwise be complex and time-consuming. The table for a \(\chi^2\) distribution is a prime example. It is essentially a ready reckoner that tells you the critical values of the \(\chi^2\) statistic for different areas under the curve and degrees of freedom.

Using the \(\chi^2\) Table

To utilize these tables, you first identify the degrees of freedom for your statistical test - in our problem, it was 40. Next, you determine the area to the right of the critical value you're interested in. This area corresponds to the significance level 'a' in this scenario, which is 0.025. By cross-referencing the degrees of freedom with the area or probability, we locate the critical \(\chi^2\) value in the table, which, as given in the problem, turns out to be 58.12 for our example.
Note that without statistical tables, we would need to use complex mathematical methods or computer software to find this value.
Probability Distribution
A probability distribution is a mathematical description of a random phenomenon in terms of the probabilities of various possible outcomes. In the context of the \(\chi^2\) distribution, it is specifically a way to describe the variability of squared standardized deviations from the mean. The \(\chi^2\) distribution itself is applicable in situations where you're looking at independent random variables that have been squared and summed, which is often the case when we're testing hypotheses about variances or when conducting \(\chi^2\) tests for independence or goodness of fit.

Characteristics of the \(\chi^2\) Distribution

What's unique about the \(\chi^2\) distribution is that it is always skewed right and its shape depends on the degrees of freedom. As the degrees of freedom increase, the distribution becomes more symmetrical and closely resembles a normal distribution. Understanding the shape and behavior of the \(\chi^2\) distribution is important for interpreting the results of statistical tests, as it affects the probabilities and critical values that determine the outcomes of these tests.

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Most popular questions from this chapter

Pollution Control The EPA limit on the allowable discharge of suspended solids into rivers and streams is 60 milligrams per liter \((\mathrm{mg} / \mathrm{L})\) per day. A study of water samples selected from the discharge at a phosphate mine shows that over a long period, the mean daily discharge of suspended solids is \(48 \mathrm{mg} / \mathrm{L},\) but day-to-day discharge readings are variable. State inspectors measured the discharge rates of suspended solids for \(n=20\) days and found \(s^{2}=39(\mathrm{mg} / \mathrm{L})^{2}\). Find a \(90 \%\) confidence interval for \(\sigma^{2}\). Interpret your results.

An experiment was conducted to compare the densities (in ounces per cubic inch) of cakes prepared from two different cake mixes. Six cake pans were filled with batter \(A\), and six were filled with batter B. Expecting a variation in oven temperature, the experimenter placed a pan filled with batter \(A\) and another with batter \(B\) side by side at six different locations in the oven. The six paired observations of densities are as follows: $$\begin{array}{lcccccc}\hline \text { Location } & 1 & 2 & 3 & 4 & 5 & 6 \\\\\hline \text { Batter A } & .135 & .102 & .098 & .141 & .131 & .144 \\ \text { Batter B } & .129 & .120 & .112 & .152 & .135 & .163 \\\\\hline\end{array}$$ a. Do the data present sufficient evidence to indicate a difference between the average densities of cakes prepared using the two types of batter? b. Construct a \(95 \%\) confidence interval for the difference between the average densities for the two mixes.

Use the information given in Exercises \(8-11\) to bound the \(p\) -value of the \(F\) statistic for a one-tailed test with the indicated degrees of freedom. \(F=6.16, d f_{1}=4, d f_{2}=13\)

Find the tabled value of \(t\left(t_{a}\right)\) corresponding to a right-tail area a and degrees of freedom given. $$ a=.01, d f=18 $$

What assumptions are made about the populations from which random samples are drawn when the \(t\) distribution is used to make small-sample inferences about the difference in population means?

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