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Chapter 7: Problem 43

Determine the following: a. The 95 th percentile of the chi-squared distribution with \(v=10\) b. The 5 th percentile of the chi-squared distribution with \(v=10\) c. \(P\left(10.98 \leq \chi^{2} \leq 36.78\right)\), where \(\chi^{2}\) is a chi- squared rv with \(v=22\) d. \(P\left(\chi^{2}<14.611\right.\) or \(\left.\chi^{2}>37.652\right)\), where \(\chi^{2}\) is a chisquared rv with \(v=25\)

Short Answer

Expert verified
a. 18.31 b. 3.94 c. 0.90 d. 0.2

Step by step solution

01

Finding the 95th Percentile with v=10

To find the 95th percentile of the chi-squared distribution with 10 degrees of freedom, we need to refer to a chi-squared distribution table or use statistical software. The value that corresponds to the 95th percentile is approximately 18.31.
02

Finding the 5th Percentile with v=10

Similarly, to determine the 5th percentile for the chi-squared distribution with 10 degrees of freedom, we use a chi-squared table or software. The 5th percentile is approximately 3.94.
03

Calculating P(10.98 鈮� 蠂虏 鈮� 36.78) with v=22

Use a chi-squared distribution table to find the cumulative probability at 10.98 and 36.78 for 22 degrees of freedom. For 蠂虏 = 10.98, the cumulative probability is approximately 0.05. For 蠂虏 = 36.78, the cumulative probability is approximately 0.95. Therefore, the probability P(10.98 鈮� 蠂虏 鈮� 36.78) is 0.95 - 0.05 = 0.90.
04

Calculating P(蠂虏 < 14.611 or 蠂虏 > 37.652) with v=25

Use a chi-squared distribution table for 25 degrees of freedom to find the cumulative probability at 14.611 and 37.652. The cumulative probability at 14.611 is approximately 0.1. The cumulative probability at 37.652 is approximately 0.9. Thus, P(蠂虏 < 14.611) = 0.1 and P(蠂虏 > 37.652) = 1 - 0.9 = 0.1. Therefore, the combined probability is P(蠂虏 < 14.611 or 蠂虏 > 37.652) = 0.1 + 0.1 = 0.2.

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Key Concepts

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

Percentile Calculation
When dealing with the chi-squared distribution, understanding percentiles is essential. A percentile represents a value below which a certain percentage of observations fall. For example, the 95th percentile of any given distribution tells us that 95% of the values lie below that point.
In order to calculate the 95th or 5th percentile using a chi-squared distribution table, we look for the chi-square value which corresponds to the desired percentile. For instance:
  • For a chi-squared distribution with 10 degrees of freedom, the 95th percentile is approximately 18.31. This means 95% of the data values will be below 18.31.
  • Similarly, the 5th percentile is approximately 3.94, indicating that only 5% of the data values are below 3.94.
Using statistical software can also make finding these values much quicker and often more accurate.
Cumulative Probability
Cumulative probability is another important concept when working with chi-squared distributions. It refers to the probability that a chi-squared random variable will take a value less than or equal to a particular number.
This concept is crucial for calculating probabilities within a range, such as the probability that a chi-squared value falls between two points. To determine this, we calculate the cumulative probability at the lower point and subtract it from the cumulative probability at the higher point. Here's an example:
  • With 22 degrees of freedom, the cumulative probability is about 0.05 for a value of 10.98 and about 0.95 for a value of 36.78.
  • The probability that the chi-squared value falls between 10.98 and 36.78 is obtained by calculating 0.95 - 0.05, resulting in 0.90.
Cumulative probability tables or statistical software can help in obtaining these precise probabilities.
Degrees of Freedom
Degrees of freedom (\( v \)), within the context of chi-squared distributions, is a parameter that influences the shape of the distribution. It is related to the number of independent values or quantities that can vary within an analysis.
In chi-squared distributions, the degrees of freedom usually correspond to the number of categories or groups being compared minus one.
  • For instance, in a case with 22 degrees of freedom, the shape of the chi-squared distribution would look different compared to one with 10 degrees.
  • This parameter shifts and stretches the distribution, affecting how probabilities at certain values鈥攕uch as specific percentiles鈥攁re interpreted.
Understanding degrees of freedom is crucial not only for looking up values on a chi-squared table but also for comprehending how these values may differ as the parameter changes. More degrees of freedom generally result in a distribution that is more symmetric and closer to the normal distribution.

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Most popular questions from this chapter

The following observations were made on fracture toughness of a base plate of \(18 \%\) nickel maraging steel ["Fracture Testing of Weldments," ASTM Special Publ. No. 381, 1965: \(328-356\) (in ksi \(\sqrt{\text { in., }}\) given in increasing order)]: \(\begin{array}{llllllll}69.5 & 71.9 & 72.6 & 73.1 & 73.3 & 73.5 & 75.5 & 75.7 \\\ 75.8 & 76.1 & 76.2 & 76.2 & 77.0 & 77.9 & 78.1 & 79.6 \\ 79.7 & 79.9 & 80.1 & 82.2 & 83.7 & 93.7 & & \end{array}\) Calculate a \(99 \%\) CI for the standard deviation of the fracture toughness distribution. Is this interval valid whatever the nature of the distribution? Explain.

A sample of 25 pieces of laminate used in the manufacture of circuit boards was selected and the amount of warpage (in.) under particular conditions was determined for each piece, resulting in a sample mean warpage of \(.0635\) and a sample standard deviation of .0065. a. Calculate a prediction for the amount of warpage of a single piece of laminate in a way that provides information about precision and reliability. b. Calculate an interval for which you can have a high degree of confidence that at least \(95 \%\) of all pieces of laminate result in amounts of warpage that are between the two limits of the interval.

The article "An Evaluation of Football Helmets Under Impact Conditions" (Amer. J. Sports Medicine, 1984: 233-237) reports that when each football helmet in a random sample of 37 suspension-type helmets was subjected to a certain impact test, 24 showed damage. Let \(p\) denote the proportion of all helmets of this type that would show damage when tested in the prescribed manner. a. Calculate a \(99 \%\) CI for \(p\). b. What sample size would be required for the width of a \(99 \%\) CI to be at most .10, irrespective of \(\hat{p}\) ?

Determine the confidence level for each of the following large-sample one- sided confidence bounds: a. Upper bound: \(\bar{x}+.84 s / \sqrt{n}\) b. Lower bound: \(\bar{x}-2.05 \mathrm{~s} / \sqrt{n}\) c. Upper bound: \(\bar{x}+.67 \mathrm{~s} / \sqrt{n}\)

Determine the \(t\) critical value for a two-sided confidence interval in each of the following situations: a. Confidence level \(=95 \%\), df \(=10\) b. Confidence level \(=95 \%\), df \(=15\) c. Confidence level \(=99 \%\), df \(=15\) d. Confidence level \(=99 \%, n=5\) e. Confidence level \(=98 \%, \mathrm{df}=24\) f. Confidence level \(=99 \%, n=38\)
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