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For a chi-square distribution having 12 degrees of freedom, find the area under the curve for chi-square values ranging from 3.57 to \(21.0 .\)

Short Answer

Expert verified
The area under the chi-square distribution curve with 12 degrees of freedom, ranging from 3.57 to 21.0, can be obtained by subtracting the cumulative chi-square value for 3.57 from that of 21.0. It's recommended to use a chi-square table or software application to find these values.

Step by step solution

01

Understanding Chi-Square Distribution

A chi-square distribution is widely used in statistics and it is defined by a certain parameter called 'degree of freedom' which in this case is 12. The probability density function of a chi-square distribution is used to calculate areas (probabilities) under the curve. To perform this task, we can use standard statistical tables or software which have these values readily available.
02

Finding the Area under the Curve from 3.57 to 21.0

To find the area under the chi-square distribution curve between any two given values, we will need to refer to a chi-square table or software for 12 degrees of freedom. From these resources, we find the cumulative chi-square values for 3.57 and 21.0. Usually, these tables will give the area to the left of the specified chi-square value. First, obtain the cumulative chi-square value for 21.0 (denoted as \( F(21.0) \)), and then for 3.57 (denoted as \( F(3.57) \)).
03

Calculating the Required Area

Finally, to compute the area under the curve between 3.57 and 21.0, subtract the cumulative chi-square value for 3.57 from that of 21.0. This is given by the formula \( Area = F(21.0) - F(3.57) \). This area represents the probability that a chi-square distributed random variable with 12 degrees of freedom falls within the range of 3.57 to 21.0.

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Key Concepts

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

Degrees of Freedom
In the context of a chi-square distribution, the term degrees of freedom (df) is pivotal to understanding the shape and spread of the distribution. Loosely speaking, degrees of freedom refer to the number of independent values or quantities that can vary in a statistical calculation without breaking any constraints. The number of degrees of freedom is often linked to the number of observations or categories in your data minus the number of necessary constraints or parameters estimated. For instance, in the given exercise, a chi-square distribution with 12 degrees of freedom implies that there are 13 independent observations to begin with, and one parameter has been estimated (such as the mean).

The degrees of freedom affect the shape of the chi-square distribution, with distributions having a lower number of degrees of freedom displaying more skewness towards the right and higher degrees showing a more symmetrical shape that approximates a normal distribution as the number increases.
  • In calculations, degrees of freedom are used to determine critical values from the chi-square distribution table that aid in hypothesis testing.
  • The higher the degrees of freedom, the more the curve resembles a normal distribution.
Probability Density Function
The probability density function (pdf) for a chi-square distribution provides the likelihood of a random variable falling at a specific value on the chi-square curve. Mathematically, the probability density function for a chi-square distribution with k degrees of freedom is denoted as: \[ f(x; k) = \frac{1}{2^{k/2}\Gamma(k/2)} x^{(k/2)-1}e^{-x/2} \] where x is the chi-square statistic, k is the number of degrees of freedom, and Γ represents the gamma function. It is crucial to know that the pdf is used to identify the probability of a variable falling within a particular range.

For example, in our exercise, the pdf would help us compute the probability of observing a chi-square value between 3.57 and 21.0 when the degrees of freedom are 12. Remember, the hump of the curve represents the most probable values of the chi-square statistic, with probabilities diminishing as you move away from the center.
  • The pdf is integral in calculating the area under the curve between two chi-square values.
  • This function is what makes it possible to infer probabilities and make statistical decisions.
Cumulative Chi-Square Values
Cumulative chi-square values are a key concept when dealing with the area under the curve of a chi-square distribution. The cumulative value at a particular chi-square statistic is the total area under the curve to the left of that statistic. This corresponds to the probability that a random variable with a chi-square distribution will have a value less than or equal to that statistic. These values are often found using chi-square distribution tables or software.

In our exercise scenario, we would look up the cumulative values for chi-square statistics of 3.57 and 21.0 for a distribution with 12 degrees of freedom. The goal is to find the probability between these values, which is the area under the curve. It's computed simply as the difference between the two cumulative chi-square values:
  • To calculate the area under the curve for the specified range, subtract the cumulative value for 3.57 from that for 21.0.
  • The result is the probability that the chi-square random variable falls within the given range.
This practical application of cumulative chi-square values is widely used in hypothesis testing to determine statistical significance.

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Most popular questions from this chapter

The uniform length of nails is very important to a carpenter-the length of the nails being used are matched to the materials being fastened together, thereby making a small standard deviation an important property of the nails. A sample of 35 randomly selected 2-inch nails is taken from a large quantity of Nails, Inc.'s, recent production run. The resulting length measurements have a mean length of 2.025 inches and a standard deviation of 0.048 inch. a. Determine whether an assumption of normality is reasonable. Explain. b. Is the sample evidence sufficient to reject the idea that the nails have a mean length of 2 inches? Use \(\alpha=0.05\). c. Is there sufficient evidence, at the 0.05 level, to show that the length of nails from this production run has a standard deviation greater than the advertised 0.040 inch? d. Write a short report outlining the findings and recommendations as to whether or not the carpenter should use these nails for an application that requires 2 -inch nails.

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How important is the assumption "The sampled population is normally distributed" to the use of Student's \(t\) -distribution? Using a computer, simulate drawing 100 samples of size 10 from each of three different types of population distributions, namely, a normal, a uniform, and an exponential. First generate 1000 data values from the population and construct a histogram to see what the population looks like. Then generate 100 samples of size 10 from the same population; each row represents a sample. Calculate the mean and standard deviation for each of the 100 samples. Calculate \(t \neq\) for each of the 100 samples. Construct histograms of the 100 sample means and the 100 t \(\star\) values. (Additional details can be found in the Student Solutions Manual.) For the samples from the normal population: a. Does the \(\bar{x}\) distribution appear to be normal? Find percentages for intervals and compare them with the normal distribution. b. Does the distribution of \(t \star\) appear to have a \(t\) -distribution with df \(=9 ?\) Find percentages for intervals and compare them with the \(t\) -distribution. For the samples from the rectangular or uniform population: c. Does the \(\bar{x}\) distribution appear to be normal? Find percentages for intervals and compare them with the normal distribution. d. Does the distribution of \(t \star\) appear to have a \(t\) -distribution with df \(=9 ?\) Find percentages for intervals and compare them with the \(t\) -distribution. For the samples from the skewed (exponential) population: e. Does the \(\bar{x}\) distribution appear to be normal? Find percentages for intervals and compare them with the normal distribution. f. Does the distribution of \(t \star\) appear to have a \(t\) -distribution with df \(=9 ?\) Find percentages for intervals and compare them with the \(t\) -distribution. In summary: g. In each of the preceding three situations, the sampling distribution for \(\bar{x}\) appears to be slightly different from the distribution of \(t \star .\) Explain why. h. Does the normality condition appear to be necessary in order for the calculated test statistic \(t \star\) to have a Student's \(t\) -distribution? Explain.

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